NCERT Solutions for Class 12 Maths Chapter 3 - Matrices Exercise 3.1

1. In the matrix \[A=\left[ \begin{matrix} 2 & 5 & 19 & -7  \\ 35 & -2 & \dfrac{5}{2} & 12  \\ \sqrt{3} & 1 & -5 & 17  \\ \end{matrix} \right]\], 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 \[{{a}_{13}},{{a}_{21}},{{a}_{33}},{{a}_{24}},{{a}_{23}}\]

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 $2 \times 2$matrix, $A\, = \,\left[ {{a_{ij}}} \right]$, whose elements are given by:

(i) ${a_{ij}}\, = \,\frac{{{{\left( {i + j} \right)}^2}}}{2}$

(ii) ${a_{ij}} = \frac{i}{j}$

(iii) ${a_{ij}} = \frac{{{{\left( {i + 2j} \right)}^2}}}{2}$

Ans:

(i) ${a_{ij}}\, = \,\frac{{{{\left( {i + j} \right)}^2}}}{2}$

Elements for $2 \times 2$ matrix are: ${a_{11}},{a_{12}},{a_{21}},{a_{22}}$

${a_{11}} = \frac{{{{\left( {1 + 1} \right)}^2}}}{2}\, = \,\frac{{{{\left( 2 \right)}^2}}}{2}\, = \,2$

${a_{12}}\, = \,\frac{{{{\left( {1 + 2} \right)}^2}}}{2}\, = \,\frac{{{{\left( 3 \right)}^2}}}{2}\, = \,\frac{9}{2}$

${a_{21}}\, = \,\frac{{{{\left( {2 + 1} \right)}^2}}}{2}\, = \,\frac{{{{\left( 3 \right)}^2}}}{2}\, = \,\frac{9}{2}$

${a_{22}}\, = \,\frac{{{{\left( {2 + 2} \right)}^2}}}{2}\, = \,\frac{{{{\left( 4 \right)}^2}}}{2}\, = \,8$

So, the required matrix is:  $\begin{pmatrix} 2& {\frac{9}{2}} \\ \frac{9}{2} & 8 \\ \end{pmatrix}$.

(ii) ${a_{ij}} = \frac{i}{j}$

Elements for $2 \times 2$ matrix are: ${a_{11}},{a_{12}},{a_{21}},{a_{22}}$

${a_{11}} = \frac{1}{1}\,\, = \,1$

${a_{12}}\, = \,\frac{1}{2}\,$

${a_{21}}\, = \,\frac{2}{1}\, = \,2$

${a_{22}}\, = \,\frac{2}{2}\, = \,1$

So, the required matrix is: $\begin{pmatrix} 1 & \frac{1}{2}\\ 2& 1\\ \end{pmatrix}$.

(iii) ${a_{ij}} = \frac{{{{\left( {i + 2j} \right)}^2}}}{2}$

Elements for $2 \times 2$ matrix are: ${a_{11}},{a_{12}},{a_{21}},{a_{22}}$

${a_{11}} = \frac{{{{\left( {1 + 2} \right)}^2}}}{2}\, = \,\frac{{{{\left( 3 \right)}^2}}}{2}\, = \,\frac{9}{2}$

${a_{12}}\, = \,\frac{{{{\left( {1 + 4} \right)}^2}}}{2}\, = \,\frac{{{{\left( 5 \right)}^2}}}{2}\, = \,\frac{{25}}{2}$

${a_{21}}\, = \,\frac{{{{\left( {2 + 2} \right)}^2}}}{2}\, = \,\frac{{{{\left( 4 \right)}^2}}}{2}\, = 8$

${a_{22}}\, = \,\frac{{{{\left( {2 + 4} \right)}^2}}}{2}\, = \,\frac{{{{\left( 6 \right)}^2}}}{2}\, = 1\,8$

So, the required matrix is:$\begin{pmatrix} \frac{9}{2}& \frac{25}{2}\\ 8& 18\\ \end{pmatrix}$.

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

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

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

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

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

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

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

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

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,

\[4+z=7\]

\[\Rightarrow z=3\]

\[x+z=5\]

\[\Rightarrow x=2\]

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

7. Find the value of \[a,b,c,d\] from the equation:

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

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),

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

\[\Rightarrow d=4\]

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

8. \[A={{\left[ {{a}_{y}} \right]}_{m\times n}}\] 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.

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

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

2. Not possible to find

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

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

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).

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

1. \[27\]

2. \[18\]

3. \[81\]

4. \[512\]

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 \[{{2}^{9}}=512\].

Conclusion

Class 12 Maths Exercise 3.1 on Matrices, is fundamental for understanding the basics of matrices, which are crucial for advanced mathematics. This exercise focuses on the definition, types, and operations of matrices. It's important to grasp concepts like the order of a matrix, equality of matrices, and different types of matrices such as row, column, and square matrices. Pay special attention to solving problems related to these topics, as a strong foundation in these basics will help in understanding more complex matrix operations in later exercises.

Class 12 Maths Chapter 3: Exercises Breakdown

Other Study Materials for CBSE Class 12 Maths Chapter 3 Matrices

Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.