\displaystyle \textbf{Question 1. }\text{For any two matrices }A\text{ and }B,\text{ we have}
\displaystyle \text{(a) }AB=BA\qquad \text{(b) }AB\ne BA
\displaystyle \text{(c) }AB=O\qquad \text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle \text{For any two matrices }A\text{ and }B,\text{ matrix multiplication is not always commutative.}
\displaystyle \therefore AB\ne BA\text{ generally.}
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 2. }\text{If matrix }A=[a_{ij}]_{2\times2},\text{ where }a_{ij}=\begin{cases}1,&\text{if }i\ne j\\0,&\text{if }i=j\end{cases}\text{ Then,}
\displaystyle A^{2}\text{ is equal to}
\displaystyle \text{(a) }I\qquad \text{(b) }A
\displaystyle \text{(c) }O\qquad \text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}0&1\\1&0\end{bmatrix}
\displaystyle \therefore A^{2}=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}
\displaystyle =\begin{bmatrix}1&0\\0&1\end{bmatrix}=I
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 3. }\text{If }A=\begin{bmatrix}\alpha&\beta\\ \gamma&-\alpha\end{bmatrix}\text{ is such that }A^{2}=I,\text{ then}
\displaystyle \text{(a) }1+\alpha^{2}+\beta\gamma=0\qquad \text{(b) }1-\alpha^{2}+\beta\gamma=0
\displaystyle \text{(c) }1-\alpha^{2}-\beta\gamma=0\qquad \text{(d) }1+\alpha^{2}-\beta\gamma=0
\displaystyle \text{Answer:}
\displaystyle A^{2}=\begin{bmatrix}\alpha&\beta\\ \gamma&-\alpha\end{bmatrix}\begin{bmatrix}\alpha&\beta\\ \gamma&-\alpha\end{bmatrix}
\displaystyle =\begin{bmatrix}\alpha^{2}+\beta\gamma&0\\0&\alpha^{2}+\beta\gamma\end{bmatrix}
\displaystyle \text{Given, }A^{2}=I
\displaystyle \therefore \alpha^{2}+\beta\gamma=1
\displaystyle \therefore 1-\alpha^{2}-\beta\gamma=0
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 4. }\text{If matrix }A\text{ given by }A=\begin{bmatrix}1&-1\\0&3\\2&5\end{bmatrix},\text{ then the order of}
\displaystyle \text{the matrix }A\text{ is}
\displaystyle \text{(a) }1\times2\qquad \text{(b) }2\times3\qquad \text{(c) }3\times2\qquad \text{(d) }2\times2
\displaystyle \text{Answer:}
\displaystyle \text{The given matrix has }3\text{ rows and }2\text{ columns.}
\displaystyle \therefore \text{Order of }A=3\times2
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 5. }\text{If }A=[a_{ij}]=[\sin jx^i],\ 1\leq i\leq3,\ 1\leq j\leq3\text{ and }B=[b_{ij}]=[\cos ix^j],
\displaystyle 1\leq i\leq3,\ 1\leq j\leq3.\text{ Then, the value of }\frac{a_{22}}{b_{12}}\text{ is}
\displaystyle \text{(a) }2\cos x^2\qquad \text{(b) }2\sin^2x
\displaystyle \text{(c) }2\sin x^2\qquad \text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle a_{22}=\sin(2x^2)
\displaystyle b_{12}=\cos(x^2)
\displaystyle \therefore \frac{a_{22}}{b_{12}}=\frac{\sin(2x^2)}{\cos x^2}
\displaystyle =\frac{2\sin x^2\cos x^2}{\cos x^2}=2\sin x^2
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 6. }\text{If }A=\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix},\text{ then }A^2-4A\text{ is equal to}
\displaystyle \text{(a) }2I_3\qquad \text{(b) }3I_3\qquad \text{(c) }4I_3\qquad \text{(d) }5I_3
\displaystyle \text{Answer:}
\displaystyle A^2=\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}
\displaystyle =\begin{bmatrix}9&8&8\\8&9&8\\8&8&9\end{bmatrix}
\displaystyle 4A=4\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}=\begin{bmatrix}4&8&8\\8&4&8\\8&8&4\end{bmatrix}
\displaystyle \therefore A^2-4A=\begin{bmatrix}9&8&8\\8&9&8\\8&8&9\end{bmatrix}-\begin{bmatrix}4&8&8\\8&4&8\\8&8&4\end{bmatrix}
\displaystyle =\begin{bmatrix}5&0&0\\0&5&0\\0&0&5\end{bmatrix}=5I_3
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 7. }\text{If }a_{ij}=\frac{1}{2}(3i+2j)\text{ and }A=[a_{ij}]_{2\times2},\text{ then }a_{21}+a_{22}\text{ is}
\displaystyle \text{equal to}
\displaystyle \text{(a) }1\qquad \text{(b) }8\qquad \text{(c) }9\qquad \text{(d) }-1
\displaystyle \text{Answer:}
\displaystyle a_{21}=\frac{1}{2}(3\cdot2+2\cdot1)=\frac{1}{2}(8)=4
\displaystyle a_{22}=\frac{1}{2}(3\cdot2+2\cdot2)=\frac{1}{2}(10)=5
\displaystyle \therefore a_{21}+a_{22}=4+5=9
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 8. }\text{If }A=[a_{ij}]_{2\times2},\text{ where }a_{ij}=i^{2}+2j^{2},\text{ then }A\text{ is equal}
\displaystyle \text{to}
\displaystyle \text{(a) }\begin{bmatrix}3&9\\6&12\end{bmatrix}\qquad \text{(b) }\begin{bmatrix}3&6\\9&12\end{bmatrix}
\displaystyle \text{(c) }\begin{bmatrix}6&12\\3&9\end{bmatrix}\qquad \text{(d) }\begin{bmatrix}9&3\\6&1\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle a_{11}=1^{2}+2(1^{2})=3
\displaystyle a_{12}=1^{2}+2(2^{2})=9
\displaystyle a_{21}=2^{2}+2(1^{2})=6
\displaystyle a_{22}=2^{2}+2(2^{2})=12
\displaystyle \therefore A=\begin{bmatrix}3&9\\6&12\end{bmatrix}
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 9. }\text{If }A=\begin{bmatrix}\cos\alpha&-\sin\alpha\\ \sin\alpha&\cos\alpha\end{bmatrix},\text{ then for what value of }\alpha,\ A\text{ is}
\displaystyle \text{an identity matrix?}
\displaystyle \text{Answer:}
\displaystyle \text{For }A\text{ to be an identity matrix,}
\displaystyle \cos\alpha=1\text{ and }\sin\alpha=0
\displaystyle \therefore \alpha=2n\pi,\ n\in Z
\\

\displaystyle \textbf{Question 10. }\text{Show that all the diagonal elements of a skew-symmetric matrix are zero.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=[a_{ij}]\text{ be a skew-symmetric matrix.}
\displaystyle \therefore A'=-A
\displaystyle \therefore a_{ji}=-a_{ij}
\displaystyle \text{For diagonal elements, }i=j.
\displaystyle \therefore a_{ii}=-a_{ii}
\displaystyle \therefore 2a_{ii}=0
\displaystyle \therefore a_{ii}=0
\displaystyle \therefore \text{All diagonal elements of a skew-symmetric matrix are zero.}
\\

\displaystyle \textbf{Question 11. }\text{Show that }A'A\text{ and }AA'\text{ are both symmetric matrices for any matrix }A.
\displaystyle \text{Answer:}
\displaystyle \text{We know that }(AB)'=B'A'.
\displaystyle \therefore (A'A)'=A'(A')'=A'A
\displaystyle \therefore A'A\text{ is symmetric.}
\displaystyle \text{Also, }(AA')'=(A')'A'=AA'
\displaystyle \therefore AA'\text{ is symmetric.}
\displaystyle \therefore A'A\text{ and }AA'\text{ are both symmetric matrices.}
\\

\displaystyle \textbf{Question 12. }\text{Find the value of }x,\ y\text{ and }z,\text{ if } \begin{bmatrix}-2x+y\\x+y+z\\x+y\end{bmatrix}=\begin{bmatrix}-3\\3\\3\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle \text{By equating corresponding elements,}
\displaystyle -2x+y=-3 \quad ...(1)
\displaystyle x+y+z=3 \quad ...(2)
\displaystyle x+y=3 \quad ...(3)
\displaystyle \text{From (3), }y=3-x
\displaystyle \text{Substituting in (1),}
\displaystyle -2x+3-x=-3
\displaystyle -3x=-6
\displaystyle x=2
\displaystyle \therefore y=3-2=1
\displaystyle \text{From (2), }2+1+z=3
\displaystyle \therefore z=0
\displaystyle \therefore x=2,\ y=1,\ z=0
\\

\displaystyle \textbf{Question 13. }\text{If }X=\begin{bmatrix}3&1&-1\\5&-2&-3\end{bmatrix}\text{ and }Y=\begin{bmatrix}2&1&-1\\7&2&4\end{bmatrix}\text{ then find}
\displaystyle \text{matrix }Z,\text{ such that }X+Y+Z\text{ is zero matrix.}
\displaystyle \text{Answer:}
\displaystyle X+Y=\begin{bmatrix}3&1&-1\\5&-2&-3\end{bmatrix}+\begin{bmatrix}2&1&-1\\7&2&4\end{bmatrix}
\displaystyle =\begin{bmatrix}5&2&-2\\12&0&1\end{bmatrix}
\displaystyle \text{Since }X+Y+Z=O,
\displaystyle Z=-(X+Y)
\displaystyle =\begin{bmatrix}-5&-2&2\\-12&0&-1\end{bmatrix}
\\

\displaystyle \textbf{Question 14. }\text{If }A=\begin{bmatrix}2&0\\1&4\end{bmatrix}\text{ and }B=\begin{bmatrix}-1&2\\3&0\end{bmatrix}\text{ then find }(AB)'.
\displaystyle \text{Answer:}
\displaystyle AB=\begin{bmatrix}2&0\\1&4\end{bmatrix}\begin{bmatrix}-1&2\\3&0\end{bmatrix}
\displaystyle =\begin{bmatrix}-2&4\\11&2\end{bmatrix}
\displaystyle \therefore (AB)'=\begin{bmatrix}-2&11\\4&2\end{bmatrix}
\\

\displaystyle \textbf{Question 15. }\text{If }A\text{ and }B\text{ are symmetric matrices, then prove that } \\ AB+BA\text{ is a symmetric matrix.}
\displaystyle \text{Answer:}
\displaystyle \text{Since }A\text{ and }B\text{ are symmetric, }A'=A\text{ and }B'=B.
\displaystyle (AB+BA)'=(AB)'+(BA)'
\displaystyle =B'A'+A'B'
\displaystyle =BA+AB
\displaystyle =AB+BA
\displaystyle \therefore (AB+BA)'=AB+BA
\displaystyle \therefore AB+BA\text{ is a symmetric matrix.}
\displaystyle \text{Hence proved.}
\\

\displaystyle \textbf{Question 16. }\text{From the following matrix equation, find the value of }x\text{ and }y.
\displaystyle \begin{bmatrix}x+y&4\\-5&3y\end{bmatrix}=\begin{bmatrix}3&4\\-5&6\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{Since the two matrices are equal, their corresponding elements are equal.}
\displaystyle x+y=3
\displaystyle 3y=6
\displaystyle \therefore y=2
\displaystyle \therefore x+2=3
\displaystyle \therefore x=1
\displaystyle \therefore x=1,\ y=2
\\

\displaystyle \textbf{Question 17. }\text{Find the product and hence find the order of product matrices}
\displaystyle \begin{bmatrix}1\\2\\3\end{bmatrix}\begin{bmatrix}2&3&4\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \begin{bmatrix}1\\2\\3\end{bmatrix}\begin{bmatrix}2&3&4\end{bmatrix}
\displaystyle =\begin{bmatrix}1\cdot2&1\cdot3&1\cdot4\\2\cdot2&2\cdot3&2\cdot4\\3\cdot2&3\cdot3&3\cdot4\end{bmatrix}
\displaystyle =\begin{bmatrix}2&3&4\\4&6&8\\6&9&12\end{bmatrix}
\displaystyle \text{Order of the product matrix is }3\times3.
\\

\displaystyle \textbf{Question 18. }\text{If }\begin{bmatrix}1&2\\3&4\end{bmatrix}\begin{bmatrix}3&1\\2&5\end{bmatrix}=\begin{bmatrix}7&11\\k&23\end{bmatrix},\text{ then find the value of }k.
\displaystyle \text{Answer:}
\displaystyle \begin{bmatrix}1&2\\3&4\end{bmatrix}\begin{bmatrix}3&1\\2&5\end{bmatrix}
\displaystyle =\begin{bmatrix}1\cdot3+2\cdot2&1\cdot1+2\cdot5\\3\cdot3+4\cdot2&3\cdot1+4\cdot5\end{bmatrix}
\displaystyle =\begin{bmatrix}7&11\\17&23\end{bmatrix}
\displaystyle \therefore k=17
\\

\displaystyle \textbf{Question 19. }\text{For what values of }x\text{ and }y\text{ are the following matrices equal?}
\displaystyle A=\begin{bmatrix}2x+1&3y\\0&y^2-5y\end{bmatrix}\text{ and }B=\begin{bmatrix}x+3&y^2+2\\0&-6\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{Since }A=B,\text{ corresponding elements are equal.}
\displaystyle 2x+1=x+3
\displaystyle \therefore x=2
\displaystyle 3y=y^2+2
\displaystyle \therefore y^2-3y+2=0
\displaystyle \therefore (y-1)(y-2)=0
\displaystyle \therefore y=1\text{ or }y=2
\displaystyle \text{Also, }y^2-5y=-6
\displaystyle \therefore y^2-5y+6=0
\displaystyle \therefore (y-2)(y-3)=0
\displaystyle \therefore y=2\text{ or }y=3
\displaystyle \therefore \text{Common value of }y=2
\displaystyle \therefore x=2,\ y=2
\\

\displaystyle \textbf{Question 20. }\text{Find }X\text{ and }Y\text{ if }X+Y=\begin{bmatrix}7&0\\2&5\end{bmatrix}\text{ and }X-Y=\begin{bmatrix}3&0\\0&3\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle X+Y=\begin{bmatrix}7&0\\2&5\end{bmatrix}\quad ...(1)
\displaystyle X-Y=\begin{bmatrix}3&0\\0&3\end{bmatrix}\quad ...(2)
\displaystyle \text{Adding (1) and (2),}
\displaystyle 2X=\begin{bmatrix}10&0\\2&8\end{bmatrix}
\displaystyle \therefore X=\begin{bmatrix}5&0\\1&4\end{bmatrix}
\displaystyle \text{Subtracting (2) from (1),}
\displaystyle 2Y=\begin{bmatrix}4&0\\2&2\end{bmatrix}
\displaystyle \therefore Y=\begin{bmatrix}2&0\\1&1\end{bmatrix}
\\

\displaystyle \textbf{Question 21. }\text{Find the matrix }A\text{ such that }A-\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\2&4&6\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle A-\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\2&4&6\end{bmatrix}
\displaystyle \therefore A=\begin{bmatrix}-7&-8&-9\\2&4&6\end{bmatrix}+\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}
\displaystyle =\begin{bmatrix}-6&-6&-6\\6&9&12\end{bmatrix}
\\

\displaystyle \textbf{Question 22. }\text{If }A=\begin{bmatrix}2&3&-5\\0&-1&4\end{bmatrix},\text{ then verify that }(3A)'=3A'.
\displaystyle \text{Answer:}
\displaystyle 3A=3\begin{bmatrix}2&3&-5\\0&-1&4\end{bmatrix}=\begin{bmatrix}6&9&-15\\0&-3&12\end{bmatrix}
\displaystyle \therefore (3A)'=\begin{bmatrix}6&0\\9&-3\\-15&12\end{bmatrix}
\displaystyle A'=\begin{bmatrix}2&0\\3&-1\\-5&4\end{bmatrix}
\displaystyle \therefore 3A'=3\begin{bmatrix}2&0\\3&-1\\-5&4\end{bmatrix}=\begin{bmatrix}6&0\\9&-3\\-15&12\end{bmatrix}
\displaystyle \therefore (3A)'=3A'
\displaystyle \text{Hence verified.}
\\

\displaystyle \textbf{Question 23. }\text{For the following matrices }A\text{ and }B,\text{ verify that }(AB)'=B'A'.
\displaystyle A=\begin{bmatrix}1\\-4\\3\end{bmatrix},\quad B=\begin{bmatrix}-1&2&1\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle AB=\begin{bmatrix}1\\-4\\3\end{bmatrix}\begin{bmatrix}-1&2&1\end{bmatrix}
\displaystyle =\begin{bmatrix}-1&2&1\\4&-8&-4\\-3&6&3\end{bmatrix}
\displaystyle \therefore (AB)'=\begin{bmatrix}-1&4&-3\\2&-8&6\\1&-4&3\end{bmatrix}
\displaystyle B'=\begin{bmatrix}-1\\2\\1\end{bmatrix},\quad A'=\begin{bmatrix}1&-4&3\end{bmatrix}
\displaystyle \therefore B'A'=\begin{bmatrix}-1\\2\\1\end{bmatrix}\begin{bmatrix}1&-4&3\end{bmatrix}
\displaystyle =\begin{bmatrix}-1&4&-3\\2&-8&6\\1&-4&3\end{bmatrix}
\displaystyle \therefore (AB)'=B'A'
\displaystyle \text{Hence verified.}
\\

\displaystyle \textbf{Question 24. }\text{Express the following matrix as a sum of a symmetric and a} \\ \text{skew-symmetric matrix and verify your result.}
\displaystyle A=\begin{bmatrix}3&-2&-4\\3&-2&-5\\-1&1&2\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle A=\frac{1}{2}(A+A')+\frac{1}{2}(A-A')
\displaystyle A'=\begin{bmatrix}3&3&-1\\-2&-2&1\\-4&-5&2\end{bmatrix}
\displaystyle \frac{1}{2}(A+A')=\frac{1}{2}\begin{bmatrix}6&1&-5\\1&-4&-4\\-5&-4&4\end{bmatrix}
\displaystyle =\begin{bmatrix}3&\frac{1}{2}&-\frac{5}{2}\\\frac{1}{2}&-2&-2\\-\frac{5}{2}&-2&2\end{bmatrix}
\displaystyle \frac{1}{2}(A-A')=\frac{1}{2}\begin{bmatrix}0&-5&-3\\5&0&-6\\3&6&0\end{bmatrix}
\displaystyle =\begin{bmatrix}0&-\frac{5}{2}&-\frac{3}{2}\\\frac{5}{2}&0&-3\\\frac{3}{2}&3&0\end{bmatrix}
\displaystyle \therefore A=\begin{bmatrix}3&\frac{1}{2}&-\frac{5}{2}\\\frac{1}{2}&-2&-2\\-\frac{5}{2}&-2&2\end{bmatrix}+\begin{bmatrix}0&-\frac{5}{2}&-\frac{3}{2}\\\frac{5}{2}&0&-3\\\frac{3}{2}&3&0\end{bmatrix}
\displaystyle =\begin{bmatrix}3&-2&-4\\3&-2&-5\\-1&1&2\end{bmatrix}
\displaystyle \text{Hence verified.}
\\

\displaystyle \textbf{Question 25. }\text{Express the matrix }A=\begin{bmatrix}2&3\\-1&4\end{bmatrix}\text{ as the sum of a}
\displaystyle \text{symmetric matrix and the skew-symmetric matrix.}
\displaystyle \text{Answer:}
\displaystyle A=\frac{1}{2}(A+A')+\frac{1}{2}(A-A')
\displaystyle A'=\begin{bmatrix}2&-1\\3&4\end{bmatrix}
\displaystyle \frac{1}{2}(A+A')=\frac{1}{2}\begin{bmatrix}4&2\\2&8\end{bmatrix}=\begin{bmatrix}2&1\\1&4\end{bmatrix}
\displaystyle \frac{1}{2}(A-A')=\frac{1}{2}\begin{bmatrix}0&4\\-4&0\end{bmatrix}=\begin{bmatrix}0&2\\-2&0\end{bmatrix}
\displaystyle \therefore A=\begin{bmatrix}2&1\\1&4\end{bmatrix}+\begin{bmatrix}0&2\\-2&0\end{bmatrix}
\\

\displaystyle \textbf{Question 26. }\text{If }A=\begin{bmatrix}1&0&-2\\-2&-1&2\\3&4&1\end{bmatrix},\text{ then show that}
\displaystyle A^{3}-A^{2}-3A-I_{3}=0.
\displaystyle \text{Answer:}
\displaystyle A^{2}=\begin{bmatrix}1&0&-2\\-2&-1&2\\3&4&1\end{bmatrix}\begin{bmatrix}1&0&-2\\-2&-1&2\\3&4&1\end{bmatrix}
\displaystyle =\begin{bmatrix}-5&-8&-4\\6&9&4\\-2&0&3\end{bmatrix}
\displaystyle A^{3}=A^{2}A=\begin{bmatrix}-5&-8&-4\\6&9&4\\-2&0&3\end{bmatrix}\begin{bmatrix}1&0&-2\\-2&-1&2\\3&4&1\end{bmatrix}
\displaystyle =\begin{bmatrix}-1&-8&-10\\0&7&10\\7&12&7\end{bmatrix}
\displaystyle A^{3}-A^{2}-3A-I_{3}
\displaystyle =\begin{bmatrix}-1&-8&-10\\0&7&10\\7&12&7\end{bmatrix}-\begin{bmatrix}-5&-8&-4\\6&9&4\\-2&0&3\end{bmatrix}-3\begin{bmatrix}1&0&-2\\-2&-1&2\\3&4&1\end{bmatrix}-\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}
\displaystyle \therefore A^{3}-A^{2}-3A-I_{3}=0
\displaystyle \text{Hence proved.}
\\

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