NCERT Solutions for Class 12 Maths Chapter 3: Matrices

Access NCERT Solutions for Class 12 Maths  Chapter 3 – Matrices

Exercise 3.1

1. In the matrix \[\mathbf{A=\left[ \begin{matrix} 2 & 5 & 19 & -7  \\ 35 & -2 & \dfrac{5}{2} & 12  \\ \sqrt{3} & 1 & -5 & 17  \\ \end{matrix} \right]}\], write

i. The order of the matrix.

Ans: The order of a matrix is \[m\times n\] where \[m\] is the number of rows and \[n\] is the number of columns. Therefore, here the order is \[3\times 4\].

ii. The number of elements.

Ans: Since the order of the given matrix is \[3\times 4\] therefore, the number of elements in it is \[3\times 4=12\].

iii. Write the elements \[\mathbf{{{a}_{13}},{{a}_{21}},{{a}_{33}},{{a}_{24}},{{a}_{23}}}\]

Ans: The elements are given as \[{{a}_{mn}}\] . Therefore, here \[{{a}_{13}}=19\] , \[{{a}_{21}}=35\] , \[{{a}_{33}}=-5\] , \[{{a}_{24}}=12\] , \[{{a}_{23}}=\dfrac{5}{2}\].

2. If a matrix has \[24\] elements, what are the possible order it can have? What if it has \[13\] elements?

Ans: The order of a matrix is \[m\times n\] where \[m\] is the number of rows and \[n\] is the number of columns. To find the possible orders of a matrix, we have to find all the ordered pairs of natural numbers whose product is \[24\] .

\[\therefore \left( 1\times 24 \right),\left( 24\times 1 \right),\left( 2\times 12 \right),\left( 12\times 2 \right),\left( 3\times 8 \right),\left( 8\times 3 \right),\left( 4\times 6 \right),\left( 6\times 4 \right)\] are all the possible ordered pairs here.

If the matrix had \[13\] elements, then the ordered pairs would be \[\left( 1\times 13 \right)\] and \[\left( 13\times 1 \right)\].

3. If a matrix has \[18\] elements, what are the possible orders it can have? What if it has \[5\] elements?

Ans: The order of a matrix is \[m\times n\] where \[m\] is the number of rows and \[n\] is the number of columns. To find the possible orders of a matrix, we have to find all the ordered pairs of natural numbers whose product is \[18\] .

\[\therefore \left( 1\times 18 \right),\left( 18\times 1 \right),\left( 2\times 9 \right),\left( 9\times 2 \right),\left( 3\times 6 \right),\left( 6\times 3 \right)\] are all the possible ordered pairs here.

If the matrix had \[5\] elements, then the ordered pairs would be \[\left( 1\times 5 \right)\] and \[\left( 5\times 1 \right)\].

4. Construct a \[3\times 4\] matrix, whose elements are given by 

i. \[\mathbf{{{a}_{ij}}=\dfrac{1}{2}\left| -3i+j \right|}\]

Ans: Given that \[{{a}_{ij}}=\dfrac{1}{2}\left| -3i+j \right|\] ,

\[\therefore {{a}_{11}}=\dfrac{1}{2}\left| -3\times 1+1 \right|=1\]

\[{{a}_{21}}=\dfrac{1}{2}\left| -3\times 2+1 \right|=\dfrac{5}{2}\]

\[{{a}_{31}}=\dfrac{1}{2}\left| -3\times 3+1 \right|=4\]

\[{{a}_{12}}=\dfrac{1}{2}\left| -3\times 1+2 \right|=\dfrac{1}{2}\]

\[{{a}_{22}}=\dfrac{1}{2}\left| -3\times 2+2 \right|=2\]

\[{{a}_{32}}=\dfrac{1}{2}\left| -3\times 3+2 \right|=\dfrac{7}{2}\]

\[{{a}_{13}}=\dfrac{1}{2}\left| -3\times 1+3 \right|=0\]

\[{{a}_{23}}=\dfrac{1}{2}\left| -3\times 2+3 \right|=\dfrac{3}{2}\]

\[{{a}_{33}}=\dfrac{1}{2}\left| -3\times 3+3 \right|=3\]

\[{{a}_{14}}=\dfrac{1}{2}\left| -3\times 1+4 \right|=\dfrac{1}{2}\]

\[{{a}_{24}}=\dfrac{1}{2}\left| -3\times 2+4 \right|=1\]

\[{{a}_{34}}=\dfrac{1}{2}\left| -3\times 3+4 \right|=\dfrac{5}{2}\]

Thus, the required matrix is \[A=\left[ \begin{matrix} 1 & \dfrac{1}{2} & 0 & \dfrac{1}{2}  \\ \dfrac{5}{2} & 2 & \dfrac{3}{2} & 1  \\ 4 & \dfrac{7}{2} & 3 & \dfrac{5}{2}  \\ \end{matrix} \right]\].

ii. \[\mathbf{{{a}_{ij}}=2i-j}\]

Ans: A \[3\times 4\] matrix is given by \[A=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} & {{a}_{14}}  \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} & {{a}_{24}}  \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} & {{a}_{34}}  \\ \end{matrix} \right]\]

Given that \[{{a}_{ij}}=2i-j\] ,

\[\therefore {{a}_{11}}=2\times 1-1=1\]

\[{{a}_{21}}=2\times 2-1=3\]

\[{{a}_{31}}=2\times 3-1=5\]

\[{{a}_{12}}=2\times 1-2=0\]

\[{{a}_{22}}=2\times 2-2=4\]

\[{{a}_{32}}=2\times 3-2=4\]

\[{{a}_{13}}=2\times 1-3=-1\]

\[{{a}_{23}}=2\times 2-3=1\]

\[{{a}_{33}}=2\times 3-3=3\]

\[{{a}_{14}}=2\times 1-4=-2\]

\[{{a}_{24}}=2\times 2-4=0\]

\[{{a}_{34}}=2\times 3-4=2\]

Thus, the required matrix is \[A=\left[ \begin{matrix} 1 & 0 & -1 & -2  \\ 3 & 2 & 1 & 0  \\ 5 & 4 & 3 & 2  \\ \end{matrix} \right]\].

5. Find the value of \[x,y,z\] from the following equation:

i. \[\mathbf{\left[ \begin{matrix} 4 & 3  \\ x & 5  \\ \end{matrix} \right]=\left[ \begin{matrix} y & z  \\  1 & 5  \\ \end{matrix} \right]}\]

Ans: Given \[\left[ \begin{matrix} 4 & 3  \\  x & 5  \\ \end{matrix} \right]=\left[ \begin{matrix} y & z  \\ 1 & 5  \\ \end{matrix} \right]\] 

Comparing the corresponding elements we get,

\[x=1,y=4,z=3\]

ii. \[\mathbf{\left[ \begin{matrix} x+y & 2  \\ 5+z & xy  \\ \end{matrix} \right]=\left[ \begin{matrix} 6 & 2  \\ 5 & 8  \\ \end{matrix} \right]}\]

Ans: Given \[\left[ \begin{matrix} x+y & 2  \\ 5+z & xy  \\ \end{matrix} \right]=\left[ \begin{matrix} 6 & 2  \\ 5 & 8  \\ \end{matrix} \right]\]

Comparing the corresponding elements we get,

\[x+y=6,xy=8,5+z=5\]

Now, \[\because 5+z=5\]

\[\Rightarrow z=0\]

We know that, \[{{\left( x-y \right)}^{2}}={{\left( x+y \right)}^{2}}-4xy\]

\[\Rightarrow {{\left( x-y \right)}^{2}}=36-32\]

\[\Rightarrow \left( x-y \right)=\pm 2\]

When \[\left( x-y \right)=2\] and \[\left( x+y \right)=6\],

We get \[x=4,y=2\]

When \[\left( x-y \right)=-2\] and \[\left( x+y \right)=6\],

We get \[x=2,y=4\]

\[\therefore x=4,y=2,z=0\] or \[\therefore x=2,y=4,z=0\]

iii. \[\mathbf{\left[ \begin{matrix} x+y+z  \\ x+z  \\ y+z  \\ \end{matrix} \right]=\left[ \begin{matrix} 9  \\ 5  \\ 7  \\ \end{matrix} \right]}\]

Ans: Given \[\left[ \begin{matrix} x+y+z  \\ x+z  \\ y+z  \\ \end{matrix} \right]=\left[ \begin{matrix} 9  \\ 5  \\ 7  \\ \end{matrix} \right]\]

Comparing the corresponding elements we get,

\[x+y+z=9\]                      …(1)

\[x+z=5\]                              …(2)

\[y+z=7\]                              …(3)

From equation (1) and (2),

\[y+5=9\]

\[\Rightarrow y=4\]

From equation (3) we have,

\[4+z=7\]

\[\Rightarrow z=3\]

\[x+z=5\]

\[\Rightarrow x=2\]

\[\therefore x=2,y=4,z=3\]

6. Find the value of \[\mathbf{a,b,c,d}\] from the equation:

\[\mathbf{\left[ \begin{matrix} a-b & 2a+c  \\ 2a-b & 3c+d  \\ \end{matrix} \right]=\left[ \begin{matrix} -1 & 5  \\ 0 & 13  \\ \end{matrix} \right]}\]

Ans: Given \[\left[ \begin{matrix} a-b & 2a+c  \\ 2a-b & 3c+d  \\ \end{matrix} \right]=\left[ \begin{matrix} -1 & 5  \\ 0 & 13  \\ \end{matrix} \right]\]

Comparing the corresponding elements we get,

\[a-b=-1\]                              …(1)

\[2a-b=0\]                              …(2)

\[2a+c=5\]                              …(3)

\[3c+d=13\]                                        …(4)

From equation (2),

\[b=2a\]

From equation (1),

\[a-2a=-1\]

\[\Rightarrow a=1\]

\[\Rightarrow b=2\]

From equation (3),

\[2\times 1+c=5\]

\[\Rightarrow c=3\]

From equation (4),

\[3\times 3+d=13\]

\[\Rightarrow d=4\]

\[\therefore a=1,b=2,c=3,d=4\]

7. \[\mathbf{A={{\left[ {{a}_{y}} \right]}_{m\times n}}}\] is a square matrix, if

1. m<n

2. m>n

3. m=n

4. None of these

Ans: A given matrix is said to be a square matrix if the number of rows is equal to the number of columns.

\[\therefore A={{\left[ {{a}_{y}} \right]}_{m\times n}}\] is a square matrix if, \[m=n\].

Thus, option (C) is correct.

8. Which of the given values of \[x\] and \[y\] make the following pair of matrices equal \[\mathbf{\left[ \begin{matrix} 3x+7 & 5  \\ y+1 & 2-3x  \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & y-2  \\  8 & 4  \\ \end{matrix} \right]}\]

1. \[\mathbf{x=\dfrac{-1}{3},y=7}\]

2. Not possible to find

3. \[\mathbf{y=7,x=\dfrac{-2}{3}}\]

4. \[\mathbf{x=\dfrac{-1}{3},y=\dfrac{-2}{3}}\]

Ans: Given \[\left[ \begin{matrix} 3x+7 & 5  \\ y+1 & 2-3x  \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & y-2  \\  8 & 4  \\ \end{matrix} \right]\]

Comparing the corresponding elements we get,

\[3x+7=0\]

\[\Rightarrow x=-\dfrac{7}{3}\]

\[y-2=5\]

\[\Rightarrow y=7\]

\[y+1=8\]

\[\Rightarrow y=7\]

\[2-3x=4\]

\[\Rightarrow x=-\dfrac{2}{3}\]

Since we get two different values of \[x\] ,which is not possible. It is not possible to find the values of \[x\] and \[y\] for which the given matrices are equal.

Thus, the correct option is (B).

9. The number of all possible matrices of order \[\mathbf{3\times 3}\] with each entry \[0\] or \[1\] is:

1. \[\mathbf{27}\]

2. \[\mathbf{18}\]

3. \[\mathbf{81}\]

4. \[\mathbf{512}\]

Ans: Given a matrix of the order \[3\times 3\] has nine elements and each of these elements can be either \[0\] or \[1\] .

Now, each of the nine elements can be filled in two possible ways.

Therefore, the required number of possible matrices is \[{{2}^{9}}=512\].

Exercise 3.2

1. Let \[\mathbf{A=\left[ \begin{matrix} 2 & 4  \\ 3 & 2  \\ \end{matrix} \right],B=\left[ \begin{matrix} 1 & 3  \\ -2 & 5  \\ \end{matrix} \right],C=\left[ \begin{matrix} -2 & 5  \\ 3 & 4  \\ \end{matrix} \right]}\]

Find each of the following

i. \[\mathbf{A+B}\]

Ans: Given \[A=\left[ \begin{matrix}  2 & 4  \\  3 & 2  \\ \end{matrix} \right],B=\left[ \begin{matrix}  1 & 3  \\ -2 & 5  \\ \end{matrix} \right]\]

\[\therefore A+B=\left[ \begin{matrix}  2 & 4  \\  3 & 2  \\ \end{matrix} \right]+\left[ \begin{matrix} 1 & 3  \\ -2 & 5  \\ \end{matrix} \right]\]

\[\Rightarrow A+B=\left[ \begin{matrix} 2+1 & 4+3  \\ 3-2 & 2+5  \\ \end{matrix} \right]\]

\[\Rightarrow A+B=\left[ \begin{matrix}  3 & 7  \\ 1 & 7  \\ \end{matrix} \right]\]

ii. \[\mathbf{A-B}\]

Ans: Given \[A=\left[ \begin{matrix} 2 & 4  \\ 3 & 2  \\ \end{matrix} \right],B=\left[ \begin{matrix} 1 & 3  \\ -2 & 5  \\ \end{matrix} \right]\]

\[\therefore A-B=\left[ \begin{matrix}  2 & 4  \\ 3 & 2  \\ \end{matrix} \right]-\left[ \begin{matrix} 1 & 3  \\ -2 & 5  \\ \end{matrix} \right]\]

\[\Rightarrow A-B=\left[ \begin{matrix} 2-1 & 4-3  \\ 3+2 & 2-5  \\ \end{matrix} \right]\]

\[\Rightarrow A-B=\left[ \begin{matrix} 1 & 1  \\ 5 & -3  \\ \end{matrix} \right]\]

iii. \[\mathbf{3A-C}\]

Ans: Given \[A=\left[ \begin{matrix} 2 & 4  \\ 3 & 2  \\ \end{matrix} \right],C=\left[ \begin{matrix} -2 & 5  \\ 3 & 4  \\ \end{matrix} \right]\]

\[\therefore 3A-C=3\left[ \begin{matrix} 2 & 4  \\ 3 & 2  \\ \end{matrix} \right]-\left[ \begin{matrix} -2 & 5  \\ 3 & 4  \\ \end{matrix} \right]\]

\[\Rightarrow 3A-C=\left[ \begin{matrix} 6 & 12  \\ 9 & 6  \\ \end{matrix} \right]-\left[ \begin{matrix} -2 & 5  \\ 3 & 4  \\ \end{matrix} \right]\]

\[\Rightarrow 3A-C=\left[ \begin{matrix} 6+2 & 12-5  \\ 9-3 & 6-4  \\ \end{matrix} \right]\]

\[\Rightarrow 3A-C=\left[ \begin{matrix} 8 & 7  \\ 6 & 2  \\ \end{matrix} \right]\]

iv. \[\mathbf{AB}\]

Ans: Given \[A=\left[ \begin{matrix} 2 & 4  \\ 3 & 2  \\ \end{matrix} \right],B=\left[ \begin{matrix} 1 & 3  \\ -2 & 5  \\ \end{matrix} \right]\]

\[\therefore AB=\left[ \begin{matrix} 2 & 4  \\ 3 & 2  \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 3  \\ -2 & 5  \\ \end{matrix} \right]\]

\[\Rightarrow AB=\left[ \begin{matrix} 2\left( 1 \right)+4\left( -2 \right) & 2\left( 3 \right)+4\left( 5 \right)  \\ 3\left( 1 \right)+2\left( -2 \right) & 3\left( 3 \right)+2\left( 5 \right)  \\ \end{matrix} \right]\]

\[\Rightarrow AB=\left[ \begin{matrix}  -6 & 26  \\ -1 & 19  \\ \end{matrix} \right]\]

v. \[\mathbf{BA}\]

Ans: Given \[A=\left[ \begin{matrix} 2 & 4  \\ 3 & 2  \\ \end{matrix} \right],B=\left[ \begin{matrix} 1 & 3  \\ -2 & 5  \\ \end{matrix} \right]\]

\[\therefore BA=\left[ \begin{matrix} 1 & 3  \\ -2 & 5  \\ \end{matrix} \right]\left[ \begin{matrix} 2 & 4  \\ 3 & 2  \\ \end{matrix} \right]\]

\[\Rightarrow BA=\left[ \begin{matrix} 1\left( 2 \right)+3\left( 3 \right) & 1\left( 4 \right)+3\left( 2 \right)  \\ -2\left( 2 \right)+5\left( 3 \right) & -2\left( 4 \right)+5\left( 2 \right)  \\ \end{matrix} \right]\]

\[\Rightarrow BA=\left[ \begin{matrix} 11 & 10  \\ 11 & 2  \\ \end{matrix} \right]\]

2. Compute the following:

i. \[\mathbf{\left[ \begin{matrix} a & b  \\ -b & a  \\ \end{matrix} \right]+\left[ \begin{matrix} a & b  \\ b & a  \\ \end{matrix} \right]}\]

Ans: We have to find \[\left[ \begin{matrix} a & b  \\ -b & a  \\ \end{matrix} \right]+\left[ \begin{matrix} a & b  \\ b & a  \\ \end{matrix} \right]\]

\[\therefore \left[ \begin{matrix} a & b  \\ -b & a  \\ \end{matrix} \right]+\left[ \begin{matrix} a & b  \\ b & a  \\ \end{matrix} \right]=\left[ \begin{matrix} a+a & b+b  \\ -b+b & a+a  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} a & b  \\ -b & a  \\ \end{matrix} \right]+\left[ \begin{matrix} a & b  \\ b & a  \\ \end{matrix} \right]=\left[ \begin{matrix} 2a & 2b  \\ 0 & 2a  \\ \end{matrix} \right]\]

ii. \[\mathbf{\left[ \begin{matrix}  {{a}^{2}}+{{b}^{2}} & {{b}^{2}}+{{c}^{2}}  \\ {{a}^{2}}+{{c}^{2}} & {{a}^{2}}+{{b}^{2}}  \\ \end{matrix} \right]+\left[ \begin{matrix} 2ab & 2bc  \\ -2ac & -2ab  \\ \end{matrix} \right]}\]

Ans: We have to find 

\[\left[ \begin{matrix}  {{a}^{2}}+{{b}^{2}} & {{b}^{2}}+{{c}^{2}}  \\ {{a}^{2}}+{{c}^{2}} & {{a}^{2}}+{{b}^{2}}  \\ \end{matrix} \right]+\left[ \begin{matrix} 2ab & 2bc  \\ -2ac & -2ab  \\ \end{matrix} \right]\]

\[\therefore \left[ \begin{matrix} {{a}^{2}}+{{b}^{2}} & {{b}^{2}}+{{c}^{2}}  \\ {{a}^{2}}+{{c}^{2}} & {{a}^{2}}+{{b}^{2}}  \\ \end{matrix} \right]+\left[ \begin{matrix} 2ab & 2bc  \\ -2ac & -2ab  \\ \end{matrix} \right]=\left[ \begin{matrix} {{a}^{2}}+{{b}^{2}}+2ab & {{b}^{2}}+{{c}^{2}}+2bc  \\ {{a}^{2}}+{{c}^{2}}-2ac & {{a}^{2}}+{{b}^{2}}-2ab  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} {{a}^{2}}+{{b}^{2}} & {{b}^{2}}+{{c}^{2}}  \\ {{a}^{2}}+{{c}^{2}} & {{a}^{2}}+{{b}^{2}}  \\ \end{matrix} \right]+\left[ \begin{matrix} 2ab & 2bc  \\ -2ac & -2ab  \\ \end{matrix} \right]=\left[ \begin{matrix} {{\left( a+b \right)}^{2}} & {{\left( b+c \right)}^{2}}  \\ {{\left( a-c \right)}^{2}} & {{\left( a-b \right)}^{2}}  \\ \end{matrix} \right]\]

iii. \[\left[ \begin{matrix} -1 & 4 & -6  \\ 8 & 5 & 16  \\ 2 & 8 & 5  \\ \end{matrix} \right]+\left[ \begin{matrix} 12 & 7 & 6  \\ 8 & 0 & 5  \\ 3 & 2 & 4  \\ \end{matrix} \right]\]

Ans: We have to find \[\left[ \begin{matrix} -1 & 4 & -6  \\ 8 & 5 & 16  \\ 2 & 8 & 5  \\ \end{matrix} \right]+\left[ \begin{matrix} 12 & 7 & 6  \\ 8 & 0 & 5  \\ 3 & 2 & 4  \\ \end{matrix} \right]\]

\[\therefore \left[ \begin{matrix} -1 & 4 & -6  \\ 8 & 5 & 16  \\ 2 & 8 & 5  \\ \end{matrix} \right]+\left[ \begin{matrix} 12 & 7 & 6  \\ 8 & 0 & 5  \\ 3 & 2 & 4  \\ \end{matrix} \right]=\left[ \begin{matrix} -1+12 & 4+7 & -6+6  \\ 8+8 & 5+0 & 16+5  \\ 2+3 & 8+2 & 5+4  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} -1 & 4 & -6  \\ 8 & 5 & 16  \\ 2 & 8 & 5  \\ \end{matrix} \right]+\left[ \begin{matrix} 12 & 7 & 6  \\ 8 & 0 & 5  \\ 3 & 2 & 4  \\ \end{matrix} \right]=\left[ \begin{matrix} 11 & 11 & 0  \\ 16 & 5 & 21  \\ 5 & 10 & 9  \\ \end{matrix} \right]\]

iv. \[\left[ \begin{matrix} {{\cos }^{2}}x & {{\sin }^{2}}x  \\ {{\sin }^{2}}x & {{\cos }^{2}}x  \\ \end{matrix} \right]+\left[ \begin{matrix} {{\sin }^{2}}x & {{\cos }^{2}}x  \\ {{\cos }^{2}}x & {{\sin }^{2}}x  \\ \end{matrix} \right]\]

Ans: We have to find \[\left[ \begin{matrix} {{\cos }^{2}}x & {{\sin }^{2}}x  \\   {{\sin }^{2}}x & {{\cos }^{2}}x  \\ \end{matrix} \right]+\left[ \begin{matrix}   {{\sin }^{2}}x & {{\cos }^{2}}x  \\ {{\cos }^{2}}x & {{\sin }^{2}}x  \\ \end{matrix} \right]\]

\[\therefore \left[ \begin{matrix} {{\cos }^{2}}x & {{\sin }^{2}}x  \\ {{\sin }^{2}}x & {{\cos }^{2}}x  \\ \end{matrix} \right]+\left[ \begin{matrix} {{\sin }^{2}}x & {{\cos }^{2}}x  \\ {{\cos }^{2}}x & {{\sin }^{2}}x  \\ \end{matrix} \right]=\left[ \begin{matrix} {{\cos }^{2}}x+{{\sin }^{2}}x & {{\sin }^{2}}x+{{\cos }^{2}}x  \\ {{\sin }^{2}}x+{{\cos }^{2}}x & {{\cos }^{2}}x+{{\sin }^{2}}x  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} {{\cos }^{2}}x & {{\sin }^{2}}x  \\ {{\sin }^{2}}x & {{\cos }^{2}}x  \\ \end{matrix} \right]+\left[ \begin{matrix} {{\sin }^{2}}x & {{\cos }^{2}}x  \\ {{\cos }^{2}}x & {{\sin }^{2}}x  \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 1  \\ 1 & 1  \\ \end{matrix} \right]\]

3. Compute the indicated products

i. \[\mathbf{\left[ \begin{matrix} a & b  \\ -b & a  \\ \end{matrix} \right]\left[ \begin{matrix} a & -b  \\ b & a  \\ \end{matrix} \right]}\]

Ans: We have to find \[\left[ \begin{matrix} a & b  \\ -b & a  \\ \end{matrix} \right]\left[ \begin{matrix} a & -b  \\ b & a  \\ \end{matrix} \right]\]

\[\therefore \left[ \begin{matrix} a & b  \\ -b & a  \\ \end{matrix} \right]\left[ \begin{matrix} a & -b  \\ b & a  \\ \end{matrix} \right]=\left[ \begin{matrix} a\left( a \right)+b\left( b \right) & a\left( -b \right)+b\left( a \right)  \\ -b\left( a \right)+a\left( b \right) & -b\left( -b \right)+a\left( a \right)  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} a & b  \\ -b & a  \\ \end{matrix} \right]\left[ \begin{matrix} a & -b  \\ b & a  \\ \end{matrix} \right]=\left[ \begin{matrix}   {{a}^{2}}+{{b}^{2}} & 0  \\  0 & {{a}^{2}}+{{b}^{2}}  \\ \end{matrix} \right]\]

ii. \[\mathbf{\left[ \begin{matrix} 1  \\   2  \\   3  \\ \end{matrix} \right]\left[ \begin{matrix}   2 & 3 & 4  \\ \end{matrix} \right]}\]

Ans: We have to find \[\left[ \begin{matrix}    1  \\   2  \\   3  \\ \end{matrix} \right]\left[ \begin{matrix}  2 & 3 & 4  \\ \end{matrix} \right]\]

\[\therefore \left[ \begin{matrix}    1  \\   2  \\   3  \\ \end{matrix} \right]\left[ \begin{matrix}   2 & 3 & 4  \\ \end{matrix} \right]=\left[ \begin{matrix}   1\left( 2 \right) & 1\left( 3 \right) & 1\left( 4 \right)  \\   2\left( 2 \right) & 2\left( 3 \right) & 2\left( 4 \right)  \\   3\left( 2 \right) & 3\left( 3 \right) & 3\left( 4 \right)  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   1  \\   2  \\   3  \\ \end{matrix} \right]\left[ \begin{matrix}   2 & 3 & 4  \\ \end{matrix} \right]=\left[ \begin{matrix}   2 & 3 & 4  \\   4 & 6 & 8  \\   6 & 9 & 12  \\ \end{matrix} \right]\]

iii. \[\mathbf{\left[ \begin{matrix}   1 & -2  \\  2 & 3  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 2 & 3  \\   2 & 3 & 1  \\ \end{matrix} \right]}\]

Ans: We have to find \[\left[ \begin{matrix}   1 & -2  \\   2 & 3  \\\end{matrix} \right]\left[ \begin{matrix}   1 & 2 & 3  \\   2 & 3 & 1  \\ \end{matrix} \right]\]

\[\therefore \left[ \begin{matrix}   1 & -2  \\   2 & 3  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 2 & 3  \\   2 & 3 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   1\left( 1 \right)-2\left( 2 \right) & 1\left( 2 \right)-2\left( 3 \right) & 1\left( 3 \right)-2\left( 1 \right)  \\   2\left( 1 \right)+3\left( 2 \right) & 2\left( 2 \right)+3\left( 3 \right) & 2\left( 3 \right)+3\left( 1 \right)  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   1 & -2  \\   2 & 3  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 2 & 3  \\   2 & 3 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   -3 & -4 & 1  \\   8 & 13 & 9  \\ \end{matrix} \right]\]

iv. \[\mathbf{\left[ \begin{matrix}  2 & 3 & 4  \\   3 & 4 & 5  \\   4 & 5 & 6  \\ \end{matrix} \right]\left[ \begin{matrix}    1 & -3 & 5  \\   0 & 2 & 4  \\   3 & 0 & 5  \\ \end{matrix} \right]}\]

Ans: We have to find \[\left[ \begin{matrix}  2 & 3 & 4  \\   3 & 4 & 5  \\   4 & 5 & 6  \\ \end{matrix} \right]\left[ \begin{matrix}    1 & -3 & 5  \\   0 & 2 & 4  \\   3 & 0 & 5  \\ \end{matrix} \right]\]

\[\therefore \left[ \begin{matrix}   2 & 3 & 4  \\   3 & 4 & 5  \\   4 & 5 & 6  \\ \end{matrix} \right]\left[ \begin{matrix}  1 & -3 & 5  \\   0 & 2 & 4  \\   3 & 0 & 5  \\ \end{matrix} \right]=\left[ \begin{matrix}   2\left( 1 \right)+3\left( 0 \right)+4\left( 3 \right) & 2\left( -3 \right)+3\left( 2 \right)+4\left( 0 \right) & 2\left( 5 \right)+3\left( 4 \right)+4\left( 5 \right)  \\   3\left( 1 \right)+4\left( 0 \right)+5\left( 3 \right) & 3\left( -3 \right)+4\left( 2 \right)+5\left( 0 \right) & 3\left( 5 \right)+4\left( 4 \right)+5\left( 5 \right)  \\   4\left( 1 \right)+5\left( 0 \right)+5\left( 3 \right) & 4\left( -3 \right)+5\left( 2 \right)+6\left( 0 \right) & 4\left( 5 \right)+5\left( 4 \right)+6\left( 5 \right)  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   2 & 3 & 4  \\   3 & 4 & 5  \\   4 & 5 & 6  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & -3 & 5  \\   0 & 2 & 4  \\   3 & 0 & 5  \\ \end{matrix} \right]=\left[ \begin{matrix}   14 & 0 & 42  \\  18 & -1 & 56  \\   22 & -2 & 70  \\ \end{matrix} \right]\]

v. \[\mathbf{\left[ \begin{matrix}   2 & 1  \\   3 & 2  \\   -1 & 1  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 0 & 1  \\   -1 & 2 & 1  \\ \end{matrix} \right]}\]

Ans: We have to find \[\left[ \begin{matrix}   2 & 1  \\   3 & 2  \\   -1 & 1  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 0 & 1  \\   -1 & 2 & 1  \\ \end{matrix} \right]\]

\[\therefore \left[ \begin{matrix}   2 & 1  \\   3 & 2  \\   -1 & 1  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 0 & 1  \\   -1 & 2 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}  2\left( 1 \right)+1\left( -1 \right) & 2\left( 0 \right)+1\left( 2 \right) & 2\left( 1 \right)+1\left( 1 \right)  \\   3\left( 1 \right)+2\left( -1 \right) & 3\left( 0 \right)+2\left( 2 \right) & 3\left( 1 \right)+2\left( 1 \right)  \\   -1\left( 1 \right)+1\left( -1 \right) & -1\left( 0 \right)+1\left( 2 \right) & -1\left( 1 \right)+1\left( 1 \right)  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   2 & 1  \\   3 & 2  \\   -1 & 1  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 0 & 1  \\   -1 & 2 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 2 & 3  \\   1 & 4 & 5  \\   -2 & 2 & 0  \\ \end{matrix} \right]\]

vi. \[\mathbf{\left[ \begin{matrix}   3 & -1 & 3  \\   -1 & 0 & 2  \\ \end{matrix} \right]\left[ \begin{matrix}   2 & -3  \\   1 & 0  \\   3 & 1  \\ \end{matrix} \right]}\]

Ans: We have to find \[\left[ \begin{matrix}  3 & -1 & 3  \\   -1 & 0 & 2  \\ \end{matrix} \right]\left[ \begin{matrix}   2 & -3  \\   1 & 0  \\   3 & 1  \\ \end{matrix} \right]\]

\[\therefore \left[ \begin{matrix}   3 & -1 & 3  \\   -1 & 0 & 2  \\ \end{matrix} \right]\left[ \begin{matrix}   2 & -3  \\  1 & 0  \\   3 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}  3\left( 2 \right)-1\left( 1 \right)+3\left( 3 \right) & 3\left( -3 \right)-1\left( 0 \right)+3\left( 1 \right)  \\  -1\left( 2 \right)+0\left( 1 \right)+2\left( 3 \right) & -1\left( -3 \right)+0\left( 0 \right)+2\left( 1 \right)  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   3 & -1 & 3  \\   -1 & 0 & 2  \\ \end{matrix} \right]\left[ \begin{matrix}   2 & -3  \\   1 & 0  \\   3 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix} 14 & -6  \\   4 & 5  \\ \end{matrix} \right]\]

4.If,

\[\mathbf{A=\left[\begin{matrix} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1  \\ \end{matrix}\right],B=\left[ \begin{matrix}  3 & -1 & 2  \\  4 & 2 & 5  \\  2 & 0 & 3  \\ \end{matrix}\right],C=\left[ \begin{matrix}  4 & 1 & 2  \\  0 & 3 & 2  \\  1 & -2 & 3  \\ \end{matrix}\right]}\],

then Compute ( A+B ) and ( B-C ).Also, verify that A+( B-C)=( A+B )-C}.

Ans: Given  \[A=\left[ \begin{matrix}   1 & 2 & -3  \\   5 & 0 & 2  \\   1 & -1 & 1  \\ \end{matrix} \right],B=\left[ \begin{matrix}   3 & -1 & 2  \\   4 & 2 & 5  \\   2 & 0 & 3  \\ \end{matrix} \right],C=\left[ \begin{matrix}   4 & 1 & 2  \\   0 & 3 & 2  \\   1 & -2 & 3  \\ \end{matrix} \right]\]

\[\therefore A+B=\left[ \begin{matrix}   1 & 2 & -3  \\   5 & 0 & 2  \\   1 & -1 & 1  \\ \end{matrix} \right]+\left[ \begin{matrix}   3 & -1 & 2  \\   4 & 2 & 5  \\   2 & 0 & 3  \\ \end{matrix} \right]\]

\[\Rightarrow A+B=\left[ \begin{matrix}   1+3 & 2-1 & -3+2  \\   5+4 & 0+2 & 2+5 \\  1+2 & -1+0 & 1+3  \\ \end{matrix} \right]\]

\[\Rightarrow A+B=\left[ \begin{matrix}   4 & 1 & -1  \\   9 & 2 & 7  \\   3 & -1 & 4  \\ \end{matrix} \right]\]

And \[B-C=\left[ \begin{matrix}   3 & -1 & 2  \\   4 & 2 & 5  \\   2 & 0 & 3  \\ \end{matrix} \right]-\left[ \begin{matrix}   4 & 1 & 2  \\   0 & 3 & 2  \\   1 & -2 & 3  \\ \end{matrix} \right]\]

\[\Rightarrow B-C=\left[ \begin{matrix}   3-4 & -1-1 & 2-2  \\   4-0 & 2-3 & 5-2  \\   2-1 & 0+2 & 3-3  \\ \end{matrix} \right]\]

\[\therefore B-C=\left[ \begin{matrix}   -1 & -2 & 0  \\   4 & -1 & 3  \\   1 & 2 & 0  \\ \end{matrix} \right]\]

Now,

\[A+\left( B-C \right)=\left[ \begin{matrix}  1 & 2 & -3  \\   5 & 0 & 2  \\   1 & -1 & 1  \\ \end{matrix} \right]+\left[ \begin{matrix}   -1 & -2 & 0  \\   4 & -1 & 3  \\   1 & 2 & 0  \\ \end{matrix} \right]\]

\[\Rightarrow A+\left( B-C \right)=\left[ \begin{matrix}   1-1 & 2-2 & -3+0  \\   5+4 & 0-1 & 2+3  \\   1+1 & -1+2 & 1+0  \\ \end{matrix} \right]\]

\[\therefore A+\left( B-C \right)=\left[ \begin{matrix}   0 & 0 & -3  \\   9 & -1 & 5  \\   2 & 1 & 1  \\ \end{matrix} \right]\]                                                  …(1)

And \[\left( A+B \right)-C=\left[ \begin{matrix}   4 & 1 & -1  \\   9 & 2 & 7  \\   3 & -1 & 4  \\ \end{matrix} \right]-\left[ \begin{matrix}  4 & 1 & 2  \\   0 & 3 & 2  \\   1 & -2 & 3  \\ \end{matrix} \right]\]

\[\Rightarrow \left( A+B \right)-C=\left[ \begin{matrix}   4-4 & 1-1 & -1-2  \\   9-0 & 2-3 & 7-2  \\   3-1 & -1+2 & 4-3  \\ \end{matrix} \right]\]

\[\therefore \left( A+B \right)-C=\left[ \begin{matrix}   0 & 0 & -3  \\   9 & -1 & 5  \\   2 & 1 & 1  \\ \end{matrix} \right]\]                                                  …(2)

Thus, from equation (1) and (2),

\[A+\left( B-C \right)=\left( A+B \right)-C\]

Hence proved.

5. If \[A=\left[ \begin{matrix}   \dfrac{2}{3} & 1 & \dfrac{5}{3}  \\   \dfrac{1}{3} & \dfrac{2}{3} & \dfrac{4}{3}  \\   \dfrac{7}{3} & 2 & \dfrac{2}{3}  \\ \end{matrix} \right]\] and \[B=\left[ \begin{matrix}   \dfrac{2}{5} & \dfrac{3}{5} & 1  \\   \dfrac{1}{5} & \dfrac{2}{5} & \dfrac{4}{5}  \\    \dfrac{7}{5} & \dfrac{6}{5} & \dfrac{2}{5}  \\ \end{matrix} \right]\] then compute \[3A-5B\] .

Ans: Given that \[A=\left[ \begin{matrix}   \dfrac{2}{3} & 1 & \dfrac{5}{3}  \\   \dfrac{1}{3} & \dfrac{2}{3} & \dfrac{4}{3}  \\   \dfrac{7}{3} & 2 & \dfrac{2}{3}  \\ \end{matrix} \right]\] and \[B=\left[ \begin{matrix}   \dfrac{2}{5} & \dfrac{3}{5} & 1  \\   \dfrac{1}{5} & \dfrac{2}{5} & \dfrac{4}{5}  \\    \dfrac{7}{5} & \dfrac{6}{5} & \dfrac{2}{5}  \\ \end{matrix} \right]\] 

\[\therefore 3A-5B=3\left[ \begin{matrix}   \dfrac{2}{3} & 1 & \dfrac{5}{3}  \\   \dfrac{1}{3} & \dfrac{2}{3} & \dfrac{4}{3}  \\   \dfrac{7}{3} & 2 & \dfrac{2}{3}  \\ \end{matrix} \right]-5\left[ \begin{matrix}   \dfrac{2}{5} & \dfrac{3}{5} & 1  \\   \dfrac{1}{5} & \dfrac{2}{5} & \dfrac{4}{5}  \\   \dfrac{7}{5} & \dfrac{6}{5} &\dfrac{2}{5}  \\ \end{matrix} \right]\]

\[\Rightarrow 3A-5B=\left[ \begin{matrix}   2 & 3 & 5  \\   1 & 2 & 4  \\   7 & 6 & 2  \\ \end{matrix} \right]-\left[ \begin{matrix}   2 & 3 & 5  \\   1 & 2 & 4  \\   7 & 6 & 2  \\ \end{matrix} \right]\]

\[\Rightarrow 3A-5B=\left[ \begin{matrix}   0 & 0 & 0  \\   0 & 0 & 0  \\   0 & 0 & 0  \\ \end{matrix} \right]\]

6. Simplify \[\mathbf{\cos \theta \left[ \begin{matrix}   \cos \theta  & \sin \theta   \\   -\sin \theta  & \cos \theta   \\ \end{matrix} \right]+\sin \theta \left[ \begin{matrix}  \sin \theta  & -\cos \theta   \\  \cos \theta  & \sin \theta   \\ \end{matrix} \right]}\] .

Ans: We have to simplify \[\cos \theta \left[ \begin{matrix}   \cos \theta  & \sin \theta   \\   -\sin \theta  & \cos \theta   \\ \end{matrix} \right]+\sin \theta \left[ \begin{matrix}  \sin \theta  & -\cos \theta   \\  \cos \theta  & \sin \theta   \\ \end{matrix} \right]\] 

\[\therefore \cos \theta \left[ \begin{matrix} \cos \theta  & \sin \theta   \\   -\sin \theta  & \cos \theta   \\ \end{matrix} \right]+\sin \theta \left[ \begin{matrix}   \sin \theta  & -\cos \theta   \\   \cos \theta  & \sin \theta   \\ \end{matrix} \right]=\left[ \begin{matrix}   {{\cos }^{2}}\theta  & \cos \theta \sin \theta   \\   -\cos \theta \sin \theta  & {{\cos }^{2}}\theta   \\ \end{matrix} \right]+\left[ \begin{matrix}   {{\sin }^{2}}\theta  & -\sin \theta \cos \theta   \\   \sin \theta \cos \theta  & {{\sin }^{2}}\theta   \\ \end{matrix} \right]\]

\[\Rightarrow \cos \theta \left[ \begin{matrix}   \cos \theta  & \sin \theta   \\   -\sin \theta  & \cos \theta   \\ \end{matrix} \right]+\sin \theta \left[ \begin{matrix}   \sin \theta  & -\cos \theta   \\  \cos \theta  & \sin \theta   \\ \end{matrix} \right]=\left[ \begin{matrix}  {{\cos }^{2}}\theta +{{\sin }^{2}}\theta  & \cos \theta \sin \theta -\cos \theta \sin \theta   \\  -\cos \theta \sin \theta +\cos \theta \sin \theta  & {{\cos }^{2}}\theta +{{\sin }^{2}}\theta   \\ \end{matrix} \right]\]

\[\Rightarrow \cos \theta \left[ \begin{matrix}  \cos \theta  & \sin \theta   \\   -\sin \theta  & \cos \theta   \\ \end{matrix} \right]+\sin \theta \left[ \begin{matrix}  \sin \theta  & -\cos \theta   \\  \cos \theta  & \sin \theta   \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]\]

7. Find \[X\] and \[Y\] ,if

i. \[{X+Y=}\left [ \begin{matrix}   7 & 0  \\   2 & 5  \\ \end{matrix} \right]\] and \[X-Y=\left[ \begin{matrix}   3 & 0  \\   0 & 3  \\ \end{matrix} \right]\]

Ans: Given:

\[X+Y=\left[ \begin{matrix}   7 & 0  \\   2 & 5  \\ \end{matrix} \right]\]        …(1)

\[X-Y=\left[ \begin{matrix}   3 & 0  \\   0 & 3  \\ \end{matrix} \right]\]   …(2)

Adding equations (1) and (2), we get,

\[2X=\left[ \begin{matrix}   7 & 0  \\   2 & 5  \\ \end{matrix} \right]+\left[ \begin{matrix}   3 & 0  \\   0 & 3  \\ \end{matrix} \right]\]

\[\Rightarrow 2X=\left[ \begin{matrix}   10 & 0  \\   2 & 8  \\ \end{matrix} \right]\]

\[\Rightarrow X=\dfrac{1}{2}\left[ \begin{matrix}   10 & 0  \\   2 & 8  \\ \end{matrix} \right]\]

\[\therefore X=\left[ \begin{matrix}   5 & 0  \\   1 & 4  \\ \end{matrix} \right]\]

Now, since \[X+Y=\left[ \begin{matrix}    7 & 0  \\   2 & 5  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}  5 & 0  \\   1 & 4  \\ \end{matrix} \right]+Y=\left[ \begin{matrix}   7 & 0  \\   2 & 5  \\ \end{matrix} \right]\]

\[\Rightarrow Y=\left[ \begin{matrix}   7 & 0  \\   2 & 5  \\ \end{matrix} \right]-\left[ \begin{matrix}   5 & 0  \\   1 & 4  \\ \end{matrix} \right]\]

\[\therefore Y=\left[ \begin{matrix}   2 & 0  \\   1 & 1  \\ \end{matrix} \right]\]

\[\therefore X=\left[ \begin{matrix}   5 & 0  \\   1 & 4  \\ \end{matrix} \right],Y=\left[ \begin{matrix}   2 & 0  \\   1 & 1  \\ \end{matrix} \right]\]

ii. \[\mathbf{2X+3Y=\left[ \begin{matrix}   2 & 3  \\   4 & 0  \\ \end{matrix} \right]}\] and \[\mathbf{3X+2Y=\left[ \begin{matrix}   2 & -2  \\   -1 & 5  \\ \end{matrix} \right]}\]

Ans: Given:

\[2X+3Y=\left[ \begin{matrix}   2 & 3  \\   4 & 0  \\ \end{matrix} \right]\]                               …(1)

\[3X+2Y=\left[ \begin{matrix}   2 & -2  \\   -1 & 5  \\ \end{matrix} \right]\]                            …(2)

Multiplying equation (1) with two we get,

\[2\left( 2X+3Y \right)=2\left[ \begin{matrix}   2 & 3  \\   4 & 0  \\ \end{matrix} \right]\]

\[\Rightarrow \left( 4X+6Y \right)=\left[ \begin{matrix}   4 & 6  \\   8 & 0  \\ \end{matrix} \right]\]                      …(3)

Multiplying equation (2) with three we get,

\[3\left( 3X+2Y \right)=3\left[ \begin{matrix}   2 & -2  \\   -1 & 5  \\ \end{matrix} \right]\]

\[\Rightarrow \left( 9X+6Y \right)=\left[ \begin{matrix}   6 & -6  \\   -3 & 15  \\ \end{matrix} \right]\]                 …(4)

From equation (3) and (4),

\[\left( 4X+6Y \right)-\left( 9X+6Y \right)=\left[ \begin{matrix}   4 & 6  \\   8 & 0  \\ \end{matrix} \right]-\left[ \begin{matrix} 6 & -6  \\   -3 & 15  \\ \end{matrix} \right]\]

\[\Rightarrow -5X=\left[ \begin{matrix}   -2 & 12  \\   11 & -15  \\ \end{matrix} \right]\]

\[\Rightarrow X=\dfrac{-1}{5}\left[ \begin{matrix}   -2 & 12  \\   11 & -15  \\ \end{matrix} \right]\]

\[\therefore X=\left[ \begin{matrix}  \dfrac{2}{5} & \dfrac{-12}{5}  \\   \dfrac{-11}{5} & 3  \\ \end{matrix} \right]\]

Now, since \[2X+3Y=\left[ \begin{matrix}   2 & 3  \\   4 & 0  \\ \end{matrix} \right]\]

\[\therefore 2\left[ \begin{matrix}   \dfrac{2}{5} & \dfrac{-12}{5}  \\   \dfrac{-11}{5} & 3  \\ \end{matrix} \right]+3Y=\left[ \begin{matrix}   2 & 3  \\   4 & 0  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \dfrac{4}{5} & \dfrac{-24}{5}  \\  \dfrac{-22}{5} & 6  \\ \end{matrix} \right]+3Y=\left[ \begin{matrix}   2 & 3  \\   4 & 0  \\ \end{matrix} \right]\]

\[\Rightarrow 3Y=\left[ \begin{matrix}   2 & 3  \\   4 & 0  \\ \end{matrix} \right]-\left[ \begin{matrix}   \dfrac{4}{5} & \dfrac{-24}{5}  \\   \dfrac{-22}{5} & 6  \\ \end{matrix} \right]\]

\[\Rightarrow 3Y=\left[ \begin{matrix}   \dfrac{6}{5} & \dfrac{39}{5}  \\    \dfrac{42}{5} & -6  \\ \end{matrix} \right]\]

\[\therefore Y=\left[ \begin{matrix}  \dfrac{2}{5} & \dfrac{13}{5}  \\   \dfrac{14}{5} & -2  \\ \end{matrix} \right]\]

\[\therefore X=\left[ \begin{matrix}   \dfrac{2}{5} & \dfrac{-12}{5}  \\   \dfrac{-11}{5} & 3  \\ \end{matrix} \right],Y=\left[ \begin{matrix}   \dfrac{2}{5} & \dfrac{13}{5}  \\   \dfrac{14}{5} & -2  \\ \end{matrix} \right]\]

8. Find \[\mathbf{X}\] ,if \[\mathbf{Y=\left[ \begin{matrix}   3 & 2  \\   1 & 4  \\ \end{matrix} \right]}\] and \[\mathbf{2X+Y=\left[ \begin{matrix}   1 & 0  \\   -3 & 2  \\ \end{matrix} \right]}\] .

Ans: Given \[Y=\left[ \begin{matrix}   3 & 2  \\   1 & 4  \\ \end{matrix} \right]\]

And \[2X+Y=\left[ \begin{matrix}   1 & 0  \\   -3 & 2  \\ \end{matrix} \right]\]

\[\Rightarrow 2X+\left[ \begin{matrix}   3 & 2  \\   1 & 4  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0  \\   -3 & 2  \\ \end{matrix} \right]\]

\[\Rightarrow 2X=\left[ \begin{matrix}   1 & 0  \\   -3 & 2  \\ \end{matrix} \right]-\left[ \begin{matrix}   3 & 2  \\   1 & 4  \\ \end{matrix} \right]\]

\[\Rightarrow 2X=\left[ \begin{matrix}   -2 & -2  \\   -4 & -2  \\ \end{matrix} \right]\]

\[\Rightarrow X=\dfrac{1}{2}\left[ \begin{matrix}   -2 & -2  \\   -4 & -2  \\ \end{matrix} \right]\]

\[\therefore X=\left[ \begin{matrix}   -1 & -1  \\   -2 & -1  \\ \end{matrix} \right]\]

9. Find \[X\] and \[Y\] ,if \[\mathbf{2\left[ \begin{matrix}   1 & 3  \\   0 & x  \\ \end{matrix} \right]+\left[ \begin{matrix}   y & 0  \\   1 & 2  \\ \end{matrix} \right]=\left[ \begin{matrix}   5 & 6  \\   1 & 8  \\ \end{matrix} \right]}\]

Ans: Given: \[2\left[ \begin{matrix}   1 & 3  \\   0 & x  \\ \end{matrix} \right]+\left[ \begin{matrix}   y & 0  \\    1 & 2  \\ \end{matrix} \right]=\left[ \begin{matrix}   5 & 6  \\   1 & 8  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   2 & 6  \\   0 & 2x  \\ \end{matrix} \right]+\left[ \begin{matrix}   y & 0  \\   1 & 2  \\ \end{matrix} \right]=\left[ \begin{matrix}   5 & 6  \\   1 & 8  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   2+y & 6+0  \\   0+1 & 2x+2  \\ \end{matrix} \right]=\left[ \begin{matrix}   5 & 6  \\   1 & 8  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   2+y & 6  \\   1 & 2x+2  \\ \end{matrix} \right]=\left[ \begin{matrix}   5 & 6  \\   1 & 8  \\ \end{matrix} \right]\]

Comparing the corresponding elements of these two matrices, we get:

\[2+y=5\]

\[\Rightarrow y=3\]

\[2x+2=8\]

\[\Rightarrow x=3\]

\[\therefore x=3,y=3\]

10. Solve the equation for \[X,Y,Z\] and \[t\] if \[\mathbf{2\left[ \begin{matrix}   x & z  \\   y & t  \\ \end{matrix} \right]+3\left[ \begin{matrix}   1 & -1  \\   0 & 2  \\ \end{matrix} \right]=3\left[ \begin{matrix}   3 & 5  \\   4 & 6  \\ \end{matrix} \right]}\]

Ans: Given: \[2\left[ \begin{matrix}   x & z  \\   y & t  \\ \end{matrix} \right]+3\left[ \begin{matrix}   1 & -1  \\   0 & 2  \\ \end{matrix} \right]=3\left[ \begin{matrix}   3 & 5  \\   4 & 6  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   2x & 2z  \\   2y & 2t  \\ \end{matrix} \right]+\left[ \begin{matrix}   3 & -3  \\   0 & 6  \\ \end{matrix} \right]=\left[ \begin{matrix}   9 & 15  \\   12 & 18  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   2x+3 & 2z-3  \\   2y & 2t+6  \\ \end{matrix} \right]=\left[ \begin{matrix}   9 & 15  \\   12 & 18  \\ \end{matrix} \right]\]

Equating the corresponding elements of these two matrices, we get:

\[2x+3=9\]

\[\Rightarrow x=3\]

\[2y=12\]

\[\Rightarrow y=6\]

\[2z-3=15\]

\[\Rightarrow z=9\]

\[2t+6=18\]

\[\Rightarrow t=6\]

\[\therefore x=3,y=6,z=9,t=6\]

11. If \[\mathbf{x\left[ \begin{matrix}   2  \\   3  \\ \end{matrix} \right]+y\left[ \begin{matrix}   -1  \\   1  \\ \end{matrix} \right]=\left[ \begin{matrix}  10  \\   5  \\ \end{matrix} \right]}\] ,find values of \[x\] and \[y\] .

Ans: Given: \[x\left[ \begin{matrix}   2  \\   3  \\ \end{matrix} \right]+y\left[ \begin{matrix}   -1  \\   1  \\ \end{matrix} \right]=\left[ \begin{matrix}  10  \\   5  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   2x  \\   3x  \\ \end{matrix} \right]+\left[ \begin{matrix}   -y  \\   y  \\ \end{matrix} \right]=\left[ \begin{matrix}   10  \\   5  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   2x-y  \\   3x+y  \\ \end{matrix} \right]=\left[ \begin{matrix}   10  \\   5  \\ \end{matrix} \right]\]

Equating the corresponding elements of these two matrices, we get:

\[2x-y=10\] and

\[3x+y=5\]

Adding these two equations, we get:

\[5x=15\]

\[\Rightarrow x=3\]

Now, since \[3x+y=5\]

\[\Rightarrow y=5-3x\]

\[\Rightarrow y=5-9\]

\[\Rightarrow y=-4\]

\[\therefore x=3,y=-4\]

12. Given \[\mathbf{3\left[ \begin{matrix}   x & y  \\   z & w  \\ \end{matrix} \right]=\left[ \begin{matrix}   x & 6  \\   -1 & 2w  \\ \end{matrix} \right]+\left[ \begin{matrix}   4 & x+y  \\   z+w & 2w+3  \\ \end{matrix} \right]}\] ,find the values of \[x,y,z\] and \[w\] .

Ans: Given: \[3\left[ \begin{matrix}   x & y  \\   z & w  \\ \end{matrix} \right]=\left[ \begin{matrix}   x & 6  \\   -1 & 2w  \\ \end{matrix} \right]+\left[ \begin{matrix}   4 & x+y  \\   z+w & 2w+3  \\ \end{matrix} \right]\] 

\[\Rightarrow \left[ \begin{matrix}   3x & 3y  \\   3z & 3w  \\ \end{matrix} \right]=\left[ \begin{matrix}   x+4 & 6+x+y  \\   -1+z+w & 2w+3  \\ \end{matrix} \right]\]

Equating the corresponding elements of these two matrices, we get:

\[3x=x+4\]

\[\Rightarrow x=2\]

\[3y=6+x+y\]

\[\Rightarrow y=4\]

\[3w=2w+3\]

\[\Rightarrow w=3\]

\[3z=-1+z+w\]

\[\Rightarrow z=1\]

\[\therefore x=2,y=4,z=1,w=3\]

13. If \[\mathbf{F\left( x \right)=\left[ \begin{matrix}   \cos x & -\sin x & 0  \\  \sin x & \cos x & 0  \\   0 & 0 & 1  \\ \end{matrix} \right]}\] , show that \[\mathbf{F\left( x \right)F\left( y \right)=F\left( x+y \right)}\] .

Ans: Here, \[F\left( x \right)=\left[ \begin{matrix}   \cos x & -\sin x & 0  \\  \sin x & \cos x & 0  \\   0 & 0 & 1  \\ \end{matrix} \right]\]

\[F\left( y \right)=\left[ \begin{matrix}   \cos y & -\sin y & 0  \\   \sin y & \cos y & 0 \\   0 & 0 & 1  \\ \end{matrix} \right]\]

\[\therefore F\left( x \right)F\left( y \right)=\left[ \begin{matrix}   \cos x & -\sin x & 0  \\   \sin x & \cos x & 0  \\   0 & 0 & 1  \\ \end{matrix} \right]\left[ \begin{matrix}   \cos y & -\sin y & 0  \\   \sin y & \cos y & 0  \\   0 & 0 & 1  \\ \end{matrix} \right]\]

\[\Rightarrow F\left( x \right)F\left( y \right)=\left[ \begin{matrix}   \cos x\cos y-\sin x\sin y+0 & -\cos x\sin y-\sin x\cos y+0 & 0  \\   \sin x\cos y-\cos x\sin y+0 & -\sin x\sin y+\cos x\cos y+0 & 0  \\   0 & 0 & 1  \\ \end{matrix} \right]\]

\[\Rightarrow F\left( x \right)F\left( y \right)=\left[ \begin{matrix}   \cos \left( x+y \right) & -\sin \left( x+y \right) & 0  \\   \sin \left( x+y \right) & \cos \left( x+y \right) & 0  \\   0 & 0 & 1  \\ \end{matrix} \right]\]                            …(1)

And \[F\left( x+y \right)=\left[ \begin{matrix}   \cos \left( x+y \right) & -\sin \left( x+y \right) & 0  \\   \sin \left( x+y \right) & \cos \left( x+y \right) & 0  \\   0 & 0 & 1  \\ \end{matrix} \right]\]                               …(2)

From equation (1) and (2),

\[F\left( x \right)F\left( y \right)=F\left( x+y \right)\]

Hence proved.

14. Show that

i. \[\mathbf{\left[ \begin{matrix}   5 & -1  \\   6 & 7  \\ \end{matrix} \right]\left[ \begin{matrix}   2 & 1  \\   3 & 4  \\ \end{matrix} \right]\ne \left[ \begin{matrix}   2 & 1  \\   3 & 4  \\ \end{matrix} \right]\left[ \begin{matrix}   5 & -1  \\   6 & 7  \\ \end{matrix} \right]}\]

Ans: LHS: \[\left[ \begin{matrix}   5 & -1  \\   6 & 7  \\ \end{matrix} \right]\left[ \begin{matrix}   2 & 1  \\   3 & 4  \\ \end{matrix} \right]=\left[ \begin{matrix}   5\left( 2 \right)-1\left( 3 \right) & 5\left( 1 \right)-1\left( 4 \right)  \\   6\left( 2 \right)+7\left( 3 \right) & 6\left( 1 \right)+7\left( 4 \right)  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   5 & -1  \\   6 & 7  \\ \end{matrix} \right]\left[ \begin{matrix}   2 & 1  \\   3 & 4  \\ \end{matrix} \right]=\left[ \begin{matrix}   7 & 1  \\   33 & 34  \\ \end{matrix} \right]\]                …(1)

RHS: \[\left[ \begin{matrix}   2 & 1  \\   3 & 4  \\ \end{matrix} \right]\left[ \begin{matrix}   5 & -1  \\   6 & 7  \\ \end{matrix} \right]=\left[ \begin{matrix}   2\left( 5 \right)+1\left( 6 \right) & 2\left( -1 \right)+1\left( 7 \right)  \\   3\left( 5 \right)+4\left( 6 \right) & 3\left( -1 \right)+4\left( 7 \right)  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   2 & 1  \\   3 & 4  \\ \end{matrix} \right]\left[ \begin{matrix}   5 & -1  \\   6 & 7  \\ \end{matrix} \right]=\left[ \begin{matrix}   16 & 5  \\   39 & 25  \\ \end{matrix} \right]\]        …(2)

From equation (1) and (2),

\[LHS\ne RHS\]

Hence proved.

ii. \[\mathbf{\left[ \begin{matrix}   1 & 2 & 3  \\   0 & 1 & 0  \\   1 & 1 & 0  \\ \end{matrix} \right]\left[ \begin{matrix}   -1 & 1 & 0  \\   0 & -1 & 1  \\   2 & 3 & 4  \\ \end{matrix} \right]\ne \left[ \begin{matrix}   -1 & 1 & 0  \\   0 & -1 & 1  \\   2 & 3 & 4  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 2 & 3  \\   0 & 1 & 0  \\   1 & 1 & 0  \\ \end{matrix} \right]}\]

Ans: LHS: \[\left[ \begin{matrix}   1 & 2 & 3  \\   0 & 1 & 0  \\   1 & 1 & 0  \\ \end{matrix} \right]\left[ \begin{matrix}   -1 & 1 & 0  \\   0 & -1 & 1  \\   2 & 3 & 4  \\ \end{matrix} \right]=\left[ \begin{matrix}   1\left( -1 \right)+2\left( 0 \right)+3\left( 2 \right) & 1\left( 1 \right)+2\left( -1 \right)+3\left( 3 \right) & 1\left( 0 \right)+2\left( 1 \right)+3\left( 4 \right)  \\   0\left( -1 \right)+1\left( 0 \right)+0\left( 2 \right) & 0\left( 1 \right)+1\left( -1 \right)+0\left( 3 \right) & 0\left( 0 \right)+1\left( 1 \right)+0\left( 4 \right)  \\   1\left( -1 \right)+1\left( 0 \right)+0\left( 2 \right) & 1\left( 1 \right)+1\left( -1 \right)+0\left( 3 \right) & 1\left( 0 \right)+1\left( 1 \right)+0\left( 4 \right)  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   1 & 2 & 3  \\   0 & 1 & 0  \\   1 & 1 & 0  \\ \end{matrix} \right]\left[ \begin{matrix}   -1 & 1 & 0  \\   0 & -1 & 1  \\   2 & 3 & 4  \\ \end{matrix} \right]=\left[ \begin{matrix}   5 & 8 & 14  \\   0 & -1 & 1  \\   2 & 3 & 4  \\ \end{matrix} \right]\]                            …(1)

RHS: \[\left[ \begin{matrix}   -1 & 1 & 0  \\   0 & -1 & 1  \\   2 & 3 & 4  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 2 & 3  \\   0 & 1 & 0  \\   1 & 1 & 0  \\ \end{matrix} \right]=\left[ \begin{matrix}  -1\left( 1 \right)+1\left( 0 \right)+0\left( 1 \right) & -1\left( 2 \right)+1\left( 1 \right)+0\left( 1 \right) & -1\left( 3 \right)+1\left( 0 \right)+0\left( 0 \right)  \\   0\left( 1 \right)-1\left( 0 \right)+1\left( 1 \right) & 0\left( 2 \right)-1\left( 1 \right)+1\left( 1 \right) & 0\left( 3 \right)-1\left( 0 \right)+1\left( 0 \right)  \\  2\left( 1 \right)+3\left( 0 \right)+4\left( 1 \right) & 2\left( 2 \right)+3\left( 1 \right)+4\left( 1 \right) & 2\left( 3 \right)+3\left( 0 \right)+4\left( 0 \right)  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   -1 & 1 & 0  \\   0 & -1 & 1  \\   2 & 3 & 4  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 2 & 3  \\   0 & 1 & 0  \\   1 & 1 & 0  \\ \end{matrix} \right]=\left[ \begin{matrix}   -1 & -1 & 3  \\   1 & 0 & 0  \\   6 & 11 & 6  \\ \end{matrix} \right]\]                          …(2)

From equation (1) and (2),

\[LHS\ne RHS\]

Hence proved.

15. Find \[\mathbf{{{A}^{2}}-5A+6I}\] if \[\mathbf{A=\left[ \begin{matrix}   2 & 0 & 1  \\   2 & 1 & 3  \\   1 & -1 & 0  \\ \end{matrix} \right]}\] .

Ans: Given: \[A=\left[ \begin{matrix}   2 & 0 & 1  \\   2 & 1 & 3  \\   1 & -1 & 0  \\ \end{matrix} \right]\]

\[{{A}^{2}}=AA\]

\[\therefore {{A}^{2}}=\left[ \begin{matrix}   2 & 0 & 1  \\   2 & 1 & 3  \\   1 & -1 & 0  \\ \end{matrix} \right]\left[ \begin{matrix}   2 & 0 & 1  \\   2 & 1 & 3  \\   1 & -1 & 0  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   2\left( 2 \right)+0\left( 2 \right)+1\left( 1 \right) & 2\left( 0 \right)+0\left( 1 \right)+1\left( -1 \right) & 2\left( 1 \right)+0\left( 3 \right)+1\left( 0 \right)  \\   2\left( 2 \right)+1\left( 2 \right)+3\left( 1 \right) & 2\left( 0 \right)+1\left( 1 \right)+3\left( -1 \right) & 2\left( 1 \right)+1\left( 3 \right)+3\left( 0 \right)  \\   1\left( 2 \right)-1\left( 2 \right)+0\left( 1 \right) & 1\left( 0 \right)-1\left( 1 \right)+0\left( -1 \right) & 1\left( 1 \right)-1\left( 3 \right)+0\left( 0 \right)  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   5 & -1 & 2  \\   9 & -2 & 5  \\   0 & -1 & -2  \\ \end{matrix} \right]\]

\[\therefore {{A}^{2}}-5A+6I=\left[ \begin{matrix}   5 & -1 & 2  \\   9 & -2 & 5  \\   0 & -1 & -2  \\ \end{matrix} \right]-5\left[ \begin{matrix}   2 & 0 & 1  \\   2 & 1 & 3  \\   1 & -1 & 0  \\ \end{matrix} \right]+6\left[ \begin{matrix}   1 & 0 & 0  \\   0 & 1 & 0  \\   0 & 0 & 1  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}-5A+6I=\left[ \begin{matrix}   5 & -1 & 2  \\   9 & -2 & 5  \\   0 & -1 & -2  \\ \end{matrix} \right]-\left[ \begin{matrix}   10 & 0 & 5  \\   10 & 5 & 15  \\   5 & -5 & 0  \\ \end{matrix} \right]+\left[ \begin{matrix}   6 & 0 & 0  \\   0 & 6 & 0  \\   0 & 0 & 6  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}-5A+6I=\left[ \begin{matrix}   5-10 & -1-0 & 2-5  \\   9-10 & -2-5 & 5-15  \\   0-5 & -1+5 & -2-0  \\ \end{matrix} \right]+\left[ \begin{matrix}   6 & 0 & 0  \\   0 & 6 & 0  \\   0 & 0 & 6  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}-5A+6I=\left[ \begin{matrix}   -5 & -1 & -3  \\   -1 & -7 & -10 \\   -5 & 4 & -2  \\ \end{matrix} \right]+\left[ \begin{matrix}   6 & 0 & 0  \\   0 & 6 & 0  \\   0 & 0 & 6  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}-5A+6I=\left[ \begin{matrix}   1 & -1 & -3  \\   -1 & -1 & -10 \\   -5 & 4 & 4  \\ \end{matrix} \right]\]

16. If \[\mathbf{A=\left[ \begin{matrix}   1 & 0 & 2  \\   0 & 2 & 1  \\   2 & 0 & 3  \\ \end{matrix} \right]}\] , prove that \[\mathbf{{{A}^{3}}-6{{A}^{2}}+7A+21=0}\] .

Ans: Given: \[A=\left[ \begin{matrix}   1 & 0 & 2  \\   0 & 2 & 1  \\   2 & 0 & 3  \\ \end{matrix} \right]\]

\[{{A}^{2}}=AA\]

\[\therefore {{A}^{2}}=\left[ \begin{matrix}   1 & 0 & 2  \\   0 & 2 & 1  \\   2 & 0 & 3  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 0 & 2  \\   0 & 2 & 1  \\   2 & 0 & 3  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   1+0+4 & 0+0+0 & 2+0+6  \\   0+0+2 & 0+4+0 & 0+2+3  \\   2+0+6 & 0+0+0 & 4+0+9  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   5 & 0 & 8  \\   2 & 4 & 5  \\   8 & 0 & 13  \\ \end{matrix} \right]\]

Now, \[{{A}^{3}}={{A}^{2}}\cdot A\]

\[\therefore {{A}^{3}}=\left[ \begin{matrix}   5 & 0 & 8  \\   2 & 4 & 5  \\   8 & 0 & 13  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 0 & 2  \\   0 & 2 & 1  \\   2 & 0 & 3  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{3}}=\left[ \begin{matrix}   5+0+16 & 0+0+0 & 10+0+24  \\   2+0+10 & 0+8+0 & 4+4+15  \\   8+0+26 & 0+0+0 & 16+0+39  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{3}}=\left[ \begin{matrix}   21 & 0 & 34  \\   12 & 8 & 23  \\   34 & 0 & 55  \\ \end{matrix} \right]\]

\[\therefore {{A}^{3}}-6{{A}^{2}}+7A+21=\left[ \begin{matrix}   21 & 0 & 34  \\   12 & 8 & 23  \\   34 & 0 & 55  \\ \end{matrix} \right]-6\left[ \begin{matrix}   5 & 0 & 8  \\   2 & 4 & 5  \\   8 & 0 & 13  \\ \end{matrix} \right]+7\left[ \begin{matrix}   1 & 0 & 2  \\   0 & 2 & 1  \\   2 & 0 & 3  \\ \end{matrix} \right]+2\left[ \begin{matrix}   1 & 0 & 0  \\   0 & 1 & 0  \\   0 & 0 & 1  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{3}}-6{{A}^{2}}+7A+21=\left[ \begin{matrix}   21 & 0 & 34  \\   12 & 8 & 23  \\   34 & 0 & 55  \\ \end{matrix} \right]-\left[ \begin{matrix}   30 & 0 & 48  \\   12 & 24 & 30  \\   48 & 0 & 78  \\ \end{matrix} \right]+\left[ \begin{matrix}   7 & 0 & 14  \\   0 & 14 & 7  \\   14 & 0 & 21  \\ \end{matrix} \right]+\left[ \begin{matrix}   2 & 0 & 0  \\   0 & 2 & 0  \\   0 & 0 & 2  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{3}}-6{{A}^{2}}+7A+21=\left[ \begin{matrix}   21-30+7+2 & 0-0+0+0 & 34-48+14+0  \\   12-12+0+0 & 8-24+14+2 & 23-30+7+0  \\   34-48+14+0 & 0-0+0+0 & 55-78+21+2  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{3}}-6{{A}^{2}}+7A+21=\left[ \begin{matrix}   0 & 0 & 0  \\   0 & 0 & 0  \\   0 & 0 & 0  \\ \end{matrix} \right]\]

\[\therefore {{A}^{3}}-6{{A}^{2}}+7A+21=0\]

Hence proved.

17.  If \[\mathbf{A=\left[ \begin{matrix}   3 & -2  \\   4 & -2  \\ \end{matrix} \right]}\] and \[\mathbf{I=\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]}\] , find \[\mathbf{k}\] so that \[\mathbf{{{A}^{2}}=kA-21}\] .

Ans: Given:\[A=\left[ \begin{matrix}   3 & -2  \\   4 & -2  \\ \end{matrix} \right]\]

\[{{A}^{2}}=AA\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   3 & -2  \\   4 & -2  \\ \end{matrix} \right]\left[ \begin{matrix}   3 & -2  \\   4 & -2  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   9-8 & -6+4  \\   12-8 & -8+4  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   1 & -2  \\   4 & -4  \\ \end{matrix} \right]\]

Now, \[{{A}^{2}}=kA-21\]

\[\Rightarrow \left[ \begin{matrix}   1 & -2  \\   4 & -4  \\ \end{matrix} \right]=k\left[ \begin{matrix}   3 & -2  \\   4 & -2  \\ \end{matrix} \right]-2\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   1 & -2  \\   4 & -4  \\ \end{matrix} \right]=\left[ \begin{matrix}   3k & -2k  \\   4k & -2k  \\ \end{matrix} \right]-\left[ \begin{matrix}   2 & 0  \\   0 & 2  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   1 & -2  \\   4 & -4  \\ \end{matrix} \right]=\left[ \begin{matrix}   3k-2 & -2k  \\   4k & -2k-2  \\ \end{matrix} \right]\]

Equating the corresponding elements, we have:

\[3k-2=1\]

\[\Rightarrow k=1\]

Thus, the value of \[k\] is \[1\].

18. If \[\mathbf{A=\left[ \begin{matrix}   0 & -\tan \dfrac{\alpha }{2}  \\   \tan \dfrac{\alpha }{2} & 0  \\ \end{matrix} \right]}\] and \[I\] is the identity matrix of order \[2\] , show that \[\mathbf{I+A=\left( I-A \right)\left[ \begin{matrix}    \cos \alpha  & -\sin \alpha   \\   \sin \alpha  & \cos \alpha   \\ \end{matrix} \right]}\]

Ans: Given: \[A=\left[ \begin{matrix}   0 & -\tan \dfrac{\alpha }{2}  \\   \tan \dfrac{\alpha }{2} & 0  \\ \end{matrix} \right]\]

LHS: \[I+A=\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]+\left[ \begin{matrix}   0 & -\tan \dfrac{\alpha }{2}  \\   \tan \dfrac{\alpha }{2} & 0  \\ \end{matrix} \right]\]

\[\Rightarrow I+A=\left[ \begin{matrix}   1 & -\tan \dfrac{\alpha }{2}  \\   \tan \dfrac{\alpha }{2} & 1  \\ \end{matrix} \right]\]                                           …(1)

RHS: \[\left( I-A \right)\left[ \begin{matrix}   \cos \alpha  & -\sin \alpha   \\   \sin \alpha  & \cos \alpha   \\ \end{matrix} \right]=\left( \left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]-\left[ \begin{matrix}   0 & -\tan \dfrac{\alpha }{2}  \\   \tan \dfrac{\alpha }{2} & 0  \\ \end{matrix} \right] \right)\left[ \begin{matrix}   \cos \alpha  & -\sin \alpha   \\   \sin \alpha  & \cos \alpha   \\ \end{matrix} \right]\]

\[\Rightarrow \left( I-A \right)\left[ \begin{matrix}   \cos \alpha  & -\sin \alpha   \\   \sin \alpha  & \cos \alpha   \\ \end{matrix} \right]=\left[ \begin{matrix}  1 & \tan \dfrac{\alpha }{2}  \\   -\tan \dfrac{\alpha }{2} & 1  \\ \end{matrix} \right]\left[ \begin{matrix}   \cos \alpha  & -\sin \alpha   \\   \sin \alpha  & \cos \alpha   \\ \end{matrix} \right]\]

\[\Rightarrow \left( I-A \right)\left[ \begin{matrix}   \cos \alpha  & -\sin \alpha   \\    \sin \alpha  & \cos \alpha   \\ \end{matrix} \right]=\left[ \begin{matrix}   \cos \alpha +\sin \alpha \tan \dfrac{\alpha }{2} & -\sin \alpha +\cos \alpha \tan \dfrac{\alpha }{2}  \\   -\cos \alpha \tan \dfrac{\alpha }{2}+\sin \alpha  & \sin \alpha \tan \dfrac{\alpha }{2}+\cos \alpha   \\ \end{matrix} \right]\]

\[\Rightarrow \left( I-A \right)\left[ \begin{matrix}   \cos \alpha  & -\sin \alpha   \\   \sin \alpha  & \cos \alpha   \\ \end{matrix} \right]=\left[ \begin{matrix}   1-2{{\sin }^{2}}\dfrac{\alpha }{2}+2\sin \dfrac{\alpha }{2}-\cos \dfrac{\alpha }{2}\tan \dfrac{\alpha }{2} & -2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}+\left( 2{{\cos }^{2}}\dfrac{\alpha }{2}-1 \right)\tan \dfrac{\alpha }{2}  \\   -\left( 2{{\cos }^{2}}\dfrac{\alpha }{2}-1 \right)\tan \dfrac{\alpha }{2}+2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2} & 2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}\tan \dfrac{\alpha }{2}+1-2{{\sin }^{2}}\dfrac{\alpha }{2}  \\ \end{matrix} \right]\]

\[\Rightarrow \left( I-A \right)\left[ \begin{matrix}   \cos \alpha  & -\sin \alpha   \\   \sin \alpha  & \cos \alpha   \\ \end{matrix} \right]=\left[ \begin{matrix}   1-2{{\sin }^{2}}\dfrac{\alpha }{2}+2{{\sin }^{2}}\dfrac{\alpha }{2} & -2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}+2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}-\tan \dfrac{\alpha }{2}  \\   -\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}+\tan \dfrac{\alpha }{2}+2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2} & 2{{\sin }^{2}}\dfrac{\alpha }{2}+1-2{{\sin }^{2}}\dfrac{\alpha }{2}  \\ \end{matrix} \right]\]

\[\Rightarrow \left( I-A \right)\left[ \begin{matrix}   \cos \alpha  & -\sin \alpha   \\    \sin \alpha  & \cos \alpha   \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & -\tan \dfrac{\alpha }{2}  \\    \tan \dfrac{\alpha }{2} & 1  \\ \end{matrix} \right]\]                                        …(2)

Thus, from equation (1) and (2).

\[LHS=RHS\]

Hence proved.

19. A trust fund has Rs \[30,000\] that must be invested in two different types of bonds. The first bond pays \[5%\] interest per year, and the second bond pays \[7%\] interest per year. Using matrix multiplication, determine how to divide Rs \[30,000\] among the two types of bonds. If the trust fund must obtain an annual total interest of:

i. Rs \[\mathbf{1,800}\]

Ans: Let Rs \[x\] be invested in the first round.

Then, the sum of money invested in the second bond pays Rs \[\left( 30000-x \right)\]

It is given that the first bond pays \[5%\] interest per year, and the second bond pays \[7%\] interest per year.

We know that,

Simple interest for one year is \[\dfrac{principle\times rate}{100}\] .

Therefore, in order to obtain an annual total interest of Rs \[1,800\] ,

\[\left[ \begin{matrix}   x & \left( 30000-x \right)  \\ \end{matrix} \right]\left[ \begin{matrix}   \dfrac{5}{100}  \\   \dfrac{7}{100}  \\ \end{matrix} \right]=1800\]

\[\Rightarrow \dfrac{5x}{100}+\dfrac{7\left( 30000-x \right)}{100}=1800\]

\[\Rightarrow 210000-2x=180000\]

\[\Rightarrow x=15000\]

Thus, in order to obtain an annual total interest of Rs \[1,800\] , the trust fund should invest Rs \[15000\] in the first bond and the remaining Rs \[15000\] in the second bond.

ii. Rs \[\mathbf{2,000}\]

Ans: Let Rs \[x\] be invested in the first round.

Then, the sum of money invested in the second bond pays Rs \[\left( 30000-x \right)\]

It is given that the first bond pays \[5%\] interest per year, and the second bond pays \[7%\] interest per year.

We know that,

Simple interest for one year is \[\dfrac{principle\times rate}{100}\] .

Therefore, in order to obtain an annual total interest of Rs \[2,000\] ,

\[\left[ \begin{matrix}   x & \left( 30000-x \right)  \\ \end{matrix} \right]\left[ \begin{matrix}   \dfrac{5}{100}  \\   \dfrac{7}{100}  \\ \end{matrix} \right]=2000\]

\[\Rightarrow \dfrac{5x}{100}+\dfrac{7\left( 30000-x \right)}{100}=2000\]

\[\Rightarrow 210000-2x=200000\]

\[\Rightarrow x=5000\]

Thus, in order to obtain an annual total interest of Rs \[2,000\] , the trust fund should invest Rs \[5000\] in the first bond and the remaining Rs \[25000\] in the second bond.

20. The bookshop of a particular school has \[10\] dozen chemistry books, \[8\] dozen physics books, \[10\] dozen economics books. Their selling prices are Rs \[80\], Rs \[60\] and Rs \[40\] each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Ans: The total amount of money that will be received from the sale of all these books can be represented in the form of a matrix as:

\[12\left[ \begin{matrix}    10 & 8 & 10  \\ \end{matrix} \right]\left[ \begin{matrix}   80  \\   60  \\   40  \\ \end{matrix} \right]\]

\[\Rightarrow 12\left[ 10\times 80+8\times 60+10\times 40 \right]\]

\[\Rightarrow 12\left[ 1680 \right]\]

\[\Rightarrow 20160\]

Therefore, the bookshop will receive Rs \[20160\] from the sale.

21. Assume \[X,Y,Z,W\] and \[P\] are the matrices of order \[\mathbf{2\times n,3\times k,2\times p,n\times 3}\] and \[\mathbf{p\times k}\] respectively. The restriction on \[\mathbf{n,k,p}\] so that \[\mathbf{PY+WY}\] will be defined are:

1. \[\mathbf{k=3,p=n}\]

2. \[\mathbf{k}\] is arbitrary, \[\mathbf{p=2}\]

3. \[\mathbf{p}\] is arbitrary, \[\mathbf{k=3}\]

4. \[\mathbf{k=2,p=3}\]

Ans: Matrices \[P\] and \[Y\] are of the orders \[p\times k\] and \[3\times k\] respectively.

Therefore, the matrix \[PY\] will be defined if \[k=3\] .

Also, \[PY\] will be of order \[p\times k\] .

Since the number of columns in matrix \[W\] is equal to the number of rows in matrix \[Y\] .

Therefore, matrix \[WY\] is well-defined and is of the order \[n\times k\] .

Moreover, Matrices \[PY\] and \[WY\] can be added only when their orders are the same.

But \[PY\] is of the order \[p\times k\] and \[WY\] is of the order \[n\times k\] .

Therefore, we must have \[p=n\] .

\[\therefore p=n,k=3\] are the restrictions on \[n,k,p\] so that \[PY+WY\] will be defined.

Thus, option (A) is correct.

22. Assume \[\mathbf{X,Y,Z,W}\] and \[\mathbf{P}\] are matrices of order \[\mathbf{2\times n,3\times k,2\times p,n\times 3}\] and \[\mathbf{p\times k}\] respectively. If \[\mathbf{p=n}\] , then the order of the matrix \[\mathbf{7X-5Z}\] is:

1. \[\mathbf{p\times 2}\]

2. \[\mathbf{2\times n}\]

3. \[\mathbf{n\times 3}\]

4. \[\mathbf{p\times n}\]

Ans: Matrix \[X\] is of the order \[2\times n\] .

Thus, matrix \[7X\] is also of the same order.

Since, \[p=n\]

Matrix \[Z\] is of the order \[2\times p\] or \[2\times n\] .

Thus, matrix \[5Z\] is also of the same order.

Since, both the matrices \[7X\] and \[5Z\] are of the same order \[2\times n\].

Therefore, \[7X-5Z\] is well defined and is of the order \[2\times n\] .

Thus, option (B) is correct.

Exercise 3.3

1. Find the transpose of each of the following matrices:

i. \[\left[ \begin{matrix}   5  \\   \dfrac{1}{2}  \\   -1  \\ \end{matrix} \right]\]

Ans: The transpose of a matrix is obtained by changing its rows into columns and its columns into rows.

Thus, if \[A=\left[ \begin{matrix}  5  \\   \dfrac{1}{2}  \\   -1  \\ \end{matrix} \right]\] , then \[{{A}^{T}}=\left[ \begin{matrix}   5 & \dfrac{1}{2} & -1  \\ \end{matrix} \right]\]

ii. \[\left[ \begin{matrix}   1 & -1  \\   2 & 3  \\ \end{matrix} \right]\]

Ans: The transpose of a matrix is obtained by changing its rows into columns and its columns into rows.

Thus, if \[A=\left[ \begin{matrix}  1 & -1  \\   2 & 3  \\ \end{matrix} \right]\] , then \[{{A}^{T}}=\left[ \begin{matrix}   1 & 2  \\   -1 & 3  \\ \end{matrix} \right]\]

iii. \[\left[ \begin{matrix}   -1 & 5 & 6  \\   \sqrt{3} & 5 & 6  \\    2 & 3 & -1  \\ \end{matrix} \right]\]

Ans: The transpose of a matrix is obtained by changing its rows into columns and its columns into rows.

Thus, if \[A=\left[ \begin{matrix}   -1 & 5 & 6  \\   \sqrt{3} & 5 & 6  \\   2 & 3 & -1  \\ \end{matrix} \right]\] , then \[{{A}^{T}}=\left[ \begin{matrix}   -1 & \sqrt{3} & 2  \\   5 & 5 & 3  \\   6 & 6 & -1  \\ \end{matrix} \right]\]

2. If \[A=\left[ \begin{matrix}   -1 & 2 & 3  \\   5 & 7 & 9  \\   -2 & 1 & 1  \\ \end{matrix} \right]\] and \[B=\left[ \begin{matrix}   -4 & 1 & -5  \\   1 & 2 & 0  \\   1 & 3 & 1  \\ \end{matrix} \right]\] , then verify that

i. \[\left( A+B \right)'=A'+B'\]

Ans: Given that: \[A=\left[ \begin{matrix}   -1 & 2 & 3  \\   5 & 7 & 9  \\   -2 & 1 & 1  \\ \end{matrix} \right]\] and \[B=\left[ \begin{matrix}   -4 & 1 & -5  \\   1 & 2 & 0  \\   1 & 3 & 1  \\ \end{matrix} \right]\] 

Thus, we have \[A'=\left[ \begin{matrix}   -1 & 5 & -2  \\   2 & 7 & 1  \\   3 & 9 & 1  \\ \end{matrix} \right]\] and \[B'=\left[ \begin{matrix}   -4 & 1 & 1  \\   1 & 2 & 3  \\   -5 & 0 & 1  \\ \end{matrix} \right]\]

\[\left( A+B \right)=\left[ \begin{matrix}   -1 & 2 & 3  \\   5 & 7 & 9  \\   -2 & 1 & 1  \\ \end{matrix} \right]+\left[ \begin{matrix}   -4 & 1 & -5  \\   1 & 2 & 0  \\   1 & 3 & 1  \\ \end{matrix} \right]\]

\[\therefore \left( A+B \right)=\left[ \begin{matrix}   -5 & 3 & -2  \\   6 & 9 & 9  \\   -1 & 4 & 2  \\ \end{matrix} \right]\]

\[\Rightarrow \left( A+B \right)'=\left[ \begin{matrix}   -5 & 6 & -1  \\   3 & 9 & 4  \\   -2 & 9 & 2  \\ \end{matrix} \right]\]                                   …(1)

And \[A'+B'=\left[ \begin{matrix}   -1 & 5 & -2  \\   2 & 7 & 1  \\   3 & 9 & 1  \\ \end{matrix} \right]+\left[ \begin{matrix}   -4 & 1 & 1  \\   1 & 2 & 3  \\   -5 & 0 & 1  \\ \end{matrix} \right]\]

\[\therefore A'+B'=\left[ \begin{matrix}   -5 & 6 & -1  \\   3 & 9 & 4  \\   -2 & 9 & 2  \\ \end{matrix} \right]\]                                         …(2)

Thus, from equation (1) and (2),

\[\left( A+B \right)'=A'+B'\]

Hence proved.

ii. \[\left( A-B \right)'=A'-B'\]

Ans: Given that: \[A=\left[ \begin{matrix}   -1 & 2 & 3  \\   5 & 7 & 9  \\   -2 & 1 & 1  \\ \end{matrix} \right]\] and \[B=\left[ \begin{matrix}   -4 & 1 & -5  \\   1 & 2 & 0  \\   1 & 3 & 1  \\ \end{matrix} \right]\] 

Thus, we have \[A'=\left[ \begin{matrix}   -1 & 5 & -2  \\   2 & 7 & 1  \\   3 & 9 & 1  \\ \end{matrix} \right]\] and \[B'=\left[ \begin{matrix}   -4 & 1 & 1  \\   1 & 2 & 3  \\   -5 & 0 & 1  \\ \end{matrix} \right]\]

\[\left( A-B \right)=\left[ \begin{matrix}   -1 & 2 & 3  \\   5 & 7 & 9  \\   -2 & 1 & 1  \\ \end{matrix} \right]-\left[ \begin{matrix}  -4 & 1 & -5  \\   1 & 2 & 0  \\   1 & 3 & 1  \\ \end{matrix} \right]\]

\[\therefore \left( A-B \right)=\left[ \begin{matrix}   3 & 1 & 8  \\   4 & 5 & 9  \\   -3 & -2 & 0  \\ \end{matrix} \right]\]

\[\Rightarrow \left( A-B \right)'=\left[ \begin{matrix}   3 & 4 & -3  \\   1 & 5 & -2 \\   8 & 9 & 0  \\ \end{matrix} \right]\]                                   …(1)

And \[A'-B'=\left[ \begin{matrix}   -1 & 5 & -2  \\   2 & 7 & 1  \\   3 & 9 & 1  \\ \end{matrix} \right]-\left[ \begin{matrix}   -4 & 1 & 1  \\   1 & 2 & 3  \\   -5 & 0 & 1  \\ \end{matrix} \right]\]

\[\therefore A'-B'=\left[ \begin{matrix}   3 & 4 & -3  \\   1 & 5 & -2  \\   8 & 9 & 0  \\ \end{matrix} \right]\]                                         …(2)

Thus, from equation (1) and (2),

\[\left( A-B \right)'=A'-B'\]

Hence proved.

3. If \[A'=\left[ \begin{matrix}   3 & 4  \\   -1 & 2  \\   0 & 1  \\ \end{matrix} \right]\] and \[B=\left[ \begin{matrix}   -1 & 2 & 1  \\   1 & 2 & 3  \\ \end{matrix} \right]\] , then verify that

i. \[\left( A+B \right)'=A'+B'\]

Ans: We know that, \[A=\left( A' \right)'\]

Thus, we have: \[A=\left[ \begin{matrix}   3 & -1 & 0  \\   4 & 2 & 1  \\ \end{matrix} \right]\]

And \[B'=\left[ \begin{matrix}   -1 & 1  \\   2 & 2  \\   1 & 3  \\ \end{matrix} \right]\]

\[\therefore A+B=\left[ \begin{matrix}   3 & -1 & 0  \\   4 & 2 & 1  \\ \end{matrix} \right]+\left[ \begin{matrix}   -1 & 2 & 1  \\   1 & 2 & 3  \\ \end{matrix} \right]\]

\[\Rightarrow A+B=\left[ \begin{matrix}   2 & 1 & 1  \\   5 & 4 & 4  \\ \end{matrix} \right]\]

\[\therefore \left( A+B \right)'=\left[ \begin{matrix}   2 & 5  \\   1 & 4  \\   1 & 4  \\ \end{matrix} \right]\]                                …(1)

\[A'+B'=\left[ \begin{matrix}   3 & 4  \\   -1 & 2  \\   0 & 1  \\ \end{matrix} \right]+\left[ \begin{matrix}   -1 & 1  \\   2 & 2  \\   1 & 3  \\ \end{matrix} \right]\]

\[\therefore A'+B'=\left[ \begin{matrix}   2 & 5  \\   1 & 4  \\   1 & 4  \\ \end{matrix} \right]\]                                      …(2)

Thus, from equation (1) and (2),

\[\left( A+B \right)'=A'+B'\]

Hence proved.

ii. \[\left( A-B \right)'=A'-B'\]

Ans: We know that, \[A=\left( A' \right)'\]

Thus, we have: \[A=\left[ \begin{matrix}  3 & -1 & 0  \\   4 & 2 & 1  \\ \end{matrix} \right]\]

And \[B'=\left[ \begin{matrix}   -1 & 1  \\   2 & 2  \\   1 & 3  \\ \end{matrix} \right]\]

\[\therefore A-B=\left[ \begin{matrix}   3 & -1 & 0  \\   4 & 2 & 1  \\ \end{matrix} \right]-\left[ \begin{matrix}   -1 & 2 & 1  \\   1 & 2 & 3  \\ \end{matrix} \right]\]

\[\Rightarrow A-B=\left[ \begin{matrix}   4 & -3 & -1  \\   3 & 0 & -2  \\ \end{matrix} \right]\]

\[\therefore \left( A-B \right)'=\left[ \begin{matrix}   4 & 3  \\   -3 & 0  \\   -1 & -2  \\ \end{matrix} \right]\]                                …(1)

\[A'-B'=\left[ \begin{matrix}   3 & 4  \\   -1 & 2  \\   0 & 1  \\ \end{matrix} \right]-\left[ \begin{matrix}   -1 & 1  \\   2 & 2  \\   1 & 3  \\ \end{matrix} \right]\]

\[\therefore A'-B'=\left[ \begin{matrix}   4 & 3  \\   -3 & 0  \\   -1 & -2  \\ \end{matrix} \right]\]                                      …(2)

Thus, from equation (1) and (2),

\[\left( A-B \right)'=A'-B'\]

Hence proved.

4. If \[A'=\left[ \begin{matrix}   -2 & 3  \\   1 & 2  \\ \end{matrix} \right]\] and \[B=\left[ \begin{matrix}   -1 & 0  \\   1 & 2  \\ \end{matrix} \right]\] , then ( A+2B \right)' .

Ans: We know that, \[A=\left( A' \right)'\]

\[\therefore A=\left[ \begin{matrix}   -2 & 1  \\   3 & 2  \\ \end{matrix} \right]\]

\[\therefore A+2B=\left[ \begin{matrix}   -2 & 1  \\   3 & 2  \\ \end{matrix} \right]+2\left[ \begin{matrix}   -1 & 0  \\   1 & 2  \\ \end{matrix} \right]\]

\[\Rightarrow A+2B=\left[ \begin{matrix}   -2 & 1  \\   3 & 2  \\ \end{matrix} \right]+\left[ \begin{matrix}   -2 & 0  \\   2 & 4  \\ \end{matrix} \right]\]

\[\Rightarrow A+2B=\left[ \begin{matrix}   -4 & 1  \\   5 & 6  \\ \end{matrix} \right]\]

\[\therefore \left( A+2B \right)'=\left[ \begin{matrix}   -4 & 5  \\   1 & 6  \\ \end{matrix} \right]\]

5. For the matrices \[A\] and \[B\] , verify that \[\left( AB \right)'=B'A'\] where

i. \[A=\left[ \begin{matrix}   1  \\   -4  \\   3  \\ \end{matrix} \right],B=\left[ \begin{matrix}    -1 & 2 & 1  \\ \end{matrix} \right]\]

Ans: LHS: \[AB=\left[ \begin{matrix}   1  \\   -4  \\   3  \\ \end{matrix} \right]\left[ \begin{matrix}   -1 & 2 & 1  \\ \end{matrix} \right]\]

\[\Rightarrow AB=\left[ \begin{matrix}   -1 & 2 & 1  \\   4 & -8 & -4  \\   -3 & 6 & 3  \\ \end{matrix} \right]\]

\[\therefore \left( AB \right)'=\left[ \begin{matrix}   -1 & 4 & -3  \\   2 & -8 & 6 \\   1 & -4 & 3  \\ \end{matrix} \right]\]                                 …(1)

Now, \[A'=\left[ \begin{matrix}   1 & -4 & 3  \\ \end{matrix} \right]\]

And \[B'=\left[ \begin{matrix}   -1  \\   2  \\   1  \\ \end{matrix} \right]\]

RHS: \[\therefore B'A'=\left[ \begin{matrix}   -1  \\   2  \\   1  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & -4 & 3  \\ \end{matrix} \right]\]                      

 \[\therefore B'A'=\left[ \begin{matrix}   -1 & 4 & -3  \\   2 & -8 & 6  \\   1 & -4 & 3  \\ \end{matrix} \right]\]                                     …(2)

Thus, from equation (1) and (2),

\[\left( AB \right)'=B'A'\]

Hence proved.

ii. \[A=\left[ \begin{matrix}   0  \\   1  \\ 2  \\ \end{matrix} \right], B=\left[ \begin{matrix}   1 & 5 & 7  \\ \end{matrix} \right]\]

Ans: LHS: \[AB=\left[ \begin{matrix}   0  \\   1  \\   2  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 5 & 7  \\ \end{matrix} \right]\]

\[\Rightarrow AB=\left[ \begin{matrix}   0 & 1 & 2  \\   0 & 5 & 10  \\   0 & 7 & 14  \\ \end{matrix} \right]\]

\[\therefore \left( AB \right)'=\left[ \begin{matrix}   0 & 1 & 2  \\   0 & 5 & 10  \\   0 & 7 & 14  \\ \end{matrix} \right]\]                                 …(1)

Now, \[A'=\left[ \begin{matrix}   0 & 1 & 2  \\ \end{matrix} \right]\]

And \[B'=\left[ \begin{matrix}   1  \\   5  \\   7  \\ \end{matrix} \right]\]

RHS: \[\therefore B'A'=\left[ \begin{matrix}   1  \\   5  \\   7  \\ \end{matrix} \right]\left[ \begin{matrix}   0 & 1 & 2  \\ \end{matrix} \right]\]

\[\therefore B'A'=\left[ \begin{matrix}   0 & 1 & 2  \\   0 & 5 & 10  \\   0 & 7 & 14  \\ \end{matrix} \right]\]                                  …(2)

Thus, from equation (1) and (2),

\[\left( AB \right)'=B'A'\]

Hence proved.

6. If 

i. \[A=\left[ \begin{matrix}   \cos \alpha  & \sin \alpha   \\   -\sin \alpha  & \cos \alpha   \\ \end{matrix} \right]\] , then verify that \[A'A=I\]

Ans: Here, \[A'=\left[ \begin{matrix}   \cos \alpha  & -\sin \alpha   \\   \sin \alpha  & \cos \alpha   \\ \end{matrix} \right]\]

\[\therefore A'A=\left[ \begin{matrix}   \cos \alpha  & -\sin \alpha   \\   \sin \alpha  & \cos \alpha   \\ \end{matrix} \right]\left[ \begin{matrix}   \cos \alpha  & \sin \alpha   \\   -\sin \alpha  & \cos \alpha   \\ \end{matrix} \right]\]

\[\Rightarrow A'A=\left[ \begin{matrix}   {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha  & \sin \alpha \cos \alpha -\sin \alpha \cos \alpha   \\   \sin \alpha \cos \alpha -\sin \alpha \cos \alpha  & {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha   \\ \end{matrix} \right]\]

\[\therefore A'A=\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]\]

\[\therefore A'A=I\]

Hence proved.

ii. \[A=\left[ \begin{matrix}   \sin \alpha  & \cos \alpha   \\   -\cos \alpha  & \sin \alpha   \\ \end{matrix} \right]\] , then verify that \[A'A=I\]

Ans: Here, \[A'=\left[ \begin{matrix}   \sin \alpha  & -\cos \alpha   \\   \cos \alpha  & \sin \alpha   \\ \end{matrix} \right]\]

\[\therefore A'A=\left[ \begin{matrix}   \sin \alpha  & -\cos \alpha   \\    \cos \alpha  & \sin \alpha   \\ \end{matrix} \right]\left[ \begin{matrix}   \sin \alpha  & \cos \alpha   \\   -\cos \alpha  & \sin \alpha   \\ \end{matrix} \right]\]

\[\Rightarrow A'A=\left[ \begin{matrix}   {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha  & \sin \alpha \cos \alpha -\sin \alpha \cos \alpha   \\   \sin \alpha \cos \alpha -\sin \alpha \cos \alpha  & {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha   \\ \end{matrix} \right]\]

\[\therefore A'A=\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]\]

\[\therefore A'A=I\]

Hence proved.

7. Show that

i. The matrix \[A=\left[ \begin{matrix}   1 & -1 & 5  \\   -1 & 2 & 1  \\   5 & 1 & 3  \\ \end{matrix} \right]\] is a symmetric matrix.

Ans: Given \[A=\left[ \begin{matrix}   1 & -1 & 5  \\   -1 & 2 & 1  \\   5 & 1 & 3  \\ \end{matrix} \right]\]

\[\therefore A'=\left[ \begin{matrix}   1 & -1 & 5  \\   -1 & 2 & 1  \\   5 & 1 & 3  \\ \end{matrix} \right]\]

Here, we have \[A'=A\]

Thus, \[A\] is a symmetric matrix.

ii. The matrix \[A=\left[ \begin{matrix}   0 & 1 & -1  \\   -1 & 0 & 1  \\   1 & -1 & 0  \\ \end{matrix} \right]\] is a skew symmetric matrix.

Ans: Given \[A=\left[ \begin{matrix}   0 & 1 & -1  \\   -1 & 0 & 1  \\   1 & -1 & 0  \\ \end{matrix} \right]\]

\[\therefore A'=\left[ \begin{matrix}   0 & -1 & 1  \\   1 & 0 & -1  \\   -1 & 1 & 0 \\ \end{matrix} \right]\]

\[\Rightarrow A'=-\left[ \begin{matrix}   0 & 1 & -1  \\   -1 & 0 & 1  \\   1 & -1 & 0  \\ \end{matrix} \right]\]

Here, we have \[A'=-A\]

Thus, \[A\] is a skew symmetric matrix.

8. For the matrix \[A=\left[ \begin{matrix}   1 & 5  \\   6 & 7  \\ \end{matrix} \right]\] , verify that

i. \[\left( A+A' \right)\] is a symmetric matrix.

Ans: Given \[A=\left[ \begin{matrix}   1 & 5  \\   6 & 7  \\ \end{matrix} \right]\]

\[\therefore A'=\left[ \begin{matrix}   1 & 6  \\   5 & 7  \\ \end{matrix} \right]\]

\[A+A'=\left[ \begin{matrix}   1 & 5  \\   6 & 7  \\ \end{matrix} \right]+\left[ \begin{matrix}   1 & 6  \\   5 & 7  \\ \end{matrix} \right]\]

\[\Rightarrow A+A'=\left[ \begin{matrix}   2 & 11  \\   11 & 14  \\ \end{matrix} \right]\]

\[\left( A+A' \right)'=\left[ \begin{matrix}   2 & 11  \\   11 & 14  \\ \end{matrix} \right]\]

Here, we have \[A+A'=\left( A+A' \right)'\]

Thus, \[A+A'\] is a symmetric matrix.

ii. \[\left( A-A' \right)\] is a skew symmetric matrix.

Ans: Given \[A=\left[ \begin{matrix}   1 & 5  \\   6 & 7  \\ \end{matrix} \right]\]

\[\therefore A'=\left[ \begin{matrix}   1 & 6  \\   5 & 7  \\ \end{matrix} \right]\]

\[A-A'=\left[ \begin{matrix}   1 & 5  \\   6 & 7  \\ \end{matrix} \right]-\left[ \begin{matrix}   1 & 6  \\   5 & 7  \\ \end{matrix} \right]\]

\[\Rightarrow A-A'=\left[ \begin{matrix}   0 & -1  \\   1 & 0  \\ \end{matrix} \right]\]

\[\left( A-A' \right)'=\left[ \begin{matrix}   0 & 1  \\   -1 & 0  \\ \end{matrix} \right]\]

\[\Rightarrow \left( A-A' \right)'=-\left[ \begin{matrix}   0 & -1  \\   1 & 0  \\ \end{matrix} \right]\]

Here, we have \[A-A'=-\left( A-A' \right)'\]

Thus, \[A-A'\] is a skew symmetric matrix.

9. Find \[\dfrac{1}{2}\left( A+A' \right)\] and \[\dfrac{1}{2}\left( A-A' \right)\] , when \[A=\left[ \begin{matrix}   0 & a & b  \\   -a & 0 & c  \\   -b & -c & 0  \\ \end{matrix} \right]\] .

Ans: Given \[A=\left[ \begin{matrix}   0 & a & b  \\   -a & 0 & c  \\   -b & -c & 0  \\ \end{matrix} \right]\]

\[\therefore A'=\left[ \begin{matrix}   0 & -a & -b  \\   a & 0 & -c  \\   b & c & 0  \\ \end{matrix} \right]\]

\[A+A'=\left[ \begin{matrix}   0 & a & b  \\   -a & 0 & c  \\   -b & -c & 0  \\ \end{matrix} \right]+\left[ \begin{matrix}   0 & -a & -b  \\   a & 0 & -c  \\   b & c & 0 \\ \end{matrix} \right]\]

\[\Rightarrow A+A'=\left[ \begin{matrix}   0 & 0 & 0  \\   0 & 0 & 0  \\   0 & 0 & 0  \\ \end{matrix} \right]\]

\[\therefore \dfrac{1}{2}\left( A+A' \right)=\left[ \begin{matrix}   0 & 0 & 0  \\   0 & 0 & 0  \\   0 & 0 & 0  \\ \end{matrix} \right]\]

Now, \[A-A'=\left[ \begin{matrix}   0 & a & b  \\   -a & 0 & c  \\   -b & -c & 0  \\ \end{matrix} \right]-\left[ \begin{matrix}   0 & -a & -b  \\   a & 0 & -c  \\   b & c & 0  \\ \end{matrix} \right]\]

\[\left( A-A' \right)=\left[ \begin{matrix}   0 & 2a & 2b  \\  -2a & 0 & 2c  \\   -2b & -2c & 0  \\ \end{matrix} \right]\]

\[\therefore \dfrac{1}{2}\left( A-A' \right)=\left[ \begin{matrix}   0 & a & b  \\   -a & 0 & c  \\   -b & -c & 0  \\ \end{matrix} \right]\]

10. Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

i. \[\left[ \begin{matrix}   3 & 5  \\   1 & -1  \\ \end{matrix} \right]\]

Ans: Let \[A=\left[ \begin{matrix}  3 & 5  \\   1 & -1  \\ \end{matrix} \right]\] and \[A'=\left[ \begin{matrix}   3 & 1  \\   5 & -1  \\ \end{matrix} \right]\]

Now, \[A+A'=\left[ \begin{matrix}   3 & 5  \\   1 & -1  \\ \end{matrix} \right]+\left[ \begin{matrix}   3 & 1  \\   5 & -1  \\ \end{matrix} \right]\]

\[\Rightarrow A+A'=\left[ \begin{matrix}   6 & 6  \\   6 & -2  \\ \end{matrix} \right]\]

Let \[P=\dfrac{1}{2}\left( A+A' \right)\]

\[\Rightarrow P=\dfrac{1}{2}\left[ \begin{matrix}  6 & 6  \\   6 & -2  \\ \end{matrix} \right]\]

\[\Rightarrow P=\left[ \begin{matrix}   3 & 3  \\   3 & -1  \\ \end{matrix} \right]\]

Now, \[P'=\left[ \begin{matrix}   3 & 3  \\   3 & -1  \\ \end{matrix} \right]\]

Here, \[P=P'\]

\[\therefore P=\dfrac{1}{2}\left( A+A' \right)\] is a symmetric matrix.

Now, \[A-A'=\left[ \begin{matrix}   3 & 5  \\   1 & -1  \\ \end{matrix} \right]-\left[ \begin{matrix}   3 & 1  \\   5 & -1  \\ \end{matrix} \right]\]

\[\Rightarrow A-A'=\left[ \begin{matrix}   0 & 4  \\   -4 & 0  \\ \end{matrix} \right]\]

Let \[Q=\dfrac{1}{2}\left( A-A' \right)\]

\[\Rightarrow Q=\dfrac{1}{2}\left[ \begin{matrix}   0 & 4  \\   -4 & 0  \\ \end{matrix} \right]\]

\[\Rightarrow Q=\left[ \begin{matrix}   0 & 2  \\   -2 & 0  \\ \end{matrix} \right]\]

Now, \[Q'=\left[ \begin{matrix}   0 & 2  \\   -2 & 0  \\ \end{matrix} \right]\]

Here, \[Q=-Q'\]

\[\therefore Q=\dfrac{1}{2}\left( A-A' \right)\] is a skew symmetric matrix.

Thus, \[A\]is the sum of matrices \[P=\left[ \begin{matrix}   3 & 3  \\   3 & -1  \\ \end{matrix} \right]\] and \[Q=\left[ \begin{matrix}   0 & 2  \\   -2 & 0  \\ \end{matrix} \right]\].

ii. \[\left[ \begin{matrix}   6 & -2 & 2  \\   -2 & 3 & -1  \\   2 & -1 & 3  \\ \end{matrix} \right]\]

Ans: Let \[A=\left[ \begin{matrix}   6 & -2 & 2  \\   -2 & 3 & -1  \\   2 & -1 & 3  \\ \end{matrix} \right]\] and \[A'=\left[ \begin{matrix}   6 & -2 & 2  \\   -2 & 3 & -1 \\   2 & -1 & 3  \\ \end{matrix} \right]\]

Now, \[A+A'=\left[ \begin{matrix}   6 & -2 & 2  \\   -2 & 3 & -1  \\   2 & -1 & 3  \\ \end{matrix} \right]+\left[ \begin{matrix}   6 & -2 & 2  \\   -2 & 3 & -1  \\   2 & -1 & 3  \\ \end{matrix} \right]\]

Let \[P=\dfrac{1}{2}\left( A+A' \right)\]

\[\Rightarrow P=\dfrac{1}{2}\left[ \begin{matrix}   12 & -4 & 4  \\   -4 & 6 & -2 \\   4 & -2 & 6  \\ \end{matrix} \right]\]

\[\Rightarrow P=\left[ \begin{matrix}   6 & -2 & 2  \\   -2 & 3 & -1  \\   2 & -1 & 3  \\ \end{matrix} \right]\]

Now, \[P'=\left[ \begin{matrix}   6 & -2 & 2  \\   -2 & 3 & -1  \\   2 & -1 & 3  \\ \end{matrix} \right]\]

Here, \[P=P'\]

\[\therefore P=\dfrac{1}{2}\left( A+A' \right)\] is a symmetric matrix.

Now, \[A-A'=\left[ \begin{matrix}  6 & -2 & 2  \\   -2 & 3 & -1  \\   2 & -1 & 3  \\ \end{matrix} \right]-\left[ \begin{matrix}   6 & -2 & 2  \\   -2 & 3 & -1  \\   2 & -1 & 3  \\ \end{matrix} \right]\]

\[\Rightarrow A-A'=\left[ \begin{matrix}   0 & 0 & 0  \\   0 & 0 & 0  \\   0 & 0 & 0  \\ \end{matrix} \right]\]

Let \[Q=\dfrac{1}{2}\left( A-A' \right)\]

\[\Rightarrow Q=\dfrac{1}{2}\left[ \begin{matrix}   0 & 0 & 0  \\   0 & 0 & 0  \\   0 & 0 & 0  \\ \end{matrix} \right]\]

\[\Rightarrow Q=\left[ \begin{matrix}   0 & 0 & 0  \\   0 & 0 & 0  \\   0 & 0 & 0 \\ \end{matrix} \right]\]

Now, \[Q'=\left[ \begin{matrix}   0 & 0 & 0  \\   0 & 0 & 0  \\   0 & 0 & 0  \\ \end{matrix} \right]\]

Here, \[Q=-Q'\]

\[\therefore Q=\dfrac{1}{2}\left( A-A' \right)\] is a skew symmetric matrix.

Thus, \[A\]is the sum of matrices \[P=\left[ \begin{matrix}  6 & -2 & 2  \\   -2 & 3 & -1  \\   2 & -1 & 3  \\ \end{matrix} \right]\] and \[Q=\left[ \begin{matrix}   0 & 0 & 0  \\  0 & 0 & 0  \\   0 & 0 & 0  \\ \end{matrix} \right]\].

iii. \[\left[ \begin{matrix}   3 & 3 & -1  \\   -2 & -2 & 1  \\   -4 & -5 & 2  \\ \end{matrix} \right]\]

Ans: Let \[A=\left[ \begin{matrix}   3 & 3 & -1  \\   -2 & -2 & 1  \\   -4 & -5 & 2  \\ \end{matrix} \right]\] and \[A'=\left[ \begin{matrix}   3 & -2 & -4  \\   3 & -2 & -5  \\   -1 & 1 & 2  \\ \end{matrix} \right]\]

Now, \[A+A'=\left[ \begin{matrix}   3 & 3 & -1  \\   -2 & -2 & 1  \\   -4 & -5 & 2  \\ \end{matrix} \right]+\left[ \begin{matrix}   3 & -2 & -4  \\   3 & -2 & -5  \\   -1 & 1 & 2  \\ \end{matrix} \right]\]

\[\Rightarrow A+A'=\left[ \begin{matrix}   6 & 1 & -5  \\   1 & -4 & -4  \\   -5 & -4 & 4  \\ \end{matrix} \right]\]

Let \[P=\dfrac{1}{2}\left( A+A' \right)\]

\[\Rightarrow P=\dfrac{1}{2}\left[ \begin{matrix}   6 & 1 & -5  \\   1 & -4 & -4 \\   -5 & -4 & 4  \\ \end{matrix} \right]\]

\[\Rightarrow P=\left[ \begin{matrix}   3 & \dfrac{1}{2} & \dfrac{-5}{2}  \\   \dfrac{1}{2} & -2 & -2  \\   \dfrac{-5}{2} & -2 & 2  \\ \end{matrix} \right]\]

Now, \[P'=\left[ \begin{matrix}   3 & \dfrac{1}{2} & \dfrac{-5}{2}  \\  \dfrac{1}{2} & -2 & -2  \\   \dfrac{-5}{2} & -2 & 2  \\ \end{matrix} \right]\]

Here, \[P=P'\]

\[\therefore P=\dfrac{1}{2}\left( A+A' \right)\] is a symmetric matrix.

Now, \[A-A'=\left[ \begin{matrix}   3 & 3 & -1  \\   -2 & -2 & 1  \\   -4 & -5 & 2  \\ \end{matrix} \right]-\left[ \begin{matrix}   3 & -2 & -4  \\   3 & -2 & -5  \\   -1 & 1 & 2  \\ \end{matrix} \right]\]

\[\Rightarrow A-A'=\left[ \begin{matrix}   0 & 5 & 3  \\   -5 & 0 & 6  \\   -3 & -6 & 0  \\ \end{matrix} \right]\]

Let \[Q=\dfrac{1}{2}\left( A-A' \right)\]

\[\Rightarrow Q=\dfrac{1}{2}\left[ \begin{matrix}   0 & 5 & 3  \\   -5 & 0 & 6  \\   -3 & -6 & 0  \\ \end{matrix} \right]\]

\[\Rightarrow Q=\left[ \begin{matrix}   0 & \dfrac{5}{2} & \dfrac{3}{2}  \\   \dfrac{-5}{2} & 0 & 3  \\   \dfrac{-3}{2} & -3 & 0  \\ \end{matrix} \right]\]

Now, \[Q'=\left[ \begin{matrix}   0 & -\dfrac{5}{2} & -\dfrac{3}{2}  \\ \dfrac{5}{2} & 0 & -3  \\   \dfrac{3}{2} & 3 & 0  \\ \end{matrix} \right]\]

Here, \[Q=-Q'\]

\[\therefore Q=\dfrac{1}{2}\left( A-A' \right)\] is a skew symmetric matrix.

Thus, \[A\]is the sum of matrices \[P=\left[ \begin{matrix}   3 & \dfrac{1}{2} & \dfrac{-5}{2}  \\   \dfrac{1}{2} & -2 & -2  \\  \dfrac{-5}{2} & -2 & 2  \\ \end{matrix} \right]\] and \[Q=\left[ \begin{matrix}  0 & \dfrac{5}{2} & \dfrac{3}{2}  \\    \dfrac{-5}{2} & 0 & 3  \\  \dfrac{-3}{2} & -3 & 0  \\ \end{matrix} \right]\].

iv. \[\left[ \begin{matrix}   1 & 5  \\   -1 & 2  \\ \end{matrix} \right]\]

Ans: Let \[A=\left[ \begin{matrix}   1 & 5  \\   -1 & 2  \\ \end{matrix} \right]\] and \[A'=\left[ \begin{matrix}   1 & -1  \\   5 & 2  \\ \end{matrix} \right]\]

Now, \[A+A'=\left[ \begin{matrix}   1 & 5  \\   -1 & 2  \\ \end{matrix} \right]+\left[ \begin{matrix}   1 & -1  \\   5 & 2  \\ \end{matrix} \right]\]

\[\Rightarrow A+A'=\left[ \begin{matrix}   2 & 4  \\   4 & 4  \\ \end{matrix} \right]\]

Let \[P=\dfrac{1}{2}\left( A+A' \right)\]

\[\Rightarrow P=\dfrac{1}{2}\left[ \begin{matrix}   2 & 4  \\   4 & 4  \\ \end{matrix} \right]\]

\[\Rightarrow P=\left[ \begin{matrix}   1 & 2  \\   2 & 2  \\ \end{matrix} \right]\]

Now, \[P'=\left[ \begin{matrix}   1 & 2  \\   2 & 2  \\ \end{matrix} \right]\]

Here, \[P=P'\]

\[\therefore P=\dfrac{1}{2}\left( A+A' \right)\] is a symmetric matrix.

Now, \[A-A'=\left[ \begin{matrix}   1 & 5  \\   -1 & 2  \\ \end{matrix} \right]-\left[ \begin{matrix}   1 & -1  \\   5 & 2  \\ \end{matrix} \right]\]

\[\Rightarrow A-A'=\left[ \begin{matrix}   0 & 6  \\   -6 & 0  \\ \end{matrix} \right]\]

Let \[Q=\dfrac{1}{2}\left( A-A' \right)\]

\[\Rightarrow Q=\dfrac{1}{2}\left[ \begin{matrix}   0 & 6  \\   -6 & 0  \\ \end{matrix} \right]\]

\[\Rightarrow Q=\left[ \begin{matrix}   0 & 3  \\   -3 & 0  \\ \end{matrix} \right]\]

Now, \[Q'=\left[ \begin{matrix}    0 & -3  \\   3 & 0  \\ \end{matrix} \right]\]

Here, \[Q=-Q'\]

\[\therefore Q=\dfrac{1}{2}\left( A-A' \right)\] is a skew symmetric matrix.

Thus, \[A\]is the sum of matrices \[P=\left[ \begin{matrix}   1 & 2  \\   2 & 2  \\ \end{matrix} \right]\] and \[Q=\left[ \begin{matrix}   0 & 3  \\   -3 & 0  \\ \end{matrix} \right]\].

11. If \[A,B\] are symmetric matrices of same order, then \[AB-BA\] is a:

1. Skew symmetric matrix

2. Symmetric matrix

3. Zero matrix

4. Identity matrix

Ans: \[A\] and \[B\] are symmetric matrix, therefore, we have:

\[A'=A\] and \[B'=B\]                                                    …(1)

Here, \[\left( AB-BA \right)'=\left( AB \right)'-\left( BA \right)'\]

\[\Rightarrow \left( AB-BA \right)'=B'A'-A'B'\]

From (1),

\[\Rightarrow \left( AB-BA \right)'=BA-AB\]

\[\Rightarrow \left( AB-BA \right)'=-\left( AB-BA \right)\]                           …(2)

From equation (2),

\[AB-BA\] is a skew symmetric matrix.

12. If \[A=\left[ \begin{matrix}   \cos \alpha  & -\sin \alpha   \\   \sin \alpha  & \cos \alpha   \\ \end{matrix} \right]\] , then \[\mathbf{A+A'=I}\] , if the value of \[\mathbf{\alpha}\] is

1. \[\dfrac{\pi }{6}\]

2. \[\dfrac{\pi }{3}\]

3. \[n\]

4. \[\dfrac{3\pi }{2}\]

Ans: Given \[A=\left[ \begin{matrix}   \cos \alpha  & -\sin \alpha   \\   \sin \alpha  & \cos \alpha   \\ \end{matrix} \right]\]

\[\Rightarrow A'=\left[ \begin{matrix}   \cos \alpha  & \sin \alpha   \\   -\sin \alpha  & \cos \alpha   \\ \end{matrix} \right]\]

Now, \[\because A+A'=I\]

\[\therefore \left[ \begin{matrix}   \cos \alpha  & -\sin \alpha   \\   \sin \alpha  & \cos \alpha   \\ \end{matrix} \right]+\left[ \begin{matrix}   \cos \alpha  & \sin \alpha   \\    -\sin \alpha  & \cos \alpha   \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   2\cos \alpha  & 0  \\   0 & 2\cos \alpha   \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]\]

Equating the corresponding elements of the two matrices, we have:

\[\cos \alpha =\dfrac{1}{2}\]

\[\Rightarrow \alpha ={{\cos }^{-1}}\left( \dfrac{1}{2} \right)\]

\[\Rightarrow \alpha =\dfrac{\pi }{3}\]

Thus, option (B) is correct.

Exercise 3.4

1. Find the inverse of each of the matrices, if it exists.

\[\left[ \begin{matrix}   1 & -1  \\   2 & 3  \\ \end{matrix} \right]\]

Ans: Let \[A=\left[ \begin{matrix}   1 & -1  \\   2 & 3  \\ \end{matrix} \right]\]

Since \[A=IA\]

\[\therefore \left[ \begin{matrix}   1 & -1  \\   0 & 5  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{2}}\to {{R}_{2}}-2{{R}_{1}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & -1  \\   0 & 5  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0  \\   -2 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{2}}\to \dfrac{1}{5}{{R}_{2}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & -1  \\   0 & 5  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0  \\   \dfrac{-2}{5} & \dfrac{1}{5}  \\ \end{matrix} \right]A\]

Applying \[{{R}_{1}}\to {{R}_{1}}+{{R}_{2}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & -1  \\   0 & 5  \\ \end{matrix} \right]=\left[ \begin{matrix}   \dfrac{3}{5} & \dfrac{1}{5}  \\   \dfrac{-2}{5} & \dfrac{1}{5}  \\ \end{matrix} \right]A\]

\[\therefore {{A}^{-1}}=\left[ \begin{matrix}   \dfrac{3}{5} & \dfrac{1}{5}  \\   \dfrac{-2}{5} & \dfrac{1}{5}  \\ \end{matrix} \right]\]

2. Find the inverse of each of the matrices, if it exists.

\[\left[ \begin{matrix}   2 & 1  \\   1 & 1  \\ \end{matrix} \right]\]

Ans: Let \[A=\left[ \begin{matrix}   2 & 1  \\   1 & 1  \\ \end{matrix} \right]\]

Since \[A=IA\]

\[\therefore \left[ \begin{matrix}   2 & 1  \\   1 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{1}}\to {{R}_{1}}-{{R}_{2}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & -1  \\   0 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & -1  \\   0 & 5  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & -1  \\   -1 & 2  \\ \end{matrix} \right]A\]

\[\therefore {{A}^{-1}}=\left[ \begin{matrix}   1 & -1  \\   -1 & 2  \\ \end{matrix} \right]\]

3. Find the inverse of each of the matrices, if it exists.

\[\left[ \begin{matrix}   1 & 3  \\   2 & 7  \\ \end{matrix} \right]\]

Ans: Let \[A=\left[ \begin{matrix}   1 & 3  \\   2 & 7  \\ \end{matrix} \right]\]

Since \[A=IA\]

\[\therefore \left[ \begin{matrix}   1 & 3  \\   2 & 7  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{2}}\to {{R}_{2}}-2{{R}_{1}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 3  \\   0 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0  \\   -2 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{1}}\to {{R}_{1}}-3{{R}_{2}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   7 & -3  \\   -2 & 1  \\ \end{matrix} \right]A\]

\[\therefore {{A}^{-1}}=\left[ \begin{matrix}   7 & -3  \\   -2 & 1  \\ \end{matrix} \right]\]

4. Find the inverse of each of the matrices, if it exists.

\[\left[ \begin{matrix}   2 & 3  \\  5 & 7  \\ \end{matrix} \right]\]

Ans: Let \[A=\left[ \begin{matrix}   2 & 3  \\   5 & 7  \\ \end{matrix} \right]\]

Since \[A=IA\]

\[\therefore \left[ \begin{matrix}   2 & 3  \\  5 & 7  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{2}}\to {{R}_{2}}-2{{R}_{1}}\]

\[\Rightarrow \left[ \begin{matrix}   2 & 3  \\   1 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0  \\   -2 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{1}}\leftrightarrow {{R}_{2}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 1  \\   2 & 3  \\ \end{matrix} \right]=\left[ \begin{matrix}   -2 & 1  \\   1 & 0  \\ \end{matrix} \right]A\]

Applying \[{{R}_{2}}\leftrightarrow {{R}_{2}}-2{{R}_{1}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 1  \\   0 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   -2 & 1  \\   5 & -2  \\ \end{matrix} \right]A\]

Applying \[{{R}_{1}}\to {{R}_{1}}-{{R}_{2}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   -7 & 3  \\   5 & -2  \\ \end{matrix} \right]A\]

\[\therefore {{A}^{-1}}=\left[ \begin{matrix}   7 & 3  \\   5 & -2  \\ \end{matrix} \right]\]

5. Find the inverse of each of the matrices, if it exists.

\[\left[ \begin{matrix}   2 & 1  \\   7 & 4  \\ \end{matrix} \right]\]

Ans: Let \[A=\left[ \begin{matrix}   2 & 1  \\   7 & 4  \\ \end{matrix} \right]\]

Since \[A=IA\]

\[\therefore \left[ \begin{matrix}   2 & 1  \\   7 & 4  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{2}}\to {{R}_{2}}-3{{R}_{1}}\]

\[\Rightarrow \left[ \begin{matrix}   2 & 1  \\   1 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0  \\   -3 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{1}}\to {{R}_{1}}-{{R}_{2}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   4 & -1  \\   -7 & 2  \\ \end{matrix} \right]A\]

\[\therefore {{A}^{-1}}=\left[ \begin{matrix}   4 & -1  \\   -7 & 2  \\ \end{matrix} \right]\]

6. Find the inverse of each of the matrices, if it exists.

\[\left[ \begin{matrix}  2 & 5  \\   1 & 3  \\ \end{matrix} \right]\]

Ans: Let \[A=\left[ \begin{matrix}  2 & 5  \\   1 & 3  \\ \end{matrix} \right]\]

Since \[A=IA\]

\[\therefore \left[ \begin{matrix}   2 & 5  \\  1 & 3  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{1}}\to \dfrac{1}{2}{{R}_{1}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & \dfrac{5}{2}  \\  0 & 3  \\ \end{matrix} \right]=\left[ \begin{matrix}   \dfrac{1}{2} & 0  \\   0 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\]

\[\Rightarrow \left[ \begin{matrix}  1 & \dfrac{5}{2}  \\   0 & \dfrac{1}{2}  \\ \end{matrix} \right]=\left[ \begin{matrix}   \dfrac{1}{2} & 0  \\   \dfrac{-1}{2} & 1 \\ \end{matrix} \right]A\]

Applying \[{{R}_{1}}\to {{R}_{1}}-5{{R}_{2}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 0  \\   0 & \dfrac{1}{2}  \\ \end{matrix} \right]=\left[ \begin{matrix}   3 & -5  \\   \dfrac{-1}{2} & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{2}}\to 2{{R}_{2}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   3 & -5  \\   -1 & 2  \\ \end{matrix} \right]A\]

\[\therefore {{A}^{-1}}=\left[ \begin{matrix}   3 & -5  \\   -1 & 2  \\ \end{matrix} \right]\]

7. Find the inverse of each of the matrices, if it exists.

\[\left[ \begin{matrix}   3 & 1  \\   5 & 2  \\ \end{matrix} \right]\]

Ans: Let \[A=\left[ \begin{matrix}   3 & 1  \\   5 & 2  \\ \end{matrix} \right]\]

Since \[A=IA\]

\[\therefore \left[ \begin{matrix}   3 & 1  \\   5 & 2  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{1}}\to 2{{R}_{1}}\]

\[\Rightarrow \left[ \begin{matrix}   6 & 2  \\   5 & 2  \\ \end{matrix} \right]=\left[ \begin{matrix}   2 & 0  \\   0 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{1}}\to {{R}_{1}}-{{R}_{2}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 0  \\   5 & 2  \\ \end{matrix} \right]=\left[ \begin{matrix}   2 & -1  \\   0 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{2}}\to {{R}_{2}}-5{{R}_{1}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 0  \\   0 & 2  \\ \end{matrix} \right]=\left[ \begin{matrix}   2 & -1  \\   -10 & 6  \\ \end{matrix} \right]A\]

Applying \[{{R}_{2}}\to \dfrac{1}{2}{{R}_{2}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   2 & -1  \\  -5 & 3  \\ \end{matrix} \right]A\]

\[\therefore {{A}^{-1}}=\left[ \begin{matrix}   2 & -1  \\   -5 & 3  \\ \end{matrix} \right]\]

8. Find the inverse of each of the matrices, if it exists.

\[\left[ \begin{matrix}   4 & 5  \\   3 & 4  \\ \end{matrix} \right]\]

Ans: Let \[A=\left[ \begin{matrix}   4 & 5  \\   3 & 4  \\ \end{matrix} \right]\]

Since \[A=IA\]

\[\therefore \left[ \begin{matrix}   4 & 5  \\  3 & 4  \\ \end{matrix} \right]=\left[ \begin{matrix}  1 & 0  \\  0 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{1}}\to {{R}_{1}}-{{R}_{2}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 1  \\   3 & 4  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & -1  \\   0 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{2}}\to {{R}_{2}}-3{{R}_{1}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 1  \\   0 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   4 & -5  \\   -3 & 4  \\ \end{matrix} \right]A\]

Applying \[{{R}_{1}}\to {{R}_{1}}-{{R}_{2}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   4 & -5  \\   -3 & 4  \\ \end{matrix} \right]A\]

\[\therefore {{A}^{-1}}=\left[ \begin{matrix}   4 & -5  \\   -3 & 4  \\ \end{matrix} \right]\]

9. Find the inverse of each of the matrices, if it exists.

\[\left[ \begin{matrix}   3 & 10  \\   2 & 7  \\ \end{matrix} \right]\]

Ans: Let \[A=\left[ \begin{matrix}   3 & 10  \\   2 & 7  \\ \end{matrix} \right]\]

Since \[A=IA\]

\[\therefore \left[ \begin{matrix}   3 & 10  \\   2 & 7  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0  \\  0 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{1}}\to {{R}_{1}}-{{R}_{2}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 3  \\   2 & 7  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & -1  \\   0 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{2}}\to {{R}_{2}}-2{{R}_{1}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 3  \\  0 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & -1  \\  -2 & 3  \\ \end{matrix} \right]A\]

Applying \[{{R}_{1}}\to {{R}_{1}}-3{{R}_{2}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 0  \\  0 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   7 & -10  \\   -2 & 3  \\ \end{matrix} \right]A\]

\[\therefore {{A}^{-1}}=\left[ \begin{matrix}  7 & -10  \\   -2 & 3  \\ \end{matrix} \right]\]

10. Find the inverse of each of the matrices, if it exists.

\[\left[ \begin{matrix}   3 & -1  \\   -4 & 2  \\ \end{matrix} \right]\]

Ans: Let \[A=\left[ \begin{matrix}   3 & -1  \\   -4 & 2  \\ \end{matrix} \right]\]

Since \[A=IA\]

\[\therefore \left[ \begin{matrix}   3 & -1  \\   -4 & 2  \\ \end{matrix} \right]=\left[ \begin{matrix}  1 & 0  \\  0 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{1}}\to {{R}_{1}}+{{R}_{2}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & -1  \\  -4 & 2  \\ \end{matrix} \right]=\left[ \begin{matrix}    1 & 1  \\   0 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{1}}\to -{{R}_{1}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & -1  \\   0 & -2  \\ \end{matrix} \right]=\left[ \begin{matrix}   -1 & -1  \\   0 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{2}}\to {{R}_{2}}+4{{R}_{1}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & -1  \\   0 & -2  \\ \end{matrix} \right]=\left[ \begin{matrix}   -1 & -1  \\   -4 & -3  \\ \end{matrix} \right]A\]

Applying \[{{R}_{2}}\to -\dfrac{1}{2}{{R}_{2}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & -1  \\   0 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   -1 & -1  \\   2 & \dfrac{3}{2}  \\ \end{matrix} \right]A\]

Applying \[{{R}_{1}}\to {{R}_{1}}+{{R}_{2}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & \dfrac{1}{2}  \\   2 & \dfrac{3}{2}  \\ \end{matrix} \right]A\]

\[\therefore {{A}^{-1}}=\left[ \begin{matrix}   1 & \dfrac{1}{2}  \\   2 & \dfrac{3}{2}  \\ \end{matrix} \right]\]

11. Find the inverse of each of the matrices, if it exists.

\[\left[ \begin{matrix}   2 & -6  \\   1 & -2  \\ \end{matrix} \right]\]

Ans: Let \[A=\left[ \begin{matrix}   2 & -6  \\   1 & -2  \\ \end{matrix} \right]\]

Since \[A=IA\]

\[\therefore \left[ \begin{matrix}   2 & -6  \\  1 & -2  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{1}}\leftrightarrow {{R}_{2}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & -2  \\   2 & -6  \\ \end{matrix} \right]=\left[ \begin{matrix}   0 & 1  \\   1 & 0  \\ \end{matrix} \right]A\]

Applying \[{{R}_{2}}\to {{R}_{2}}-2{{R}_{1}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & -2  \\   0 & -2  \\ \end{matrix} \right]=\left[ \begin{matrix}   0 & 1  \\   1 & -2  \\ \end{matrix} \right]A\]

Applying \[{{R}_{2}}\to -\dfrac{1}{2}{{R}_{2}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & -2  \\   0 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   0 & 1  \\   \dfrac{-1}{2} & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{1}}\to {{R}_{1}}+2{{R}_{2}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   -1 & 3  \\   \dfrac{-1}{2} & 1  \\ \end{matrix} \right]A\]

\[\therefore {{A}^{-1}}=\left[ \begin{matrix}   -1 & 3  \\   \dfrac{-1}{2} & 1  \\ \end{matrix} \right]\]

12. Find the inverse of each of the matrices, if it exists.

\[\left[ \begin{matrix}   6 & -3  \\   -2 & 1  \\ \end{matrix} \right]\]

Ans: Let \[A=\left[ \begin{matrix}   6 & -3  \\   -2 & 1  \\ \end{matrix} \right]\]

Since \[A=IA\]

\[\therefore \left[ \begin{matrix}   6 & -3  \\   -2 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{1}}\to \dfrac{1}{6}{{R}_{1}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & \dfrac{-1}{2}  \\  -2 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   \dfrac{1}{6} & 0  \\   0 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{2}}\to {{R}_{2}}+2{{R}_{1}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & -\dfrac{1}{2}  \\   0 & 0  \\ \end{matrix} \right]=\left[ \begin{matrix}   \dfrac{1}{6} & 3  \\   \dfrac{1}{3} & 1  \\ \end{matrix} \right]A\]

Since, we can see all the zeros in the second row of the matrix on the L.H.S, \[\therefore {{A}^{-1}}\] does not exist.

13. Find the inverse of each of the matrices, if it exists.

\[\left[ \begin{matrix}  2 & -3  \\   -1 & 2  \\ \end{matrix} \right]\]

Ans: Let \[A=\left[ \begin{matrix}   2 & -3  \\   -1 & 2  \\ \end{matrix} \right]\]

Since \[A=IA\]

\[\therefore \left[ \begin{matrix}   2 & -3  \\  -1 & 2  \\ \end{matrix} \right]=\left[ \begin{matrix}  1 & 0  \\   0 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{1}}\to {{R}_{1}}+{{R}_{2}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & -1  \\  -1 & 2  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 1  \\   0 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{2}}\to {{R}_{2}}+{{R}_{1}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & -1  \\   0 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 1  \\   1 & 2  \\ \end{matrix} \right]A\]

Applying \[{{R}_{1}}\to {{R}_{1}}+{{R}_{2}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   2 & 3  \\   1 & 2  \\ \end{matrix} \right]A\]

\[\therefore {{A}^{-1}}=\left[ \begin{matrix}   2 & 3  \\   1 & 2  \\ \end{matrix} \right]\]

14. Find the inverse of each of the matrices, if it exists.

\[\left[ \begin{matrix}   2 & 1  \\   4 & 2  \\ \end{matrix} \right]\]

Ans: Let \[A=\left[ \begin{matrix}   2 & 1  \\   4 & 2  \\ \end{matrix} \right]\]

Since \[A=IA\]

\[\therefore \left[ \begin{matrix}   2 & 1  \\   4 & 2  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{1}}\to {{R}_{1}}-\dfrac{1}{2}{{R}_{2}}\]

\[\Rightarrow \left[ \begin{matrix}   0 & 0  \\   4 & 2  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & \dfrac{-1}{2}  \\   0 & 1  \\ \end{matrix} \right]A\]

Since, we can see all the zeros in the first row of the matrix on the L.H.S, \[\therefore {{A}^{-1}}\] does not exist.

15.  Find the inverse of each of the matrices, if it exists.

\[\left[ \begin{matrix}   1 & 3 & -2  \\   -3 & 0 & -5  \\   2 & 5 & 0  \\ \end{matrix} \right]\]

Ans: Let \[A=\left[ \begin{matrix}   1 & 3 & -2  \\   -3 & 0 & -5  \\   2 & 5 & 0  \\ \end{matrix} \right]\]

Since \[A=IA\]

\[\therefore \left[ \begin{matrix}   1 & 3 & -2  \\   -3 & 0 & -5  \\   2 & 5 & 0  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0 & 0  \\   0 & 1 & 0  \\   0 & 0 & 1 \\ \end{matrix} \right]A\]

Applying \[{{R}_{2}}\to {{R}_{2}}+3{{R}_{1}}\] and \[{{R}_{3}}\to {{R}_{3}}-2{{R}_{1}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 3 & -2  \\   0 & 9 & -11  \\   0 & -1 & 4  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0 & 0  \\   3 & 1 & 0  \\   -2 & 0 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{1}}\to {{R}_{1}}+3{{R}_{3}}\] and \[{{R}_{2}}\to {{R}_{2}}+8{{R}_{3}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 0 & 10  \\   0 & 1 & 21  \\   0 & -1 & 4  \\ \end{matrix} \right]=\left[ \begin{matrix}   -5 & 0 & 3  \\   -13 & 1 & 8  \\   -2 & 0 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{3}}\to {{R}_{3}}+{{R}_{2}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 0 & 10  \\   0 & 1 & 21  \\   0 & 0 & 25  \\ \end{matrix} \right]=\left[ \begin{matrix}   -5 & 0 & 3  \\   -13 & 1 & 8  \\   -15 & 1 & 9  \\ \end{matrix} \right]A\]

Applying \[{{R}_{3}}\to \dfrac{1}{25}{{R}_{3}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 0 & 10  \\   0 & 1 & 21  \\   0 & 0 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   -5 & 0 & 3  \\   -13 & 1 & 8  \\   -\dfrac{3}{5} & \dfrac{1}{25} & \dfrac{9}{25}  \\ \end{matrix} \right]A\]

Applying \[{{R}_{1}}\to {{R}_{1}}-10{{R}_{3}}\] and \[{{R}_{2}}\to {{R}_{2}}-21{{R}_{3}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 0 & 0  \\   0 & 1 & 0  \\   0 & 0 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & -\dfrac{2}{5} & -\dfrac{3}{5}  \\  -\dfrac{2}{5} & \dfrac{4}{25} & \dfrac{11}{25}  \\   -\dfrac{3}{5} & \dfrac{1}{25} & \dfrac{9}{25}  \\ \end{matrix} \right]A\]

\[\therefore {{A}^{-1}}=\left[ \begin{matrix}   1 & -\dfrac{2}{5} & -\dfrac{3}{5}  \\   -\dfrac{2}{5} & \dfrac{4}{25} & \dfrac{11}{25}  \\   -\dfrac{3}{5} & \dfrac{1}{25} & \dfrac{9}{25}  \\ \end{matrix} \right]\]

16. Find the inverse of each of the matrices, if it exists.

\[\left[ \begin{matrix}   2 & 0 & -1  \\   5 & 1 & 0  \\   0 & 1 & 3  \\ \end{matrix} \right]\]

Ans: Let \[A=\left[ \begin{matrix}   2 & 0 & -1  \\   5 & 1 & 0  \\   0 & 1 & 3  \\ \end{matrix} \right]\]

Since \[A=IA\]

\[\therefore \left[ \begin{matrix}   2 & 0 & -1  \\  5 & 1 & 0  \\   0 & 1 & 3  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0 & 0  \\   0 & 1 & 0  \\   0 & 0 & 1 \\ \end{matrix} \right]A\]

Applying \[{{R}_{1}}\to \dfrac{1}{2}{{R}_{1}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 0 & -\dfrac{1}{2}  \\   5 & 1 & 0  \\   0 & 1 & 3  \\ \end{matrix} \right]=\left[ \begin{matrix}   \dfrac{1}{2} & 0 & 0  \\   0 & 1 & 0  \\   0 & 0 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{2}}\to {{R}_{2}}-5{{R}_{1}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 0 & -\dfrac{1}{2}  \\   0 & 1 & \dfrac{5}{2}  \\   0 & 1 & 3  \\ \end{matrix} \right]=\left[ \begin{matrix}   \dfrac{1}{2} & 0 & 0  \\   -\dfrac{5}{2} & 1 & 0  \\   0 & 0 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{3}}\to {{R}_{3}}-{{R}_{2}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 0 & -\dfrac{1}{2}  \\   0 & 1 & \dfrac{5}{2} \\   0 & 0 & \dfrac{1}{2}  \\ \end{matrix} \right]=\left[ \begin{matrix}   \dfrac{1}{2} & 0 & 0  \\   -\dfrac{5}{2} & 1 & 0  \\   \dfrac{5}{2} & -1 & 1  \\ \end{matrix} \right]A\]

Applying \[{{R}_{3}}\to 2{{R}_{3}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 0 & -\dfrac{1}{2}  \\   0 & 1 & \dfrac{5}{2} \\   0 & 0 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   \dfrac{1}{2} & 0 & 0  \\   -\dfrac{5}{2} & 1 & 0  \\   5 & -2 & 2  \\ \end{matrix} \right]A\]

Applying \[{{R}_{1}}\to {{R}_{1}}+\dfrac{1}{2}{{R}_{3}}\] and \[{{R}_{2}}\to {{R}_{2}}-\dfrac{5}{2}{{R}_{3}}\]

\[\Rightarrow \left[ \begin{matrix}   1 & 0 & 0  \\   0 & 1 & 0  \\   0 & 0 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}   3 & -1 & 1  \\   -15 & 6 & -5  \\   5 & -2 & 2  \\ \end{matrix} \right]A\]

\[\therefore {{A}^{-1}}=\left[ \begin{matrix}   3 & -1 & 1  \\   -15 & 6 & -5  \\   5 &-2 & 2  \\ \end{matrix} \right]\]

17. Matrices \[A\] and \[B\] will be inverse of each other only if

1. \[AB=BA\]

2. \[AB=0,BA=I\]

3. \[AB=BA=0\]

4. \[AB=BA=I\]

Ans: Since, if \[A\] is a square matrix of order \[m\] , and if there exists another square matrix \[B\] of the same order \[m\] , such that \[AB=BA=I\] , then \[B\] is said to be the inverse of \[A\]. In such a case, it is clear that \[A\] is the inverse of \[B\].

Thus, matrices \[A\] and \[B\] will be inverse of each other only if \[AB=BA=I\].

Thus, option (D) is correct.

Miscellaneous Solutions

1. Let \[A=\left[ \begin{matrix}   0 & 1  \\   0 & 0  \\ \end{matrix} \right]\] , show that \[{{\left( aI+bA \right)}^{n}}={{a}^{n}}I+n{{a}^{n-1}}bA\] , where \[I\] is the identity matrix of order \[2}\] and \[n\in N\].

Ans: Given \[A=\left[ \begin{matrix}   0 & 1  \\   0 & 0  \\ \end{matrix} \right]\]

By using the principle of mathematical induction.

For \[n=1\] ,

\[P\left( 1 \right):{{\left( aI+bA \right)}^{1}}={{a}^{1}}I+{{a}^{0}}bA\]

\[\Rightarrow P\left( 1 \right):\left( aI+bA \right)=aI+bA\]

Therefore, the result is true for \[n=1\] .

Let the result be true for \[n=k\] .

That is, \[P\left( k \right):{{\left( aI+bA \right)}^{k}}={{a}^{k}}I+k{{a}^{k-1}}bA\]

Now, we have to prove that the result is true for \[n=k+1\] .

\[{{\left( aI+bA \right)}^{k-1}}={{\left( aI+bA \right)}^{k}}\left( aI+bA \right)\]

\[\Rightarrow {{\left( aI+bA \right)}^{k-1}}=\left( {{a}^{k}}I+k{{a}^{k-1}}bA \right)\left( aI+bA \right)\]

\[\Rightarrow {{\left( aI+bA \right)}^{k-1}}={{a}^{k}}I+\left( k+1 \right){{a}^{k}}bA+k{{a}^{k-1}}{{b}^{2}}{{A}^{2}}\]                        …(1)

Now, 

\[{{A}^{2}}=\left[ \begin{matrix}   0 & 1  \\   0 & 0  \\ \end{matrix} \right]\left[ \begin{matrix}   0 & 1  \\   0 & 0  \\ \end{matrix} \right]\]

\[{{A}^{2}}=\left[ \begin{matrix}   0 & 0  \\  0 & 0  \\ \end{matrix} \right]\]

\[{{A}^{2}}=0\]

\[\Rightarrow {{\left( aI+bA \right)}^{k-1}}={{a}^{k}}I+\left( k+1 \right){{a}^{k}}bA\]

Thus, the result is true for \[n=k+1\]

Therefore, by the principal of mathematical induction, we have:

\[{{\left( aI+bA \right)}^{n}}={{a}^{n}}I+n{{a}^{n-1}}bA\] where \[A=\left[ \begin{matrix}   0 & 1  \\   0 & 0  \\ \end{matrix} \right]\] , \[n\in N\].

2. If \[A=\left[ \begin{matrix}  1 & 1 & 1  \\   1 & 1 & 1  \\   1 & 1 & 1  \\ \end{matrix} \right]\] , prove that \[{{A}^{n}}=\left[ \begin{matrix}   {{3}^{n-1}} & {{3}^{n-1}} & {{3}^{n-1}}  \\   {{3}^{n-1}} & {{3}^{n-1}} & {{3}^{n-1}}  \\   {{3}^{n-1}} & {{3}^{n-1}} & {{3}^{n-1}}  \\ \end{matrix} \right]\] , \[n\in N\].

Ans: Given \[A=\left[ \begin{matrix}   1 & 1 & 1  \\   1 & 1 & 1  \\   1 & 1 & 1  \\ \end{matrix} \right]\]

By using the principles of mathematical induction.

For \[n=1\] , we have

\[P\left( 1 \right)=\left[ \begin{matrix}   {{3}^{1-1}} & {{3}^{1-1}} & {{3}^{1-1}}  \\   {{3}^{1-1}} & {{3}^{1-1}} & {{3}^{1-1}}  \\   {{3}^{1-1}} & {{3}^{1-1}} & {{3}^{1-1}}  \\ \end{matrix} \right]\]

\[\Rightarrow P\left( 1 \right)=\left[ \begin{matrix}   {{3}^{0}} & {{3}^{0}} & {{3}^{0}}  \\   {{3}^{0}} & {{3}^{0}} & {{3}^{0}}  \\   {{3}^{0}} & {{3}^{0}} & {{3}^{0}}  \\ \end{matrix} \right]\]

\[\Rightarrow P\left( 1 \right)=\left[ \begin{matrix}   1 & 1 & 1  \\   1 & 1 & 1  \\   1 & 1 & 1  \\ \end{matrix} \right]\]

\[\Rightarrow P\left( 1 \right)=A\]

Therefore, the result is true for \[n=1\] .

Let the result be true for \[n=k\] .

\[P\left( k \right):{{A}^{k}}=\left[ \begin{matrix}   {{3}^{k-1}} & {{3}^{k-1}} & {{3}^{k-1}}  \\   {{3}^{k-1}} & {{3}^{k-1}} & {{3}^{k-1}}  \\   {{3}^{k-1}} & {{3}^{k-1}} & {{3}^{k-1}}  \\ \end{matrix} \right]\]

Now, we have to prove that the result is true for \[n=k+1\] .

Now, \[{{A}^{k+1}}=A\cdot {{A}^{k}}\]

\[\Rightarrow {{A}^{k+1}}=\left[ \begin{matrix}   1 & 1 & 1  \\   1 & 1 & 1  \\   1 & 1 & 1  \\ \end{matrix} \right]\left[ \begin{matrix}   {{3}^{k-1}} & {{3}^{k-1}} & {{3}^{k-1}}  \\   {{3}^{k-1}} & {{3}^{k-1}} & {{3}^{k-1}}  \\   {{3}^{k-1}} & {{3}^{k-1}} & {{3}^{k-1}}  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{k+1}}=\left[ \begin{matrix}   3\cdot {{3}^{k-1}} & 3\cdot {{3}^{k-1}} & 3\cdot {{3}^{k-1}}  \\   3\cdot {{3}^{k-1}} & 3\cdot {{3}^{k-1}} & 3\cdot {{3}^{k-1}}  \\   3\cdot {{3}^{k-1}} & 3\cdot {{3}^{k-1}} & 3\cdot {{3}^{k-1}}  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{k+1}}=\left[ \begin{matrix}   {{3}^{\left( k+1 \right)-1}} & {{3}^{\left( k+1 \right)-1}} & {{3}^{\left( k+1 \right)-1}}  \\   {{3}^{\left( k+1 \right)-1}} & {{3}^{\left( k+1 \right)-1}} & {{3}^{\left( k+1 \right)-1}}  \\   {{3}^{\left( k+1 \right)-1}} & {{3}^{\left( k+1 \right)-1}} & {{3}^{\left( k+1 \right)-1}}  \\ \end{matrix} \right]\]

Thus, the result is true for \[n=k+1\]

Therefore, by the principal of mathematical induction, we have:

 \[{{A}^{n}}=\left[ \begin{matrix}   {{3}^{n-1}} & {{3}^{n-1}} & {{3}^{n-1}}  \\  {{3}^{n-1}} & {{3}^{n-1}} & {{3}^{n-1}}  \\   {{3}^{n-1}} & {{3}^{n-1}} & {{3}^{n-1}}  \\ \end{matrix} \right]\] where \[A=\left[ \begin{matrix}   0 & 1  \\   0 & 0  \\ \end{matrix} \right]\] , \[n\in N\].

3. If \[A=\left[ \begin{matrix}   3 & -4  \\   1 & -1  \\ \end{matrix} \right]\] , then prove \[{{A}^{n}}=\left[ \begin{matrix}   1+2n & -4n  \\   n & 1-2n  \\ \end{matrix} \right]\] where \[n\] is any positive integer.

Ans: Given \[A=\left[ \begin{matrix}   3 & -4  \\   1 & -1  \\ \end{matrix} \right]\]

By using the principle of mathematical induction.

For \[n=1\] ,

\[P\left( 1 \right):A=\left[ \begin{matrix}   3 & -4  \\   1 & -1  \\ \end{matrix} \right]\]

Therefore, the result is true for \[n=1\] .

Let the result be true for \[n=k\] .

That is, \[P\left( k \right):{{A}^{k}}=\left[ \begin{matrix}   1+2k & -4k  \\   k & 1-2k  \\ \end{matrix} \right]\] , \[n\in N\]

Now, we have to prove that the result is true for \[n=k+1\] .

\[{{A}^{k+1}}=A\cdot {{A}^{k}}\]

\[\Rightarrow {{A}^{k+1}}=\left[ \begin{matrix}   1+2k & -4k  \\   k & 1-2k  \\ \end{matrix} \right]\left[ \begin{matrix}   3 & -4  \\   1 & -1  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{k+1}}=\left[ \begin{matrix}   3+2k & -4-4k  \\  1+k & -1-2k  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{k+1}}=\left[ \begin{matrix}   1+2\left( k+1 \right) & -4\left( k+1 \right)  \\   1+k & 1-2\left( k+1 \right)  \\ \end{matrix} \right]\]

Thus, the result is true for \[n=k+1\]

Therefore, by the principal of mathematical induction, we have:

\[{{A}^{n}}=\left[ \begin{matrix}   1+2n & -4n  \\   n & 1-2n  \\ \end{matrix} \right]\] , \[n\in N\].

4. If \[A\] and \[B\] are symmetric matrices, prove that \[AB-BA\] is a skew symmetric matrix.

Ans: \[A\] and \[B\] are symmetric matrix, therefore, we have:

\[A'=A\] and \[B'=B\]                                                    …(1)

Here, \[\left( AB-BA \right)'=\left( AB \right)'-\left( BA \right)'\]

\[\Rightarrow \left( AB-BA \right)'=B'A'-A'B'\]

From (1),

\[\Rightarrow \left( AB-BA \right)'=BA-AB\]

\[\Rightarrow \left( AB-BA \right)'=-\left( AB-BA \right)\]                           …(2)

From equation (2),

\[AB-BA\] is a skew symmetric matrix.

5. Show that the matrix \[B'AB\] is symmetric or skew symmetric accordingly as \[A\] is symmetric or skew symmetric.

Ans: Let \[A\] be a symmetric matrix, then \[A'=A\]               …(1)

\[\left( B'AB \right)'=\left\{ B'\left( AB \right) \right\}'\]

\[\Rightarrow \left( B'AB \right)'=\left( AB \right)'B'\]

\[\Rightarrow \left( B'AB \right)'=B'\left( A'B \right)\]

From (1),

\[\Rightarrow \left( B'AB \right)'=B'\left( AB \right)\]

Thus, if \[A\] is a symmetric matrix, then \[B'AB\] is a symmetric matrix.

Let \[A\] be a skew symmetric matrix, then \[A'=-A\]               …(2)

\[\left( B'AB \right)'=\left\{ B'\left( AB \right) \right\}'\]

\[\Rightarrow \left( B'AB \right)'=\left( AB \right)'B'\]

\[\Rightarrow \left( B'AB \right)'=B'\left( A'B \right)\]

From (2),

\[\Rightarrow \left( B'AB \right)'=B'\left( -AB \right)\]

\[\Rightarrow \left( B'AB \right)'=-B'AB\]

Thus, if \[A\] is a skew-symmetric matrix, then \[B'AB\] is a skew-symmetric matrix.

Therefore, the matrix \[B'AB\] is symmetric or skew symmetric accordingly as \[A\] is symmetric or skew symmetric.

6. Solve the system of linear equations, using matrix method.

\[2x-y=-2\]

\[3x+4y=3\]

Ans: The given system of equation can be written in the form of \[AX=B\] where,

\[A=\left[ \begin{matrix}   2 & -1  \\   3 & 4  \\ \end{matrix} \right],X=\left[ \begin{matrix}   x  \\   y  \\ \end{matrix} \right],B=\left[ \begin{matrix}   -2  \\  3  \\ \end{matrix} \right]\]

Now, \[\left| A \right|=8+3=11\]

\[\because \left| A \right|\ne 0\] ,therefore its inverse exists.

We know that,

\[{{A}^{-1}}=\dfrac{1}{\left| A \right|}adjA\]

\[\therefore {{A}^{-1}}=\dfrac{1}{11}adj\left[ \begin{matrix}   4 & 1  \\   -3 & 2  \\ \end{matrix} \right]\]

We can write \[X={{A}^{-1}}B\]

\[\therefore X=\dfrac{1}{11}\left[ \begin{matrix}   4 & 1  \\   -3 & 2  \\ \end{matrix} \right]\left[ \begin{matrix}   -2  \\   3  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   x  \\   y  \\ \end{matrix} \right]=\dfrac{1}{11}\left[ \begin{matrix}   -5  \\   12  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   x  \\   y  \\ \end{matrix} \right]=\left[ \begin{matrix}   \dfrac{-5}{11}  \\   \dfrac{12}{11}  \\ \end{matrix} \right]\]

Thus, \[x=\dfrac{-5}{11}\] and \[y=\dfrac{12}{11}\].

7. For what values of \[x\] , \[\left[ \begin{matrix}   1 & 2 & 1  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 2 & 0  \\   2 & 0 & 1  \\   1 & 0 & 2  \\ \end{matrix} \right]\left[ \begin{matrix}   0  \\   2  \\   x  \\ \end{matrix} \right]=O\] ?

Ans: Given \[\left[ \begin{matrix}   1 & 2 & 1  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 2 & 0  \\   2 & 0 & 1  \\   1 & 0 & 2  \\ \end{matrix} \right]\left[ \begin{matrix}   0  \\   2  \\   x  \\ \end{matrix} \right]=O\]

\[\Rightarrow \left[ \begin{matrix}   1+4+1 & 2+0+0 & 0+2+2  \\ \end{matrix} \right]\left[ \begin{matrix}   0  \\   2  \\   x  \\ \end{matrix} \right]=O\]

\[\Rightarrow \left[ \begin{matrix}   6 & 2 & 4  \\\end{matrix} \right]\left[ \begin{matrix}   0  \\   2  \\   x  \\ \end{matrix} \right]=O\]

\[\Rightarrow \left[ 6\left( 0 \right)+2\left( 2 \right)+4\left( x \right) \right]=O\]

\[\Rightarrow \left[ 4+4x \right]=\left[ 0 \right]\]

\[\Rightarrow 4+4x=0\]

\[\therefore x=-1\]

Thus, the required value of \[x\] is \[-1\] .

8. If \[A=\left[ \begin{matrix}   3 & 1  \\   -1 & 2  \\ \end{matrix} \right]\] , show that \[{{A}^{2}}-5A+7I=O\] .

Ans: Given \[A=\left[ \begin{matrix}   3 & 1  \\   -1 & 2  \\ \end{matrix} \right]\]

\[\therefore {{A}^{2}}=A\cdot A\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   3 & 1  \\   -1 & 2  \\ \end{matrix} \right]\left[ \begin{matrix}   3 & 1  \\   -1 & 2  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   3\left( 3 \right)+1\left( -1 \right) & 3\left( 1 \right)+1\left( 2 \right)  \\   -1\left( 3 \right)+2\left( -1 \right) & -1\left( 1 \right)+2\left( 2 \right)  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   9-1 & 3+2  \\   -3-2 & -1+4  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   8 & 5  \\   -5 & 3  \\ \end{matrix} \right]\]

LHS: \[{{A}^{2}}-5A+7I\]

\[\therefore \left[ \begin{matrix}   8 & 5  \\   -5 & 3  \\ \end{matrix} \right]-5\left[ \begin{matrix}  3 & 1  \\   -1 & 2  \\ \end{matrix} \right]+7\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   8 & 5  \\   -5 & 3  \\ \end{matrix} \right]-\left[ \begin{matrix}   15 & 5  \\   -5 & 10  \\ \end{matrix} \right]+\left[ \begin{matrix}   7 & 0  \\   0 & 7  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   0 & 0  \\   0 & 0  \\ \end{matrix} \right]\] 

\[\Rightarrow O\]

RHS: \[\Rightarrow O\]

\[LHS=RHS\]

\[\therefore {{A}^{2}}-5A+7I=O\] hence proved.

9. Find \[X\] , if \[\left[ \begin{matrix}   x & -5 & -1  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 0 & 2  \\   0 & 2 & 1  \\   2 & 0 & 3  \\ \end{matrix} \right]\left[ \begin{matrix}   x  \\   4  \\   1  \\ \end{matrix} \right]=O\] .

Ans: Given \[\left[ \begin{matrix}   x & -5 & -1  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 0 & 2  \\   0 & 2 & 1  \\   2 & 0 & 3  \\ \end{matrix} \right]\left[ \begin{matrix}   x  \\   4  \\   1  \\ \end{matrix} \right]=O\]

\[\Rightarrow \left[ \begin{matrix}   x-2 & -10 & 2x-8  \\ \end{matrix} \right]\left[ \begin{matrix}   x  \\   4  \\   1  \\ \end{matrix} \right]=O\]

\[\Rightarrow \left[ x\left( x-2 \right)-40+2x-8 \right]=O\]

\[\Rightarrow \left[ {{x}^{2}}-48 \right]=\left[ 0 \right]\]

\[\Rightarrow {{x}^{2}}-48=0\]

\[\therefore x=\pm 4\sqrt{3}\]

Thus, the required value of \[x\] is \[\pm 4\sqrt{3}\] .

10. A manufacture produces three products \[X,Y,Z\] which he sells in two markets.

Annual sales are indicated below:

1. If unit sale prices of \[X,Y,Z\] are Rs \[2.50\] , Rs \[1.50\] and Rs \[1.00\] , respectively, find the total revenue in each market with the help of matrix algebra.

Ans: Here the total revenue in market \[I\] can be represented in the form of a matrix as:

\[\left[ \begin{matrix}   10000 & 2000 & 18000  \\ \end{matrix} \right]\left[ \begin{matrix}   2.50  \\   1.50  \\   1.00  \\ \end{matrix} \right]\]

\[\Rightarrow 1000\times 2.50+2000\times 1.50+18000\times 1.00\]

\[\Rightarrow 46000\]

And, the total revenue in market \[II\] can be represented in the form of a matrix as:

\[\left[ \begin{matrix}   6000 & 20000 & 8000  \\ \end{matrix} \right]\left[ \begin{matrix}   2.50  \\   1.50  \\   1.00  \\ \end{matrix} \right]\]

\[\Rightarrow 6000\times 2.50+20000\times 1.50+8000\times 1.00\]

\[\Rightarrow 53000\]

Thus, the total revenue in market \[I\] is Rs \[46000\] and the same in market \[II\] is Rs \[53000\].

2. If the unit costs of the above three commodities are Rs \[2.00\] , Rs \[1.00\] and \[50\] paise respectively. Find the gross profit.

Ans: Here, the total cost prices of all the products in the market \[I\] can be represented in the form of a matrix as:

\[\left[ \begin{matrix}   10000 & 2000 & 18000  \\ \end{matrix} \right]\left[ \begin{matrix}   2.00  \\   1.00  \\   0.50  \\ \end{matrix} \right]\]

\[\Rightarrow 1000\times 2.00+2000\times 1.00+18000\times 0.50\]

\[\Rightarrow 31000\]

As the total revenue in market \[I\] is Rs \[46000\] , the gross profit in this market is \[Rs46000-Rs31000=Rs15000\] .

And, the total revenue in market \[II\] can be represented in the form of a matrix as:

\[\left[ \begin{matrix}   6000 & 20000 & 8000  \\ \end{matrix} \right]\left[ \begin{matrix}   2.00  \\   1.00  \\   0.50  \\ \end{matrix} \right]\]

\[\Rightarrow 6000\times 2.00+20000\times 1.00+8000\times 0.50\]

\[\Rightarrow 36000\]

As the total revenue in market \[II\] is Rs \[53000\] , the gross profit in this market is \[Rs53000-Rs36000=Rs17000\] .

11. Find the matrix \[X\] so that \[X\left[ \begin{matrix}   1 & 2 & 3  \\   4 & 5 & 6  \\ \end{matrix} \right]=\left[ \begin{matrix}   -7 & -8 & -9  \\   2 & 4 & 6  \\ \end{matrix} \right]\].

Ans: Given \[X\left[ \begin{matrix}   1 & 2 & 3  \\   4 & 5 & 6  \\ \end{matrix} \right]=\left[ \begin{matrix}   -7 & -8 & -9  \\   2 & 4 & 6  \\ \end{matrix} \right]\]

Here, \[X\] has to be a \[2\times 2\] matrix.

Now, let \[X=\left[ \begin{matrix}   a & c  \\   b & d  \\ \end{matrix} \right]\]

Thus, we have:

\[\left[ \begin{matrix}   a & c  \\   b & d  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 2 & 3  \\   4 & 5 & 6  \\ \end{matrix} \right]=\left[ \begin{matrix}   -7 & -8 & -9 \\   2 & 4 & 6  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   a+4c & 2a+5c & 3a+6c  \\   b+4d & 2b+5d & 3b+6d  \\ \end{matrix} \right]=\left[ \begin{matrix}   -7 & -8 & -9  \\   2 & 4 & 6  \\ \end{matrix} \right]\]

Comparing the corresponding elements of two matrices, we have:

\[a+4c=-7\]

\[2a+5c=-8\]

\[3a+6c=-9\]

\[b+4d=2\]

\[2b+5d=4\]

\[3b+6d=6\]

Now, solving the above equations we get,

\[\therefore a=1,b=2,c=-2,d=0\]

Thus, the required matrix \[X\] is \[\left[ \begin{matrix}   1 & -2  \\   2 & 0  \\ \end{matrix} \right]\].

12. If \[A\] and \[B\] are square matrices of the same order such that \[AB=BA\] , then prove by induction that \[A{{B}^{n}}={{B}^{n}}A\] . Further, prove that \[{{\left( AB \right)}^{n}}={{B}^{n}}{{A}^{n}}\] for all \[n\in N\].

Ans: Given, \[A\] and \[B\] are square matrices of the same order such that \[AB=BA\] .

For \[n=1\] , we have:

\[P\left( 1 \right):AB=BA\]                          (Given)

\[\Rightarrow A{{B}^{1}}={{B}^{1}}A\]

Therefore, the result is true for \[n=1\] .

Let the result be true for \[n=k\] .

\[P\left( k \right):A{{B}^{k}}={{B}^{k}}A\]                                        …(1)

Now, we have to prove that the result is true for \[n=k+1\] .

\[A{{B}^{k+1}}=A{{B}^{k}}\cdot B\]

\[\Rightarrow A{{B}^{k+1}}=\left( {{B}^{k}}A \right)B\]

\[\Rightarrow A{{B}^{k+1}}={{B}^{k}}\left( AB \right)\]

\[\Rightarrow A{{B}^{k+1}}={{B}^{k}}\left( BA \right)\]

\[\Rightarrow A{{B}^{k+1}}=\left( {{B}^{k}}B \right)A\]

\[\Rightarrow A{{B}^{k+1}}={{B}^{k+1}}A\]

Therefore, the result is true for \[n=k+1\] .

By the principle of mathematical induction, we have \[A{{B}^{n}}={{B}^{n}}A\] , \[n\in N\].

13. Choose the correct answer in the following questions:

If \[A=\left[ \begin{matrix}   \alpha  & \beta   \\   \gamma  & -\alpha   \\ \end{matrix} \right]\] is such that \[{{A}^{2}}=I\] then 

1. \[1+{{\alpha }^{2}}+\beta \gamma =0\]

2. \[1-{{\alpha }^{2}}+\beta \gamma =0\]

3. \[1-{{\alpha }^{2}}-\beta \gamma =0\]

4. \[1+{{\alpha }^{2}}-\beta \gamma =0\]

Ans: Given \[A=\left[ \begin{matrix}   \alpha  & \beta   \\   \gamma  & -\alpha  \\ \end{matrix} \right]\] 

\[\therefore {{A}^{2}}=A\cdot A\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   \alpha  & \beta   \\   \gamma  & -\alpha   \\ \end{matrix} \right]\left[ \begin{matrix}   \alpha  & \beta   \\   \gamma  & -\alpha   \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   {{\alpha }^{2}}+\beta \gamma  & \alpha \beta -\alpha \beta   \\   \alpha \gamma -\alpha \gamma  & \beta \gamma +{{\alpha }^{2}}  \\ \end{matrix} \right]\]

Now, \[{{A}^{2}}=I\]

\[\Rightarrow \left[ \begin{matrix}   {{\alpha }^{2}}+\gamma  & 0  \\   0 & \beta \gamma +{{\alpha }^{2}}  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]\]

Equating the corresponding elements, we get:

\[\beta \gamma +{{\alpha }^{2}}=1\]

\[\Rightarrow {{\alpha }^{2}}+\beta \gamma -1=0\]

\[\Rightarrow 1-{{\alpha }^{2}}-\beta \gamma =0\]

14. If the matrix \[A\] is both symmetric and skew symmetric, then

1. \[A\] is a diagonal matrix

2. \[A\] is a zero matrix

3. \[A\] is a square matrix

4. None of these

Ans: If a matrix \[A\] is both symmetric and skew symmetric, then

\[A'=A\] and \[A'=-A\]

\[\Rightarrow A=-A\]

\[\Rightarrow A+A=O\]

\[\Rightarrow A=O\]

Thus, option (B) is correct.

15. If \[A\] is a square matrix such that \[{{A}^{2}}=A\] , then \[{{\left( I+A \right)}^{3}}-7A\] is equal to

1. \[A\]

2. \[I-A\]

3. \[I\]

4. \[3A\]

Ans: \[{{\left( I+A \right)}^{3}}-7A={{I}^{3}}+{{A}^{3}}+3A+3{{A}^{2}}-7A\]

\[\Rightarrow {{\left( I+A \right)}^{3}}-7A=I+{{A}^{3}}+3A+3{{A}^{2}}-7A\]

Given that \[{{A}^{2}}=A\] ,

\[\Rightarrow {{\left( I+A \right)}^{3}}-7A=I+{{A}^{2}}\cdot A+3A+3{{A}^{2}}-7A\]

\[\Rightarrow {{\left( I+A \right)}^{3}}-7A=I+A\cdot A-A\]

\[\Rightarrow {{\left( I+A \right)}^{3}}-7A=I\]

Thus, option (C) is correct.

NCERT Solutions for Class 12 Maths – Free PDF Download

Class 12 Maths Chapter 3 NCERT solutions are now available online in PDF format. One can find the proper answers of Matrices Class 12 here, which have been developed by accomplished mentors complimenting their dexterity. The Class 12 Maths Ch 3 NCERT solutions PDF is uploaded on the official website. Download the file and start solving it right away.

Chapter 3 – Matrices 

3.1 Introduction

In NCERT Class 12 Maths Chapter 3 PDF, students will explore concepts of matrices. The learners can clearly understand the necessary application of matrices in various branches of mathematics. This mathematical tool simplifies the work to a great extent as compared to other straight forward methods. The evolution of the concept of Matrices 12th is the result of an attempt to extract compact and simple techniques of solving systems of linear equations. In this chapter, students will be better acquainted with the basics of matrix and matrix algebra. The utility of matrices expands far beyond the representation of the coefficients in the linear equations system.

Electronic spreadsheet programs for personal computers use Matrix notation and operations.

These, in turn, are used in different areas of business and science like budgeting, sales projection, cost estimation, analyzing the results of an experiment, etc. Many physical operations, such as magnification and reflection through a plane can be presented mathematically with the help of matrices. They are also used in cryptography. Good knowledge of matrices is important in varied individual branches of sciences and also in genetics, economics, sociology, modern psychology and industrial management.

3.2 Matrix

In this section of the chapter,  the fundamental principles of matrices are discussed. With the help of examples cited by established academicians, students can understand the whole concept. Get the proper definition of Matrix Class 12 and learn the order of a matrix considering only those matrices whose elements are real numbers or functions taking actual values. Students will get to know that the horizontal and vertical lines of entries in a matrix are called rows and columns respectively. This section also talks about the size of a matrix by the number of rows and columns it contains.

3.3 Types of Matrices

This part of the chapter delves into different types of matrices illustrated with various examples. It covers column matrix, row matrix, square matrix, diagonal matrix, scalar matrix, identity matrix and zero matrices. It also explores the equality of matrices accompanied with examples and solutions. The questions that have been provided will induce the sensibility of a learner to help them delve deeper into the aspects concerning the NCERT solution of Maths Class 12.

3.4 Operations on Matrices

Matrix Class 12 NCERT Solutions introduces certain operations on matrices, namely, the addition of matrices, multiplication of a matrix by a scalar, differences and multiplication of matrices. Highlighting properties of matrix addition, scalar multiplication of a matrix, multiplication of matrices, etc., students can get a profound understanding of how matrices operate.

3.5. Transpose of a Matrix

This section will focus on transposing a matrix and unique types of matrices such as symmetric and skew-symmetric matrices. It will establish the foundations of the transpose of the matrix with compelling examples and properties. This chapter engraves the basics with the help of illustrated examples and solutions.

3.6 Symmetric and Skew Symmetric Matrices

In this section, the chapter will assist in learning symmetric and skew-symmetric matrices. One can also comprehend the difference between a symmetric matrix and a skew-symmetric matrix. It is imperative to know about the transpose of a matrix and how to find it. Moreover, it is essential to remember all the parts of Chapter 3 that have been covered up until now. This section also explores theorems proving the nature of the symmetric matrix and skew-symmetric matrix.

3.7 Elementary Operation (Transformation) of a Matrix

Above section of the chapter explains the six operations (transformations) on a matrix, three of which are due to rows and three due to columns. These are known as elementary operations or changes.

3.8 Invertible Matrices

Learn the properties of invertible matrices. This chapter highlights the process of finding a matrix's inverse is known as matrix inversion. It is essential to know that not all matrices are invertible. For a matrix to be invertible, it must be able to be multiplied by its inverse. Furthermore, one can learn how to invert a matrix by elementary operations, shaping a better idea with diverse examples.

Some Important Points to remember

1. A matrix is said to be an ordered rectangular array of numbers or functions. These numbers or functions in the array are called the elements or the entries of the matrix.

2. Order of a Matrix

The order of the matrix determines the dimension of the matrix and the number of rows and columns in the matrix. The general representation of matrix order is Amxn, where m is the number of rows and n is the number of columns in the matrix.

3. Types of Matrices

• Column Matrix: A column matrix is a matrix with only one column.

• Row Matrix: A row matrix is a matrix with only one row.

• Square Matrix: A square matrix is one that has an equal number of rows and columns.

• Diagonal Matrix: A diagonal matrix is a square matrix in which all of the entries except the diagonal elements are zeroes.

• Scalar Matrix: A scalar matrix is a diagonal matrix in which all diagonal members are the same (non-zero). 

• Identity Matrix: An identity matrix is a diagonal matrix in which all diagonal members are '1' and all non-diagonal elements are zero. It is represented by the letter I. 

• Zero or Null Matrix: If all of the elements in a matrix are 0, it is said to be a zero or null matrix.

4. Equality of Matrices: Let A and B be two matrices. These matrices will be equal, if

(i) orders of A and B will be the same

(ii) corresponding elements of two matrices are the same

5. Operations on Matrices

• Addition of Matrices: The addition of matrices is one of the operations on matrices. By adding the corresponding matrices' elements, two or more matrices of the same order can be added. 

• Subtraction of Matrices: Matrix subtraction is a matrix operation that performs element-wise subtraction of matrices of the same order, that is, matrices with the same number of rows and columns. When we subtract two matrices, we subtract the elements in each row and column from the respective elements in another matrix's row and column.

• Multiplication of Matrices: Matrix multiplication is a matrix operation done on two matrices when they are both compatible. The primary condition for multiplying two matrices is that the number of columns in the first matrix equals the number of rows in the second matrix, so the order of the matrices becomes important.

6. Properties of Multiplication of Matrices

1. Non-commutativity: 

Matrix multiplication will be not commutative i.e. if AB & BA are both defined, then it is not mandatory that AB ≠ BA.

2. Associative law:

For three matrices A, B, and C, if multiplication is defined, then we can write it as A (BC) = (AB) C.

3. Multiplicative identity: 

For any square matrix A, there will be an identity matrix of the same order in which  IA = AI = A.

NCERT Solutions For Class 12 Maths Other Chapters PDF Download

• Chapter 1 - Relations and Functions

• Chapter 2 - Inverse Trigonometric Functions

• Chapter 4 - Determinants

• Chapter 5 - Continuity and Differentiability

• Chapter 6 - Application of Derivatives

• Chapter 7 - Integrals

• Chapter 8 - Application of Integrals

• Chapter 9 - Differential Equations

• Chapter 10 - Vector Algebra

• Chapter 11 - Three Dimensional Geometry

• Chapter 12 - Linear Programming

• Chapter 13 - Probability

Key Features of NCERT Solutions for Class 12 Maths Chapter 3

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• Matrices Class 12 Solutions are crafted elaborately to assist students in the most uncomplicated fashion, easy to study and learn the chapter.

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