NCERT Exemplar for Class 12 Maths Chapter-3 (Book Solutions)

Short Answer:

Sample Questions:

1. Construct a matrix $A={\left[a_{ij}\right]}_{2\times 2}$  whose elements $a_{ij}$ are given by $a_{ij}=e^{2ix}sinjx.$

Ans: Given: Elements ${\text{a}}_{\text{ij}}\text{ = }{\text{e}}^{2\text{ix}}\sin \text{jx}\text{.}$

Form the elements by substituting values of $\text{i and j}\text{.}$Then, construct a matrix.

For the values of $\text{i and j}\text{.}$

$$\begin{array}{c}\text{i = 1, j = 1 gives }{\text{a}}_{\text{11}}\text{ = }{\text{e}}^{2\text{x}}\sin \text{x}\\ \text{i = 1, j = 2 gives }{\text{a}}_{\text{12}}\text{ = }{\text{e}}^{2\text{x}}\sin 2\text{x}\\ \text{i = 2, j = 1 gives }{\text{a}}_{\text{21}}\text{ = }{\text{e}}^{\text{4x}}\sin \text{x}\\ \text{i = 2, j = 2 gives }{\text{a}}_{\text{22}}\text{ = }{\text{e}}^{\text{4x}}\sin 2\text{x}\\ \therefore \text{ A = }\left[\begin{array}{cc}{\text{e}}^{2\text{x}}\sin \text{x}& {\text{e}}^{2\text{x}}\sin 2\text{x}\\ {\text{e}}^{\text{4x}}\sin \text{x}& {\text{e}}^{\text{4x}}\sin 2\text{x}\end{array}\right]\end{array}$$

2. If $$A=\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right],B=\left[\begin{array}{ccc}1& 3& 2\\ 4& 3& 1\end{array}\right],C=\left[\begin{array}{c}1\\ 2\end{array}\right],D=\left[\begin{array}{ccc}4& 6& 8\\ 5& 7& 9\end{array}\right]then$$ which of the sums $$A+B,B+C,C+DandB+D$$ is defined?

Ans : Given: Matrices $$\text{A, B and C}\text{.}$$

The sum of two matrices is defined only when they have same order.

Since, matrices having same order can be added. 

Order of $$\text{A = }{\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]}_{2\times \text{ 2}}\text{ and B = }{\left[\begin{array}{ccc}1& 3& 2\\ 4& 3& 1\end{array}\right]}_{2\times \text{ 3}},$$

Order of $$\text{B = }{\left[\begin{array}{ccc}1& 3& 2\\ 4& 3& 1\end{array}\right]}_{2\times \text{ 3}}\text{ and C = }{\left[\begin{array}{c}1\\ 2\end{array}\right]}_{2\times \text{ 1}},$$

Order of $$\text{C = }{\left[\begin{array}{c}1\\ 2\end{array}\right]}_{2\times \text{ 1}}\text{and D = }{\left[\begin{array}{ccc}4& 6& 8\\ 5& 7& 9\end{array}\right]}_{2\times \text{ 3}},$$

Order of $$\text{B = }{\left[\begin{array}{ccc}1& 3& 2\\ 4& 3& 1\end{array}\right]}_{2\times \text{ 3}}\text{ and D = }{\left[\begin{array}{ccc}4& 6& 8\\ 5& 7& 9\end{array}\right]}_{2\times \text{ 3}}.$$

Therefore, $$\text{B + D with order 2 }\times \text{ 3}$$ is defined.

3. Show that a matrix which is both symmetric and skew symmetric is a zero matrix.

Ans : 

Given: Symmetric and Skew symmetric matrix.

For symmetric matrix $$\text{A' = A}$$ and for skew symmetric matrix $$\text{A' = }-\text{ A}\text{.}$$

Let $$\text{A}$$ be a matrix with elements $${\text{a}}_{\text{ij}}.$$

As, $$\text{A}$$ is both symmetric and skew symmetric,

$$\begin{array}{c}\Rightarrow {\text{a}}_{\text{ij}}\text{ = }{\text{a}}_{\text{ji}}\text{ (i)}\\ \Rightarrow {\text{a}}_{\text{ij}}\text{ = }-{\text{a}}_{\text{ji}}\\ \Rightarrow {\text{a}}_{\text{ij}}\text{ + }{\text{a}}_{\text{ji}}\text{ = 0 (ii)}\\ \text{from (i) and (ii),}\\ 2{\text{a}}_{\text{ij}}\text{ = 0}\\ \therefore {\text{a}}_{\text{ij }}\text{ = 0}\end{array}$$

Therefore, $$\text{A}$$ is zero matrix.

4. If $$\left[\begin{array}{cc}2x& 3\end{array}\right]\left[\begin{array}{cc}1& 2\\ -3& 0\end{array}\right]\left[\begin{array}{c}x\\ 8\end{array}\right]=0,$$ find the value of $$\text{x}\text{.}$$

Ans:  Given: $$\left[\begin{array}{cc}2\text{x}& 3\end{array}\right]\left[\begin{array}{cc}1& 2\\ -\text{ 3}& 0\end{array}\right]\left[\begin{array}{c}\text{x}\\ 8\end{array}\right]\text{ = 0}\text{.}$$

Use matrix multiplication. Then compare the elements with the zero matrix.

Here,

$$\begin{array}{c}\left[\begin{array}{cc}2\text{x}& 3\end{array}\right]\left[\begin{array}{cc}1& 2\\ -\text{ 3}& 0\end{array}\right]\left[\begin{array}{c}\text{x}\\ 8\end{array}\right]\text{ = 0}\end{array}$$

$$\begin{array}{c}\Rightarrow \left[\begin{array}{cc}2\text{x }-\text{ 9}& 4\text{x}\end{array}\right]\left[\begin{array}{c}\text{x}\\ 8\end{array}\right]\text{ = [0]}\end{array}$$

$$\begin{array}{c}\Rightarrow \text{ [}2{\text{x}}^2-\text{ 9x + 32x] = [0]}\end{array}$$

$$\begin{array}{c}\Rightarrow 2{\text{x}}^2-\text{ 23x = 0 }\\ \Rightarrow \text{ x(2x + 23) = 0}\\ \Rightarrow \text{ x = 0, x = }\frac{-\text{ 23}}{2}\end{array}$$

5. If $$Ais3\times 3$$ invertible matrix, then show that for any scalar$$k(nonzero),kA$$ is invertible and $$(kA{)}^{-1}=\frac{1}{k}{A}^{-1}.$$

Ans :Given: $$\text{A is 3 }\times \text{ 3}$$ invertible matrix.

For invertible matrix, $$\text{A}{\text{A}}^{-\text{ 1}}\text{ = I}\text{.}$$

Here,

$$\begin{array}{c}\text{(kA)}\left(\frac{1}{\text{k}}{\text{A}}^{-\text{ 1}}\right)\text{ = }\left(\text{k}\text{.}\frac{1}{\text{k}}\right)(\text{A}{\text{A}}^{-\text{ 1}})\\ \Rightarrow \text{ 1}\text{.I = I}\\ \therefore \text{ (kA) is inverse of }\left(\frac{1}{\text{k}}{\text{A}}^{-\text{ 1}}\right).\end{array}$$

Long Answer:

6. Express the matrix $$\text{A}$$ as the sum of a symmetric and a skew symmetric matrix, where $$A=\left[\begin{array}{ccc}2& 4& -6\\ 7& 3& 5\\ 1& -2& 4\end{array}\right].$$

Ans : Given: Matrix $$\text{A}\text{.}$$

The matrix $$\text{A}$$ is expressed as $$\frac{\text{A + A'}}{2}\text{ + }\frac{\text{A }-\text{ A'}}{2}.$$Here $$\frac{\text{A + A'}}{2}\text{ is symmetric and }\frac{\text{A }-\text{ A'}}{2}$$ is skew-symmetric matrix.

Here,

$$\begin{array}{c}\text{A = }\left[\begin{array}{ccc}2& 4& -\text{ 6}\\ 7& 3& 5\\ 1& -\text{ 2}& 4\end{array}\right],\text{ then A' = }\left[\begin{array}{ccc}2& 7& 1\\ 4& 3& -\text{ 2}\\ -\text{ 6}& 5& 4\end{array}\right]\\ \text{now, }\frac{\text{A + A'}}{2}\text{ = }\frac{1}{2}\left[\begin{array}{ccc}4& 11& -\text{ 5}\\ 11& 6& 3\\ -\text{ 5}& 3& 8\end{array}\right]\end{array}$$ $$\Rightarrow \frac{\text{A + A'}}{2}\text{ = }\left[\begin{array}{ccc}2& \frac{11}{2}& \frac{-\text{ 5}}{2}\\ \frac{11}{2}& 3& \frac{3}{2}\\ \frac{-\text{ 5}}{2}& \frac{3}{2}& 4\end{array}\right]$$ $$\begin{array}{c}\text{now, }\frac{\text{A }-\text{ A'}}{2}\text{ = }\frac{1}{2}\left[\begin{array}{ccc}0& -\text{ 3}& -\text{ 7}\\ 3& 0& 7\\ 7& -\text{ 7 }& 0\end{array}\right]\\ \Rightarrow \frac{\text{A }-\text{ A'}}{2}\text{ = }\left[\begin{array}{ccc}0& \frac{-\text{ 3}}{2}& \frac{-\text{ 7}}{2}\\ \frac{3}{2}& 0& \frac{7}{2}\\ \frac{7}{2}& \frac{-\text{ 7}}{2}& 0\end{array}\right]\\ \therefore \frac{\text{A + A'}}{2}\text{ + }\frac{\text{A }-\text{ A'}}{2}\text{ = }\left[\begin{array}{ccc}2& \frac{11}{2}& \frac{-\text{ 5}}{2}\\ \frac{11}{2}& 3& \frac{3}{2}\\ \frac{-\text{ 5}}{2}& \frac{3}{2}& 4\end{array}\right]\text{ + }\left[\begin{array}{ccc}0& \frac{-\text{ 3}}{2}& \frac{-\text{ 7}}{2}\\ \frac{3}{2}& 0& \frac{7}{2}\\ \frac{7}{2}& \frac{-\text{ 7}}{2}& 0\end{array}\right]\\ \text{ = }\left[\begin{array}{ccc}2& 4& -\text{ 6}\\ 7& 3& 5\\ 1& -\text{ 2}& 4\end{array}\right]\\ \text{ = A}\text{.}\end{array}$$

7. If $$A=\left[\begin{array}{ccc}1& 3& 2\\ 2& 0& -1\\ 1& 2& 3\end{array}\right],$$then shoe that $$\text{A}$$satisfies the equation $$A^3-4A^2-3A+11I=0.$$

Ans: Given: Matrix $$\text{A}\text{.}$$

First, find $${\text{A}}^3\text{ and }{\text{A}}^2.$$ Then, substitute in the equation.

Here,

$$\begin{array}{c}{\text{A}}^2\text{ = A }\times \text{ A}\\ \Rightarrow {\text{A}}^2\text{ = }\left[\begin{array}{ccc}1& 3& 2\\ 2& 0& -\text{ 1}\\ 1& 2& 3\end{array}\right]\times \left[\begin{array}{ccc}1& 3& 2\\ 2& 0& -\text{ 1}\\ 1& 2& 3\end{array}\right]\\ \Rightarrow {\text{A}}^2\text{ = }\left[\begin{array}{ccc}1\text{ + 6 + 2}& 3\text{ + 0 + 4}& 2\text{ - 3 + 6}\\ 2\text{ + 0 }-\text{ 1}& 6\text{ + }0-\text{ 2}& 4\text{ + 0 }-\text{ 3}\\ 1\text{ + 4 + 3}& 3\text{ + 0 + 6}& 2-\text{ 2 + 9}\end{array}\right]\\ \Rightarrow {\text{A}}^2\text{ = }\left[\begin{array}{ccc}9& 7& 5\\ 1& 4& 1\\ 8& 9& 9\end{array}\right]\\ \text{Now, }{\text{A}}^3\text{ = }{\text{A}}^2\times \text{ A}\\ \Rightarrow {\text{A}}^3\text{ = }\left[\begin{array}{ccc}9& 7& 5\\ 1& 4& 1\\ 8& 9& 9\end{array}\right]\times \left[\begin{array}{ccc}1& 3& 2\\ 2& 0& -\text{ 1}\\ 1& 2& 3\end{array}\right]\\ \Rightarrow {\text{A}}^3\text{ = }\left[\begin{array}{ccc}9\text{ + 14 + 5}& 27\text{ + 0 + 10}& 18-\text{ 7 + 15}\\ 1\text{ + 8 + 1}& 3\text{ + 0 + 2}& 2-\text{ 4 + 3}\\ 8\text{ + 18 + 9}& 24\text{ + 0 + 18}& 16-9\text{ + 27}\end{array}\right]\\ {\text{A}}^3\text{ = }\left[\begin{array}{ccc}28& 37& 26\\ 10& 5& 1\\ 35& 42& 34\end{array}\right]\end{array}$$

Substitute the values in the equation, $${\text{A}}^3\text{ - 4}{\text{A}}^2\text{ - 3A + 11I = 0}\text{.}$$

$$\begin{array}{c}\text{ = }\left[\begin{array}{ccc}28& 37& 26\\ 10& 5& 1\\ 35& 42& 34\end{array}\right]-\text{ 4}\left[\begin{array}{ccc}9& 7& 5\\ 1& 4& 1\\ 8& 9& 9\end{array}\right]-\text{ 3}\left[\begin{array}{ccc}1& 3& 2\\ 2& 0& -\text{ 1}\\ 1& 2& 3\end{array}\right]\text{ + 11}\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 0\end{array}\right]\\ =\left[\begin{array}{ccc}28-\text{ 36 }-\text{ 3 + 11}& 37-\text{ 28 }-\text{ 9 + 0}& 26-\text{ 20 }-\text{ 6 + 0}\\ 10-\text{ 4 }-\text{ 6 + 0}& \text{5 }-\text{ 16 + 0 + 11}& 1-\text{ 4 + 3 + 0}\\ 35-\text{ 32 }-\text{ 3 + 0}& 42-\text{ 36 }-\text{ 6 + 0}& 34-\text{ 36 }-\text{ 9 + 11}\end{array}\right]\\ \text{ = }\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\\ \text{ = 0}\end{array}$$

8. Let $$A=\left[\begin{array}{cc}2& 3\\ -1& 2\end{array}\right].$$ Then show that $$A^2-4A+7I=0.$$Using this result calculate $$A^5$$ also.

Ans : Given: Matrix $$\text{A}\text{.}$$

First, find $${\text{A}}^2\text{ and use }{\text{A}}^5\text{ = }{\text{A}}^3{\text{A}}^2.$$

Here,

$$\begin{array}{c}{\text{A}}^2\text{ = A }\times \text{ A}\\ \Rightarrow {\text{A}}^2\text{ = }\left[\begin{array}{cc}2& 3\\ -\text{ 1}& 2\end{array}\right]\times \left[\begin{array}{cc}2& 3\\ \text{ 1}& 2\end{array}\right]\\ \Rightarrow {\text{A}}^2\text{ = }\left[\begin{array}{cc}1& 12\\ -\text{ 4}& 1\end{array}\right]\end{array}$$

Substitute in the equation, $${\text{A}}^2-\text{ 4A + 7I = 0}\text{.}$$

$$\begin{array}{c}=\left[\begin{array}{cc}1& 12\\ -\text{ 4}& 1\end{array}\right]\text{ - 4}\left[\begin{array}{cc}2& 3\\ -\text{ 1}& 2\end{array}\right]\text{ + 7}\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\\ =\left[\begin{array}{cc}1-\text{ 8 + 7}& 12-\text{ 12 + 0}\\ -\text{ 4 + 4 + 0}& 1-\text{ 8 + 7}\end{array}\right]\\ =\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\end{array}$$

$$\begin{array}{c}=\text{ 0}\\ \therefore {\text{A}}^2-\text{ 4A + 7I = 0}\\ \Rightarrow {\text{A}}^2\text{ = 4A - 7I}\\ \text{Now,}\end{array}$$

$$\begin{array}{c}{\text{A}}^3\text{ = A}\text{.}{\text{A}}^2\\ \Rightarrow {\text{A}}^3\text{ = A}.(\text{4A }-\text{ 7I})\\ \Rightarrow {\text{A}}^3\text{ = 4}(\text{4A }-\text{ 7I})-\text{ 7A}\end{array}$$

$$\begin{array}{c}\Rightarrow {\text{A}}^3\text{ = 16A - 28I }-\text{ 7A}\\ \Rightarrow {\text{A}}^3\text{ = 9A }-\text{ 28I}\\ \text{Now, }{\text{A}}^5\text{ = }{\text{A}}^3{\text{A}}^2\\ \Rightarrow {\text{A}}^5\text{ = (9A }-\text{ 28I})(\text{4A }-\text{ 7I})\\ \Rightarrow {\text{A}}^5\text{ = }36{\text{A}}^2\text{ - 63A - 112A + 196I}\\ \Rightarrow {\text{A}}^5\text{ = }36(\text{4A }-\text{ 7I})-\text{ 175A + 196I}\\ \Rightarrow {\text{A}}^5\text{ = }-\text{ 31A }-\text{ 56I}\\ \Rightarrow {\text{A}}^5\text{ = }-\text{ 31}\left[\begin{array}{cc}2& 3\\ -\text{ 1}& 2\end{array}\right]-\text{ 56}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\\ \Rightarrow {\text{A}}^5\text{ = }-\left[\begin{array}{cc}62& 93\\ -\text{ 31}& 62\end{array}\right]-\left[\begin{array}{cc}56& 0\\ 0& 56\end{array}\right]\\ \Rightarrow {\text{A}}^5\text{ = }\left[\begin{array}{cc}-\text{ 62 }-\text{ 56}& -\text{ 9}3-\text{ 0}\\ 31& -\text{ 62 }-\text{ 56}\end{array}\right]\\ \Rightarrow {\text{A}}^5\text{ = }\left[\begin{array}{cc}-\text{ 118}& -\text{ 9}3\\ 31& -\text{ 118}\end{array}\right]\end{array}$$

Objective Type Questions:

Choose the correct answer from the given four options in Examples $$9to12.$$

9. If $$AandB$$ are matrices of same order, then $$(A+B)(A-B)$$ is equal to

$$(A)A^2-B^2$$

$$(B)A^2-BA-AB-B^2$$

$$(C)A^2-B^2+BA-AB$$

$$(D)A^2-BA+B^2+AB$$

Ans : Given: $$\text{A and B}$$ are matrices.

Use the matrices algebra.

Here,

$$\begin{array}{c}(\text{A + B)(A }-\text{ B) = A(A }-\text{ B) + B(A }-\text{ B)}\\ \Rightarrow (\text{A + B)(A }-\text{ B) = }{\text{A}}^2-\text{ AB + BA }-{\text{B}}^2.\end{array}$$

Correct Answer: C

10. If $$A=\left[\begin{array}{ccc}2& -1& 3\\ -4& 5& 1\end{array}\right]andB=\left[\begin{array}{cc}2& 3\\ 4& -2\\ 1& 5\end{array}\right],then$$

(A)Only $$AB$$ is defined

(B)Only $$BA$$ is defined

(C)$$ABandBA$$ both are defined

(D)$$ABandBA$$ both are not defined

Ans: Given: Matrices $$\text{A and B}\text{.}$$

Product of two matrices is defined when column of one matrix is equal to the row of another matrix.

The order of matrix $$\text{A is 2 }\times \text{ 3 and order of B is 3 }\times \text{ 2}\text{.}$$

Therefore, both $$\text{AB and BA}$$ is defined.

Correct Answer: C

11. The matrix \[A  =  \left[ {\begin{array}{*{20}{c}}  0&0&5 \\   0&5&0 \\   5&0&0 

\end{array}} \right]\] is a

(A)Scalar matrix

(B)Diagonal matrix

(C)Unit matrix

(D)Square matrix

Ans : Given: Matrix $$\text{A}\text{.}$$

Matrix can be square, diagonal, unit depending on the order and elements of matrix.

Since, the order of matrix $$\text{A}$$ is $$3\times \text{ 3}\text{.}$$ The number of rows and columns are same.

Therefore, matrix $$\text{A}$$ is square matrix.

Correct Answer: D

12. If $$AandB$$ are symmetric matrices of the same order, then $$(A{B}^{\prime }-B{A}^{\prime })$$ is a

(A)Skew symmetric matrix

(B)Null matrix

(C)Symmetric matrix

(D)None of these

Ans: Given: $$\text{A and B}$$ are symmetric matrices.

For symmetric matrix $$\text{A' = A}$$ and for skew symmetric matrix $$\text{A' = }-\text{ A}\text{.}$$

Take transpose,

$$\begin{array}{c}\text{(AB' }-\text{ BA')' = (AB')' }-\text{ (BA')'}\\ \text{(AB' }-\text{ BA')' = (BA' }-\text{ AB')}\\ \text{(AB' }-\text{ BA')' = }-\text{ (AB' }-\text{ BA')}\text{.}\end{array}$$

Correct Answer: A

Fill in the blanks in each of the Examples $$13to15.$$

13: If $$AandB$$ are two skew symmetric matrices of same order, then $$AB$$ is symmetric matrix if ______  .

Ans: Given: $$\text{A and B}$$ are two skew symmetric matrices.

For symmetric matrix $$\text{A' = A}$$ and for skew symmetric matrix $$\text{A' = }-\text{ A}\text{.}$$

Take transpose,

$$\begin{array}{c}\text{(AB)' = B'A'}\\ \Rightarrow \text{ (AB)' = (}-\text{ B)(}-\text{ A)}\\ \text{(AB)' = BA}\\ \text{if, AB = BA, then}\\ \text{(AB)' = AB}\text{.}\end{array}$$

For $$\text{AB}\text{ = BA, AB}$$ is symmetric matrix.

14. If $$AandB$$ are matrices of same order, then $$(3A-2B{)}^{\prime }$$ is equal to ____ .

Ans: Given: $$\text{A and B}$$ are matrices.

Take transpose of $$\text{(3A }-\text{ 2B)'}\text{.}$$

Taking transpose,

$$\text{(3A }-\text{ 2B)' = 3A' }-\text{ 2B'}\text{.}$$

15. Addition of matrices is defined if order of the matrices is _______.

Ans: Same.

The order of must be same for addition of two matrices.

State whether the statements in each of the Examples $$16\text{ to 19}$$  is true or false:

16. If two matrices $$AandB$$ are of same order, then $$2A+B=B+2A.$$

Ans: Given: $$\text{A and B}$$ are matrices.

Matrix addition is defined only for the matrices with same order.

Since, $$\text{A and B}$$ are of same order. The matrix addition is defined.

Therefore, $$\text{2A + B = B + 2A}$$ is true.

17. Matrix subtraction is associative.

Ans : Given: Matrix.

Consider three matrices to verify the property of associativity.

Let $$\text{A, B and C}$$ are matrices,

$$\begin{array}{c}\text{(A }-\text{ B) }-\text{ C = A }-\text{ B }-\text{ C (i)}\\ \text{A }-\text{ (B }-\text{ C) = A }-\text{ B + C (ii)}\\ \text{(i) }\ne \text{ (ii)}\\ \therefore \text{ False}\text{.}\end{array}$$

18. For the non-singular matrix $$A,(A^{\prime }{)}^{-1}=(A^{-1}{)}^{\prime }.$$

Ans: For a non-singular matrix $$\text{| A | }\ne \text{ 0}\text{.}$$ The identity $$\text{A, (A'}{\text{)}}^{-\text{ 1}}\text{ = (}{\text{A}}^{-\text{ 1}}{)}^{\prime }$$ is always true.

19.  for any three matrices of same order.

Ans: Given: $$\text{A, B and C}$$ are matrices.

It is not given that, matrix $$\text{A}$$ is invertible.

It is not mentioned that matrix $$\text{A}$$ is invertible. 

$$\text{AB = AC }\Rightarrow \text{ B = C}$$ is false for non-invertible matrix.

EXERCISE:

Short Answer Type :

1. If a matrix has $$28$$ elements, what are the possible orders it can have? What if it has $$13$$ elements?

Ans: Given: Number of elements in matrix.

The order of a matrix is represented by $$\text{m }\times \text{ n}\text{.}$$ The number of elements of order $$\text{m }\times \text{ n will be mn}\text{.}$$

$$\begin{array}{c}\text{For, mn = 28,}\end{array}$$

Possible ordered pairs,

(m x n) = {(1,28), (2,14), (4,7), (7,4), (14,2), (28,1)}

for 28 elements possible orders 1 x 28,2 x 14,4 x 7,7 x 4,14 x 2,28 x 1

for 13 elements possible orders 1 x 13 and 13 x 1.

2. In the matrix $$A=\left[\begin{array}{ccc}a& 1& x\\ 2& \sqrt{3}& x^2-y\\ 0& 5& \frac{-2}{5}\end{array}\right],$$ write (i)The order of the matrix $$\text{A}\text{.}$$

Ans: First, find the order $$\text{m }\times \text{ n}$$ and then number of elements using $$\text{mn}\text{.}$$ General representation of element is $${\text{a}}_{\text{ij}}.$$

Order of matrix $$\text{A = 3 }\times \text{ 3}\text{.}$$

(ii)The number of elements.

Ans: First, find the order $$\text{m }\times \text{ n}$$ and then number of elements using $$\text{mn}\text{.}$$ General representation of element is $${\text{a}}_{\text{ij}}.$$

Number of elements $$\text{3 }\times \text{ 3 = 9}\text{.}$$

(iii)Elements $$a_{23},a_{31}and{a}_{12}.$$

Ans: First, find the order $$\text{m }\times \text{ n}$$ and then number of elements using $$\text{mn}\text{.}$$ General representation of element is $${\text{a}}_{\text{ij}}.$$ $${\text{a}}_{23}\text{ = }{\text{x}}^2-\text{ y},{\text{a}}_{31}\text{ = 0 and }{\text{a}}_{12}=1.$$

3.  Construct $$a_{2\times 2}$$ matrix, where $$(i)a_{ij}=\frac{{(i-2j)}^2}{2}$$

Ans: Given: Matrix $${\text{a}}_{2\times \text{ 2}}.$$

Substitute the values of $$\text{i and j for 2 }\times \text{ 2 matrix}\text{.}$$

Here,

$$\begin{array}{c}{\text{a}}_{\text{ij}}\text{ = }\frac{{(\text{i }-\text{ 2j)}}^2}{2},\\ \text{for,}\\ {\text{a}}_{11}\text{ = }\frac{{(1-\text{ 2)}}^2}{2}\text{ = }\frac{1}{2}\end{array}$$

$$\begin{array}{c}{\text{a}}_{12}\text{ = }\frac{{(1-\text{ 2 }\times \text{ 2)}}^2}{2}\text{ = }\frac{9}{2}\\ {\text{a}}_{21}\text{ = }\frac{{(2-\text{ 2 }\times \text{ 1)}}^2}{2}\text{ = 0}\\ {\text{a}}_{22}\text{ = }\frac{{(2-\text{ 2 }\times \text{ 2)}}^2}{2}\text{ = 2}\\ \therefore \text{ A = }\left[\begin{array}{cc}\frac{1}{2}& \frac{9}{2}\\ 0& 2\end{array}\right]\end{array}$$

$$(ii)a_{ij}=|-2i+3j|$$

Ans: Given: Matrix $${\text{a}}_{2\times \text{ 2}}.$$

Substitute the values of $$\text{i and j for 2 }\times \text{ 2 matrix}\text{.}$$

Here,

$$\begin{array}{c}{\text{a}}_{\text{ij}}\text{ = |}-\text{ 2i + 3j |,}\\ \text{for}\\ {\text{a}}_{11}\text{ = | }-\text{ 2 }\times \text{ 1 + 3 }\times \text{ 1 | = 1}\\ {\text{a}}_{12}\text{ = | }-\text{ 2 }\times \text{ 1 + 3 }\times \text{ 2 | = 4}\\ {\text{a}}_{21}\text{ = | }-\text{ 2 }\times \text{ 2 + 3 }\times \text{ 1 | = 1}\\ {\text{a}}_{22}\text{ = | }-\text{ 2 }\times \text{ 2 + 3 }\times \text{ 2 | = 2}\\ \therefore \text{ A = }\left[\begin{array}{cc}1& 4\\ 1& 2\end{array}\right]\end{array}$$

4. Construct a $$3\times 2$$matrix whose elements are given by $a_{ij}=e^{ix}sinjx.$

Ans: Given: Elements ${\text{a}}_{\text{ij}}\text{ = }{\text{e}}^{\text{ix}}\sin \text{jx}\text{.}$

Form the elements by substituting values of $\text{i and j}\text{.}$ Then, construct a matrix.

For the values of $\text{i and j}\text{.}$

$$\begin{array}{c}\text{i = 1, j = 1 gives }{\text{a}}_{\text{11}}\text{ = }{\text{e}}^{\text{x}}\sin \text{x}\\ \text{i = 1, j = 2 gives }{\text{a}}_{\text{12}}\text{ = }{\text{e}}^{\text{x}}\sin 2\text{x}\\ \text{i = 2, j = 1 gives }{\text{a}}_{\text{21}}\text{ = }{\text{e}}^{\text{2x}}\sin \text{x}\end{array}$$

$$\begin{array}{c}\text{i = 2, j = 2 gives }{\text{a}}_{\text{22}}\text{ = }{\text{e}}^{\text{2x}}\sin 2\text{x}\\ \text{i = 3, j = 1 gives }{\text{a}}_{\text{31}}\text{ = }{\text{e}}^{\text{3x}}\sin \text{x}\\ \text{i = 3, j = 2 gives }{\text{a}}_{\text{22}}\text{ = }{\text{e}}^{\text{3x}}\sin 2\text{x}\\ \therefore \text{ A = }\left[\begin{array}{cc}{\text{e}}^{\text{x}}\sin \text{x}& {\text{e}}^{\text{x}}\sin 2\text{x}\\ {\text{e}}^{\text{2x}}\sin \text{x}& {\text{e}}^{\text{2x}}\sin 2\text{x}\\ {\text{e}}^{\text{3x}}\sin \text{x}& {\text{e}}^{\text{3x}}\sin 2\text{x}\end{array}\right]\end{array}$$

5. Find the values of $$aandb,ifA=B,$$ where $$A=\left[\begin{array}{cc}a+4& 3b\\ 8& -6\end{array}\right]andB=\left[\begin{array}{cc}2a+2& b^2+2\\ 8& b^2-5b\end{array}\right]$$

Ans: Given: $$\text{A = B}\text{.}$$

If two matrices are equal, then each element of $$\text{A}$$ is equal to corresponding element of $$\text{B}\text{.}$$

$$\begin{array}{c}\text{Since, A = B,}\\ \left[\begin{array}{cc}\text{a + 4}& 3\text{b}\\ 8& -\text{ 6}\end{array}\right]\text{ = }\left[\begin{array}{cc}\text{2a + 2}& {\text{b}}^2\text{ + 2}\\ 8& {\text{b}}^2-\text{ 5b}\end{array}\right]\\ {\text{a}}_{11}\text{ = }{\text{b}}_{11}\\ \Rightarrow \text{ a + 4 = 2a + 2}\\ \Rightarrow \text{ a = 2}\\ \text{now, }{\text{a}}_{12}\text{ = }{\text{b}}_{12}\\ \Rightarrow \text{ 3b = }{\text{b}}^2\text{ + 2}\\ \Rightarrow {\text{b}}^2\text{ = 3b }-\text{ 2 - (i)}\\ \text{and, }{\text{a}}_{22}\text{ = }{\text{b}}_{22}\\ \Rightarrow -\text{ 6 = }{\text{b}}^2-\text{ 5b - (ii)}\\ \text{from (i) and (ii), we get}\\ -\text{ 6 = 3b }-\text{ 2 }-\text{ 5b}\\ \Rightarrow \text{ b = 2}\\ \therefore \text{ a = 2 and b = 2}\text{.}\end{array}$$

6. If possible, find the sum of matrices $$AandB,$$where$$A=\left[\begin{array}{cc}\sqrt{3}& 1\\ 2& 3\end{array}\right]andB=\left[\begin{array}{ccc}x& y& z\\ a& b& c\end{array}\right].$$

Ans:Given: Matrices $$\text{A and B}\text{.}$$

Two matrices are added, if they have same order.

Since, matrices $$\text{A and B}$$are of different order. The order of matrices $$\text{A and B}$$are$$\text{2 }\times \text{ 2 and 2 }\times \text{ 3}$$respectively.

Therefore, the sum of matrices $$\text{A and B}$$is not possible.

7. If $$X=\left[\begin{array}{ccc}3& 1& -1\\ 5& -2& -3\end{array}\right]andY=\left[\begin{array}{ccc}2& 1& -1\\ 7& 2& 4\end{array}\right],$$then find$$(i)X+Y.$$

Ans: Given: $$\text{X and Y}$$ are matrices.

Here,

$$\begin{array}{c}\text{X + Y = }\left[\begin{array}{ccc}3\text{ + 2}& 1\text{ + 1}& -\text{ 1 - 1}\\ 5\text{ + 7}& -\text{ 2 + 2}& -\text{ 3 + 4}\end{array}\right]\\ \therefore \text{ X + Y = }\left[\begin{array}{ccc}5& 2& -\text{ 2}\\ 12& 0& 1\end{array}\right]\end{array}$$

$$(ii)2X-3Y.$$

Ans: Given: $$\text{X and Y}$$ are matrices.

Here,

$$\begin{array}{c}\text{2X }-\text{ 3Y = 2}\left[\begin{array}{ccc}3& 1& -\text{ 1}\\ 5& -\text{ 2}& -\text{ 3}\end{array}\right]-\text{ 3}\left[\begin{array}{ccc}2& 1& -\text{ 1}\\ 7& \text{2}& 4\end{array}\right]\\ \text{2X }-\text{ 3Y = }\left[\begin{array}{ccc}6-\text{ 6}& 2-\text{ 3}& -\text{ 2 + 3}\\ 10-\text{ 21}& -\text{ 4 }-\text{ 6}& -\text{ 6 }-\text{ 12}\end{array}\right]\\ \therefore \text{ 2X }-\text{ 3Y = }\left[\begin{array}{ccc}0& -\text{ 1}& 1\\ -\text{ 11}& -\text{ 10}& -\text{ 18}\end{array}\right]\end{array}$$

(iii) A matrix $$\text{Z}$$ such that $$X+Y+Z$$ is a zero matrix.

Ans: Given: $$\text{X and Y}$$ are matrices.

Here,

$$\begin{array}{c}\text{X + Y = }\left[\begin{array}{ccc}3\text{ + 2}& 1\text{ + 1}& -\text{ 1 }-\text{ 1}\\ 5\text{ + 7}& -\text{ 2 + 2}& -\text{ 3 + 4}\end{array}\right]\\ \text{X + Y = }\left[\begin{array}{ccc}5& 2& -\text{ 2}\\ 12& 0& 1\end{array}\right]\\ \text{Since, X + Y + Z = 0}\\ \Rightarrow \text{ Z = }-\text{(X + Y)}\\ \therefore \text{ Z = }\left[\begin{array}{ccc}-5& -2& \text{2}\\ -12& 0& -1\end{array}\right]\end{array}$$

8. Find non-zero values of $$\text{x}$$ satisfying the matrix equation$$x\left[\begin{array}{cc}2x& 2\\ 3& x\end{array}\right]+2\left[\begin{array}{cc}8& 5x\\ 4& 4x\end{array}\right]=2\left[\begin{array}{cc}(x^2+8)& 24\\ 10& 6x\end{array}\right].$$

Ans :Given: Matrix equation

Solve the matrix operations. Then, compare the elements of matrices.

Here,

$$\begin{array}{c}\text{x}\left[\begin{array}{cc}\text{2x}& 2\\ 3& \text{x}\end{array}\right]\text{ + 2}\left[\begin{array}{cc}8& 5\text{x}\\ 4& \text{4x}\end{array}\right]\text{ = 2}\left[\begin{array}{cc}\text{(}{\text{x}}^2\text{ + 8)}& 24\\ 10& \text{6x}\end{array}\right]\\ \Rightarrow \left[\begin{array}{cc}\text{2}{\text{x}}^2& 2\text{x}\\ 3\text{x}& {\text{x}}^2\end{array}\right]\text{ + }\left[\begin{array}{cc}16& \text{10x}\\ 8& \text{8x}\end{array}\right]\text{ = }\left[\begin{array}{cc}\text{2}{\text{x}}^2\text{ + 16}& 48\\ 20& \text{12x}\end{array}\right]\\ \Rightarrow \left[\begin{array}{cc}\text{2}{\text{x}}^2\text{ + 16}& 2\text{x + 10x}\\ 3\text{x + 8}& {\text{x}}^2\text{ + 8x}\end{array}\right]\text{ = }\left[\begin{array}{cc}\text{2}{\text{x}}^2\text{ + 16}& 48\\ 20& \text{12x}\end{array}\right]\\ \Rightarrow \text{ 2x + 10x = 48}\\ \Rightarrow \text{ x = }\frac{48}{12}\\ \therefore \text{ x = 4}\end{array}$$

9. If $$A=\left[\begin{array}{cc}0& 1\\ 1& 1\end{array}\right]andB=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right],$$ then show that 

Ans: Given: Matrices $$\text{A and B}\text{.}$$

Substitute matrices in equation. Then use matrix algebra.

Here,

$$\begin{array}{c}\text{A + B = }\left[\begin{array}{cc}0& 0\\ 2& 1\end{array}\right]\\ \text{A }-\text{ B = }\left[\begin{array}{cc}0& 2\\ 0& 1\end{array}\right]\\ \therefore \text{ (A + B}{\text{)}}_{2\times \text{ 2}}\text{.(A }-\text{ B}{\text{)}}_{2\times \text{ 2}}\text{ = }\left[\begin{array}{cc}0& 0\\ 0& 5\end{array}\right].....\text{(i)}\\ \text{Now, }{\text{A}}^2\text{ = A}\text{.A}\\ \Rightarrow {\text{A}}^2\text{ = }\left[\begin{array}{cc}0& 1\\ 1& 1\end{array}\right].\left[\begin{array}{cc}0& 1\\ 0& 1\end{array}\right]\\ \Rightarrow {\text{A}}^2\text{ = }\left[\begin{array}{cc}0\text{ + 1}& 0\text{ + 1}\\ 0\text{ + 1}& 1\text{ + 1}\end{array}\right]\\ \Rightarrow {\text{A}}^2\text{ = }\left[\begin{array}{cc}\text{1}& \text{1}\\ \text{1}& 2\end{array}\right]\\ \text{Now, }{\text{B}}^2\text{ = B}\text{.B}\\ \Rightarrow {\text{B}}^2\text{ = }\left[\begin{array}{cc}0& -\text{ 1}\\ \text{1}& 0\end{array}\right].\left[\begin{array}{cc}0& -\text{ 1}\\ \text{1}& 0\end{array}\right]\\ \Rightarrow {\text{B}}^2\text{ = }\left[\begin{array}{cc}-\text{ 1}& 0\\ 0& -\text{ 1}\end{array}\right]\\ \therefore {\text{A}}^2\text{ - }{\text{B}}^2\text{ = }\left[\begin{array}{cc}2& 1\\ \text{1}& 3\end{array}\right]......\text{(ii)}\\ \text{from equation (i) and (ii), we get}\\ \text{(A + B)(A }-\text{ B) }\ne {\text{A}}^2-{\text{B}}^2\end{array}$$

10. Find the value of $$x,if\left[\begin{array}{ccc}1& x& 1\end{array}\right]\left[\begin{array}{ccc}1& 3& 2\\ 2& 5& 1\\ 15& 3& 2\end{array}\right]\left[\begin{array}{c}1\\ 2\\ x\end{array}\right]=0.$$

Ans: Given: Matrix equation.

Product of matrix is possible if number of rows of matrix is equal to number of columns of another matrix.

$$\begin{array}{c}{\left[\begin{array}{ccc}1& \text{x}& 1\end{array}\right]}_{1\times \text{ 3}}{\left[\begin{array}{ccc}1& 3& 2\\ 2& 5& 1\\ 15& 3& 2\end{array}\right]}_{3\times \text{ 3}}{\left[\begin{array}{c}1\\ 2\\ \text{x}\end{array}\right]}_{3\times \text{ 1}}\text{ = 0}\text{.}\\ \Rightarrow {\left[\begin{array}{ccc}1\text{ + 2x + 15}& \text{3 + 5x + 3}& 2\text{ + x + 2}\end{array}\right]}_{1\times \text{ 3}}{\left[\begin{array}{c}1\\ 2\\ \text{x}\end{array}\right]}_{3\times \text{ 1}}\text{ = 0}\\ \Rightarrow {\left[\begin{array}{ccc}16\text{ + 2x}& \text{5x + 6}& 2\text{ + x + 2}\end{array}\right]}_{1\times \text{ 3}}{\left[\begin{array}{c}1\\ 2\\ \text{x}\end{array}\right]}_{3\times \text{ 1}}\text{ = 0}\\ \Rightarrow {\left[16\text{ + 2x + 2}\text{.(5x + 6) + (x + 4)}\text{.x}\right]}_{1\times \text{ 1}}\text{ = 0}\\ \Rightarrow {\left[{\text{x}}^2\text{ + 16x + 28}\right]}_{1\times \text{ 1}}\text{ = 0}\\ \Rightarrow {\text{x}}^2\text{ + 16x + 28 = 0}\\ \Rightarrow \text{ (x + 2)(x + 14) = 0}\\ \Rightarrow \text{ x = }-\text{ 2, }-\text{ 14}\end{array}$$