\displaystyle \textbf{Question 1: } \text{If }\begin{bmatrix}x-y&2x+z\\2x-y&3z+w\end{bmatrix} =\begin{bmatrix}-1&5\\0&13\end{bmatrix},\ \text{find }x,y,z,w.\ \text{[CBSE 2002, 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Since the corresponding elements of two equal matrices are equal. Therefore,}
\displaystyle \begin{bmatrix}x-y&2x+z\\2x-y&3z+w\end{bmatrix} =\begin{bmatrix}-1&5\\0&13\end{bmatrix}
\displaystyle \Rightarrow\ x-y=-1,\ \ 2x+z=5,\ \ 2x-y=0,\ \ 3z+w=13.
\displaystyle \text{Solving the equations }x-y=-1\text{ and }2x-y=0\text{ as simultaneous linear equations, we get}
\displaystyle x=1,\ y=2.
\displaystyle \text{Now putting }x=1\text{ in }2x+z=5,\ \text{we get }z=3.
\displaystyle \text{Substituting }z=3\text{ in }3z+w=13,\ \text{we obtain }w=4.
\displaystyle \text{Thus, }x=1,\ y=2,\ z=3\ \text{and }w=4.

\displaystyle \textbf{Question 2: } \text{Simplify:}\ \cos\theta \begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix} +\sin\theta\begin{bmatrix}\sin\theta&-\cos\theta\\\cos\theta&\sin\theta\end{bmatrix}\ \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \cos\theta\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix} +\sin\theta\begin{bmatrix}\sin\theta&-\cos\theta\\\cos\theta&\sin\theta\end{bmatrix}
\displaystyle =\begin{bmatrix}\cos^{2}\theta&\sin\theta\cos\theta\\-\sin\theta\cos\theta&\cos^{2}\theta\end{bmatrix} +\begin{bmatrix}\sin^{2}\theta&-\sin\theta\cos\theta\\\sin\theta\cos\theta&\sin^{2}\theta\end{bmatrix}
\displaystyle =\begin{bmatrix}\cos^{2}\theta+\sin^{2}\theta& \sin\theta\cos\theta-\sin\theta\cos\theta\\-\sin\theta\cos\theta+\sin\theta\cos\theta& \cos^{2}\theta+\sin^{2}\theta\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}

\displaystyle \textbf{Question 3: } \text{Find a matrix }X\text{ such that }2A+B+X=O, \text{ where }A=\begin{bmatrix}-1&2\\3&4\end{bmatrix} \\ \text{ and }B=\begin{bmatrix}3&-2\\1&5\end{bmatrix}.\ \text{[CBSE 2000]}
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle 2A+B+X=O
\displaystyle \Rightarrow\ X=-2A-B
\displaystyle \Rightarrow\ X=-2\begin{bmatrix}-1&2\\3&4\end{bmatrix} -\begin{bmatrix}3&-2\\1&5\end{bmatrix}
\displaystyle \Rightarrow\ X=\begin{bmatrix}2&-4\\-6&-8\end{bmatrix} +\begin{bmatrix}-3&2\\-1&-5\end{bmatrix}
\displaystyle \Rightarrow\ X=\begin{bmatrix}2-3&-4+2\\-6-1&-8-5\end{bmatrix} =\begin{bmatrix}-1&-2\\-7&-13\end{bmatrix}

\displaystyle \textbf{Question 4: } \text{If }A=\begin{bmatrix}1&-1\\2&-1\end{bmatrix}, \ B=\begin{bmatrix}a&1\\b&-1\end{bmatrix}\text{ and }(A+B)^{2}=A^{2}+B^{2}, \\ \text{ find }a\text{ and }b.\ \text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle (A+B)^{2}=A^{2}+B^{2}
\displaystyle \Rightarrow\ (A+B)(A+B)=A^{2}+B^{2}
\displaystyle \Rightarrow\ (A+B)A+(A+B)B=A^{2}+B^{2}\ \text{[By distributive law]}
\displaystyle \Rightarrow\ A^{2}+BA+AB+B^{2}=A^{2}+B^{2}\ \text{[By distributive law]}
\displaystyle \Rightarrow\ BA+AB=O
\displaystyle \Rightarrow\ \begin{bmatrix}a&1\\b&-1\end{bmatrix} \begin{bmatrix}1&-1\\2&-1\end{bmatrix} +\begin{bmatrix}1&-1\\2&-1\end{bmatrix} \begin{bmatrix}a&1\\b&-1\end{bmatrix} =\begin{bmatrix}0&0\\0&0\end{bmatrix}
\displaystyle \Rightarrow\ \begin{bmatrix}a+2&-a-1\\b-2&-b+1\end{bmatrix} +\begin{bmatrix}a-b&2\\2a-b&3\end{bmatrix} =\begin{bmatrix}0&0\\0&0\end{bmatrix}
\displaystyle \Rightarrow\ \begin{bmatrix}2a-b+2&-a+1\\2a-2&-b+4\end{bmatrix} =\begin{bmatrix}0&0\\0&0\end{bmatrix}
\displaystyle \Rightarrow\ 2a-b+2=0,\ -a+1=0,\ 2a-2=0\ \text{and}\ -b+4=0
\displaystyle \Rightarrow\ a=1,\ b=4

\displaystyle \textbf{Question 5: } \text{Let }A=\begin{bmatrix}2&-1\\3&4\end{bmatrix}, \ B=\begin{bmatrix}5&2\\7&4\end{bmatrix},\ C=\begin{bmatrix}2&5\\3&8\end{bmatrix}.
\displaystyle \text{Find a matrix }D\text{ such that }CD-AB=O.\ \text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }D=\begin{bmatrix}a&b\\x&y\end{bmatrix}. \ \text{Then,}
\displaystyle CD-AB=O
\displaystyle \Rightarrow\ CD=AB
\displaystyle \Rightarrow\ \begin{bmatrix}2&5\\3&8\end{bmatrix} \begin{bmatrix}a&b\\x&y\end{bmatrix} =\begin{bmatrix}2&-1\\3&4\end{bmatrix} \begin{bmatrix}5&2\\7&4\end{bmatrix}
\displaystyle \Rightarrow\ \begin{bmatrix}2a+5x&2b+5y\\3a+8x&3b+8y\end{bmatrix} =\begin{bmatrix}10-7&4-4\\15+28&6+16\end{bmatrix}
\displaystyle \Rightarrow\ \begin{bmatrix}2a+5x&2b+5y\\3a+8x&3b+8y\end{bmatrix} =\begin{bmatrix}3&0\\43&22\end{bmatrix}
\displaystyle \Rightarrow\ 2a+5x=3,\ 3a+8x=43,\ 2b+5y=0\ \text{and}\ 3b+8y=22
\displaystyle \text{Solving }2a+5x=3\text{ and }3a+8x=43,\text{ we get: } a=-191\text{ and }x=77.
\displaystyle \text{Solving }2b+5y=0\text{ and }3b+8y=22,\text{ we get: } b=-110\text{ and }y=44.
\displaystyle \therefore\ D=\begin{bmatrix}a&b\\x&y\end{bmatrix} =\begin{bmatrix}-191&-110\\77&44\end{bmatrix}

\displaystyle \textbf{Question 6: } \text{Find the value of }x\text{ such that } [1\ x\ 1]\begin{bmatrix}1&3&2\\2&5&1\\15&3&2\end{bmatrix} \begin{bmatrix}1\\2\\x\end{bmatrix}=0.\ \text{[CBSE 2006]}
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle [1\ x\ 1]\begin{bmatrix}1&3&2\\2&5&1\\15&3&2\end{bmatrix} \begin{bmatrix}1\\2\\x\end{bmatrix}=0
\displaystyle \Rightarrow\ [1\ x\ 1]\begin{bmatrix}7+2x\\12+x\\21+2x\end{bmatrix}=0

\displaystyle \Rightarrow\ 7+2x+12x+x^{2}+21+2x=0
\displaystyle \Rightarrow\ x^{2}+16x+28=0
\displaystyle \Rightarrow\ (x+14)(x+2)=0\ \Rightarrow\ x=-2\ \text{or}\ -14.

\displaystyle \textbf{Question 7: } \text{If }A=\begin{bmatrix}1&0\\-1&7\end{bmatrix} \text{ and }I=\begin{bmatrix}1&0\\0&1\end{bmatrix},\text{ then find }k \text{ so that } \\ A^{2}=8A+kI. \hspace{5.0cm} \text{[CBSE 2005]}
\displaystyle \text{Answer:}
\displaystyle \text{We have,}\ A=\begin{bmatrix}1&0\\-1&7\end{bmatrix}
\displaystyle \therefore\ A^{2}=AA=\begin{bmatrix}1&0\\-1&7\end{bmatrix} \begin{bmatrix}1&0\\-1&7\end{bmatrix} =\begin{bmatrix}1&0\\-8&49\end{bmatrix}
\displaystyle \text{and,}\ 8A+kI=8\begin{bmatrix}1&0\\-1&7\end{bmatrix} +k\begin{bmatrix}1&0\\0&1\end{bmatrix} =\begin{bmatrix}8&0\\-8&56\end{bmatrix} +\begin{bmatrix}k&0\\0&k\end{bmatrix} =\begin{bmatrix}8+k&0\\-8&56+k\end{bmatrix}
\displaystyle \therefore\ A^{2}=8A+kI
\displaystyle \Rightarrow\ \begin{bmatrix}1&0\\-8&49\end{bmatrix} =\begin{bmatrix}8+k&0\\-8&56+k\end{bmatrix}
\displaystyle \Rightarrow\ 1=8+k\ \text{and}\ 56+k=49
\displaystyle \Rightarrow\ k=-7

\displaystyle \textbf{Question 8: } \text{If }A_{\alpha}= \begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix}, \ \text{then prove that}
\displaystyle \text{(i)}\ A_{\alpha}A_{\beta}=A_{\alpha+\beta} \qquad \text{(ii)}\ (A_{\alpha})^{n}= \begin{bmatrix}\cos n\alpha&\sin n\alpha\\-\sin n\alpha&\cos n\alpha\end{bmatrix} \ \text{for every} \\ \text{positive integer }n.\ \text{[CBSE 2004]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) We have,}
\displaystyle A_{\alpha}A_{\beta}= \begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix} \begin{bmatrix}\cos\beta&\sin\beta\\-\sin\beta&\cos\beta\end{bmatrix}
\displaystyle \Rightarrow\ A_{\alpha}A_{\beta}= \begin{bmatrix}\cos\alpha\cos\beta-\sin\alpha\sin\beta& \sin\alpha\cos\beta+\cos\alpha\sin\beta\\ -\sin\alpha\cos\beta-\cos\alpha\sin\beta& \cos\alpha\cos\beta-\sin\alpha\sin\beta\end{bmatrix}
\displaystyle \Rightarrow\ A_{\alpha}A_{\beta}= \begin{bmatrix}\cos(\alpha+\beta)&\sin(\alpha+\beta)\\ -\sin(\alpha+\beta)&\cos(\alpha+\beta)\end{bmatrix}=A_{\alpha+\beta}.
\displaystyle \text{(ii) We shall prove the result by mathematical induction on }n.
\displaystyle \text{STEP 1}\ \ \text{When }n=1,\ \text{by the definition of integral powers of a matrix, we have}
\displaystyle (A_{\alpha})^{1}=A_{\alpha}= \begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix} =\begin{bmatrix}\cos(1.\alpha)&\sin(1.\alpha)\\-\sin(1.\alpha)&\cos(1.\alpha)\end{bmatrix}
\displaystyle \text{So, the result is true for }n=1.
\displaystyle \text{STEP 2}\ \ \text{Let the result be true for }n=m.\ \text{Then,}
\displaystyle (A_{\alpha})^{m}= \begin{bmatrix}\cos m\alpha&\sin m\alpha\\-\sin m\alpha&\cos m\alpha\end{bmatrix} \qquad \ldots\text{(i)}
\displaystyle \text{Now we will show that the result is true for }n=m+1.
\displaystyle \text{i.e.}\ (A_{\alpha})^{m+1}= \begin{bmatrix}\cos(m+1)\alpha&\sin(m+1)\alpha\\-\sin(m+1)\alpha&\cos(m+1)\alpha\end{bmatrix}
\displaystyle \text{By the definition of integral powers of a square matrix, we have}
\displaystyle (A_{\alpha})^{m+1}=(A_{\alpha})^{m}A_{\alpha}
\displaystyle \Rightarrow\ (A_{\alpha})^{m+1}= \begin{bmatrix}\cos m\alpha&\sin m\alpha\\-\sin m\alpha&\cos m\alpha\end{bmatrix} \begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix} \qquad \text{[By assumption (i)]}
\displaystyle \Rightarrow\ (A_{\alpha})^{m+1}= \begin{bmatrix}\cos m\alpha\cos\alpha-\sin m\alpha\sin\alpha& \cos m\alpha\sin\alpha+\sin m\alpha\cos\alpha\\ -\sin m\alpha\cos\alpha-\cos m\alpha\sin\alpha& -\sin m\alpha\sin\alpha+\cos m\alpha\cos\alpha\end{bmatrix}
\displaystyle \Rightarrow\ (A_{\alpha})^{m+1}= \begin{bmatrix}\cos(m\alpha+\alpha)&\sin(m\alpha+\alpha)\\ -\sin(m\alpha+\alpha)&\cos(m\alpha+\alpha)\end{bmatrix} =\begin{bmatrix}\cos(m+1)\alpha&\sin(m+1)\alpha\\-\sin(m+1)\alpha&\cos(m+1)\alpha\end{bmatrix}
\displaystyle \text{This shows that the result is true for }n=m+1,\ \text{whenever it is true for }n=m.
\displaystyle \text{Hence, by the principle of mathematical induction, the result is valid for any positive integer }n.

\displaystyle \textbf{Question 9: } \text{Three schools }A,\ B\text{ and }C\ \text{organised a mela for collecting funds for} \\ \text{helping the}  \displaystyle \text{rehabilitation of flood victims. They sold hand made fans, mats and plates from} \\ \text{recycled material at a cost of}  \text{Rs }25,\ \text{Rs }100\ \text{and Rs }50\ \text{each. The number of articles sold} \\ \text{are given below:}
\displaystyle \begin{array}{c|ccc} \text{School} & A & B & C\\ \hline \text{Article} & & &\\ \hline \text{Hand-fans} & 40 & 25 & 35\\ \text{Mats} & 50 & 40 & 50\\ \text{Plates} & 20 & 30 & 40 \end{array}
\displaystyle \text{Find the funds collected by each school separately by selling the above articles. Also, find the} \\ \text{total funds}\text{collected for the purpose.}\ \text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Three items sold by three schools are represented by the following }3\times 3\text{ matrix }Q
\displaystyle \text{as given below.}
\displaystyle Q=\begin{array}{c} \\[-2mm] A\\ B\\ C \end{array} \left[\begin{array}{ccc} \text{Hand-fans} & \text{Mats} & \text{Plates}\\ 40 & 50 & 20\\ 25 & 40 & 30\\ 35 & 50 & 40 \end{array}\right]
\displaystyle \text{The price matrix representing price of three articles in Rs is a }3\times 1\text{ matrix given by}
\displaystyle P=\begin{array}{c} \text{Hand-fan}\\ \text{Mat}\\ \text{Plate} \end{array} \left[\begin{array}{c} 25\\ 100\\ 50 \end{array}\right]
\displaystyle \text{The funds collected by schools }A,\ B\text{ and }C\ \text{separately by selling three articles are given by the}
\displaystyle \text{product matrix }QP.
\displaystyle \therefore\ QP=\begin{array}{c} \\[-2mm] A\\ B\\ C \end{array} \left[\begin{array}{ccc} 40 & 50 & 20\\ 25 & 40 & 30\\ 35 & 50 & 40 \end{array}\right] \left[\begin{array}{c} 25\\ 100\\ 50 \end{array}\right]
\displaystyle \Rightarrow\ QP=\begin{array}{c} A\\ B\\ C \end{array} \left[\begin{array}{c} 40\times 25+50\times 100+20\times 50\\ 25\times 25+40\times 100+30\times 50\\ 35\times 25+50\times 100+40\times 50 \end{array}\right] =\begin{array}{c} A\\ B\\ C \end{array} \left[\begin{array}{c} \text{Rs }7000\\ \text{Rs }6125\\ \text{Rs }7875 \end{array}\right]
\displaystyle \text{Hence, the funds collected by schools }A,\ B\text{ and }C\ \text{are Rs }7000,\ \text{Rs }6125\ \text{and Rs }7875\ \text{respectively.}
\displaystyle \text{The total funds collected }=\text{Rs }(7000+6125+7875)=\text{Rs }21000.

\displaystyle \textbf{Question 10: } \text{If }A\text{ is a square matrix such that }A^{2}=I,\ \text{then find the simplified value of} 
\displaystyle (A-I)^{3}+(A+I)^{3}-7A.\ \text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{We have,}\ A^{2}=I
\displaystyle \therefore\ A^{3}=A^{2}A=IA=A \qquad \ldots\text{(i)}
\displaystyle \text{We know that}
\displaystyle (A+B)^{3}=A^{3}+3A^{2}B+3AB^{2}+B^{3}\ \text{and}\ (A-B)^{3}=A^{3}-3A^{2}B+3AB^{2}-B^{3}
\displaystyle \text{provided that }AB=BA.
\displaystyle \text{Since }AI=IA=A.
\displaystyle \therefore\ (A+I)^{3}=A^{3}+3A^{2}I+3AI^{2}+I^{3}\ \text{and}\ (A-I)^{3}=A^{3}-3A^{2}I+3AI^{2}-I^{3}
\displaystyle \Rightarrow\ (A+I)^{3}=A^{3}+3A^{2}+3A+I\ \text{and}\ (A-I)^{3}=A^{3}-3A^{2}+3A-I
\displaystyle \therefore\ (A+I)^{3}+(A-I)^{3}=2(A^{3}+3A)
\displaystyle \Rightarrow\ (A+I)^{3}+(A-I)^{3}=2(A+3A)\qquad \text{[Using (i)]}
\displaystyle \Rightarrow\ (A+I)^{3}+(A-I)^{3}=8A
\displaystyle \text{Hence, }(A-I)^{3}+(A+I)^{3}-7A=8A-7A=A.

\displaystyle \textbf{Question 11: } \text{If }A= \begin{bmatrix}-1\\2\\3\end{bmatrix} \text{ and }B=[-2\ -1\ -4], \\ \text{verify that }(AB)^{T}=B^{T}A^{T}. \hspace{3.0cm} \text{[CBSE 2002, 2005]}
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle A=\begin{bmatrix}-1\\2\\3\end{bmatrix} \ \text{and}\ B=[-2\ -1\ -4]
\displaystyle \therefore\ AB= \begin{bmatrix}-1\\2\\3\end{bmatrix} [-2\ -1\ -4] =\begin{bmatrix} 2&1&4\\ -4&-2&-8\\ -6&-3&-12 \end{bmatrix}
\displaystyle \Rightarrow\ (AB)^{T}= \begin{bmatrix} 2&-4&-6\\ 1&-2&-3\\ 4&-8&-12 \end{bmatrix} \qquad \ldots\text{(i)}
\displaystyle \text{Also,}\ B^{T}A^{T}= [-2\ -1\ -4]^{T} \begin{bmatrix}-1\\2\\3\end{bmatrix}^{T} =\begin{bmatrix}-2\\-1\\-4\end{bmatrix} [-1\ 2\ 3]
\displaystyle \Rightarrow\ B^{T}A^{T}= \begin{bmatrix} 2&-4&-6\\ 1&-2&-3\\ 4&-8&-12 \end{bmatrix} \qquad \ldots\text{(ii)}
\displaystyle \text{From (i) and (ii), we observe that }(AB)^{T}=B^{T}A^{T}.

\displaystyle \textbf{Question 12: } \text{Show that the elements on the main diagonal of a skew-symmetric} \\ \text{matrix are all zero.}\ \text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=[a_{ij}]\text{ be a skew-symmetric matrix. Then,}
\displaystyle a_{ij}=-a_{ji}\ \text{for all }i,j\qquad \text{[By definition]}
\displaystyle \Rightarrow\ a_{ii}=-a_{ii}\ \text{for all values of }i
\displaystyle \Rightarrow\ 2a_{ii}=0
\displaystyle \Rightarrow\ a_{ii}=0\ \text{for all values of }i
\displaystyle \Rightarrow\ a_{11}=a_{22}=a_{33}=\ldots=a_{nn}=0.

\displaystyle \textbf{Question 13. }\text{If }x\begin{bmatrix}1\\2\end{bmatrix}+y\begin{bmatrix}2\\5\end{bmatrix}=\begin{bmatrix}4\\9\end{bmatrix},\text{ then}
\displaystyle \text{(a) }x=1,\ y=2\qquad \text{(b) }x=2,\ y=1
\displaystyle \text{(c) }x=1,\ y=-1\qquad \text{(d) }x=3,\ y=2 \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{(b) We have, }x\begin{bmatrix}1\\2\end{bmatrix}+y\begin{bmatrix}2\\5\end{bmatrix}=\begin{bmatrix}4\\9\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}x\\2x\end{bmatrix}+\begin{bmatrix}2y\\5y\end{bmatrix}=\begin{bmatrix}4\\9\end{bmatrix}\Rightarrow \begin{bmatrix}x+2y\\2x+5y\end{bmatrix}=\begin{bmatrix}4\\9\end{bmatrix}
\displaystyle \text{On equating the corresponding elements, we get}
\displaystyle x+2y=4\qquad ...(i)
\displaystyle 2x+5y=9\qquad ...(ii)
\displaystyle \text{On solving Eqs. }(i)\text{ and }(ii),\text{ we get}
\displaystyle x=2,\ y=1
\\

\displaystyle \textbf{Question 14. }\text{If }A=\begin{bmatrix}3&4\\5&2\end{bmatrix}\text{ and }2A+B\text{ is a null matrix, then }B\text{ is}
\displaystyle \text{equal to}
\displaystyle \text{(a) }\begin{bmatrix}6&8\\10&4\end{bmatrix}\qquad \text{(b) }\begin{bmatrix}-6&-8\\-10&-4\end{bmatrix}
\displaystyle \text{(c) }\begin{bmatrix}5&8\\10&3\end{bmatrix}\qquad \text{(d) }\begin{bmatrix}-5&-8\\-10&-3\end{bmatrix}\hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{(b) Given, }A=\begin{bmatrix}3&4\\5&2\end{bmatrix}\Rightarrow 2A=\begin{bmatrix}6&8\\10&4\end{bmatrix}
\displaystyle \text{Also, given }2A+B\text{ is a null matrix.}
\displaystyle \Rightarrow 2A+B=\begin{bmatrix}0&0\\0&0\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}6&8\\10&4\end{bmatrix}+B=\begin{bmatrix}0&0\\0&0\end{bmatrix}
\displaystyle \Rightarrow B=\begin{bmatrix}0&0\\0&0\end{bmatrix}-\begin{bmatrix}6&8\\10&4\end{bmatrix}=\begin{bmatrix}-6&-8\\-10&-4\end{bmatrix}
\\

\displaystyle \textbf{Question 15. }\text{If }A\text{ is a square matrix and }A^2=A,\text{ then }(I+A)^2-3A
\displaystyle \text{is equal to}
\displaystyle \text{(a) }I\qquad \text{(b) }A\qquad \text{(c) }2A\qquad \text{(d) }3I\hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given, }A^{2}=A
\displaystyle \therefore (I+A)^{2}-3A=I^{2}+IA+AI+A^{2}-3A
\displaystyle =I+A+A+A-3A\qquad [\because A^{2}=A]
\displaystyle =I+3A-3A=I
\\

\displaystyle \textbf{Question 16. }\text{The product }\begin{bmatrix}a&b\\-b&a\end{bmatrix}\begin{bmatrix}a&-b\\b&a\end{bmatrix}\text{ is equal to}
\displaystyle \text{(a) }\begin{bmatrix}a^2+b^2&0\\0&a^2+b^2\end{bmatrix}\qquad \text{(b) }\begin{bmatrix}(a+b)^2&0\\(a+b)^2&0\end{bmatrix}
\displaystyle \text{(c) }\begin{bmatrix}a^2+b^2&0\\a^2+b^2&0\end{bmatrix}\qquad \text{(d) }\begin{bmatrix}a&0\\0&b\end{bmatrix}\hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{(a) }\begin{bmatrix}a&b\\-b&a\end{bmatrix}\begin{bmatrix}a&-b\\b&a\end{bmatrix}
\displaystyle =\begin{bmatrix}a^{2}+b^{2}&-ab+ba\\-ba+ab&b^{2}+a^{2}\end{bmatrix}
\displaystyle =\begin{bmatrix}a^{2}+b^{2}&0\\0&a^{2}+b^{2}\end{bmatrix}
\\

\displaystyle \textbf{Question 17. }A\text{ and }B\text{ are square matrices of same order.}
\displaystyle \text{If }(A+B)^2=A^2+B^2,\text{ then}
\displaystyle \text{(a) }AB=BA\qquad \text{(b) }AB=-BA
\displaystyle \text{(c) }AB=0\qquad \text{(d) }BA=0\hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{(b) We have, }(A+B)^{2}=A^{2}+B^{2}
\displaystyle \Rightarrow (A+B)(A+B)=A^{2}+B^{2}
\displaystyle \Rightarrow A^{2}+AB+BA+B^{2}=A^{2}+B^{2}
\displaystyle \Rightarrow AB+BA=0
\displaystyle \Rightarrow AB=-BA
\\

\displaystyle \textbf{Question 18. }\text{If }A=\begin{bmatrix}1&0\\2&1\end{bmatrix},\ B=\begin{bmatrix}x&0\\1&1\end{bmatrix}\text{ and }A=B^2,\text{ then }x\text{ equals}
\displaystyle \text{(a) }\pm 1\qquad \text{(b) }-1\qquad \text{(c) }1\qquad \text{(d) }2\hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{(c) We have, }B=\begin{bmatrix}x&0\\1&1\end{bmatrix}
\displaystyle \therefore B^{2}=B\cdot B=\begin{bmatrix}x&0\\1&1\end{bmatrix}\begin{bmatrix}x&0\\1&1\end{bmatrix}
\displaystyle =\begin{bmatrix}x^{2}+0&0+0\\x+1&0+1\end{bmatrix}=\begin{bmatrix}x^{2}&0\\x+1&1\end{bmatrix}
\displaystyle \text{Since, }A=B^{2}\Rightarrow \begin{bmatrix}1&0\\2&1\end{bmatrix}=\begin{bmatrix}x^{2}&0\\x+1&1\end{bmatrix}
\displaystyle \text{By equality of matrices,}
\displaystyle x^{2}=1,\ x+1=2
\displaystyle \Rightarrow x=\pm1,\ x=1\Rightarrow x=1
\\

\displaystyle \textbf{Question 19. }\text{If }\begin{bmatrix}2a+b&a-2b\\5c-d&4c+3d\end{bmatrix}=\begin{bmatrix}4&-3\\11&24\end{bmatrix},\text{ then the value of}
\displaystyle a+b-c+2d\text{ is}
\displaystyle \text{(a) }8\qquad \text{(b) }10\qquad \text{(c) }4\qquad \text{(d) }-8\hspace{2.2cm}\text{[CBSE 2022 (Term I)]}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Since, matrices are equal, then the corresponding elements are equal.}
\displaystyle 2a+b=4\qquad ...(i)
\displaystyle a-2b=-3\qquad ...(ii)
\displaystyle 5c-d=11\qquad ...(iii)
\displaystyle 4c+3d=24\qquad ...(iv)
\displaystyle \text{On multiplying }2\text{ in Eq. }(i)\text{ and adding the result with Eq. }(ii),\text{ we get}
\displaystyle 5a=5\Rightarrow a=1
\displaystyle \text{On putting the value }a=1\text{ in Eq. }(i),\text{ we get}
\displaystyle b=2
\displaystyle \text{Now, on multiplying }3\text{ in Eq. }(iii)\text{ and adding the result with Eq. }(iv),\text{ we get}
\displaystyle c=3
\displaystyle \text{On putting the value }c=3\text{ in Eq. }(iii),\text{ we get}
\displaystyle d=4
\displaystyle \therefore a+b-c+2d=1+2-3+8=8
\\

\displaystyle \textbf{Question 20. }\text{A matrix }A=[a_{ij}]_{3\times 3}\text{ is defined by}
\displaystyle a_{ij}=\begin{cases}2i+3j, & i<j\\5, & i=j\\3i-2j, & i>j\end{cases}\text{. The number of elements in }A
\displaystyle \text{which are more than }5,\text{ is}
\displaystyle \text{(a) }3\qquad \text{(b) }4\qquad \text{(c) }5\qquad \text{(d) }6\hspace{2.2cm}\text{[CBSE 2022 (Term I)]}
\displaystyle \text{Answer:}
\displaystyle \text{(b) Given, }A=[a_{ij}]_{3\times3}\text{ and }a_{ij}=\begin{cases}2i+3j,&i<j\\5,&i=j\\3i-2j,&i>j\end{cases}
\displaystyle \text{Let }A=\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}
\displaystyle a_{11}=5,\ a_{12}=2+6=8,\ a_{13}=2+9=11
\displaystyle a_{21}=6-2=4,\ a_{22}=5,\ a_{23}=4+9=13
\displaystyle a_{31}=9-2=7,\ a_{32}=9-4=5,\ a_{33}=5
\displaystyle \therefore A=\begin{bmatrix}5&8&11\\4&5&13\\7&5&5\end{bmatrix}
\\

\displaystyle \textbf{Question 21. }\text{If }A=\begin{bmatrix}0&2\\3&-4\end{bmatrix}\text{ and }kA=\begin{bmatrix}0&3a\\2b&24\end{bmatrix},\text{ then the value of }k,
\displaystyle a\text{ and }b\text{ respectively, are}
\displaystyle \text{(a) }-6,-12,-18\qquad \text{(b) }-6,-4,-9
\displaystyle \text{(c) }-6,4,9\qquad \text{(d) }-6,12,18\hspace{2.2cm}\text{[CBSE 2022 (Term I)]}
\displaystyle \text{Answer:}
\displaystyle \text{(b) We have, }A=\begin{bmatrix}0&2\\3&-4\end{bmatrix}\Rightarrow kA=\begin{bmatrix}0&2k\\3k&-4k\end{bmatrix}
\displaystyle \text{Given, }kA=\begin{bmatrix}0&3a\\2b&24\end{bmatrix}
\displaystyle \Rightarrow 2k=3a,\ 2b=3k\text{ and }-4k=24
\displaystyle \Rightarrow k=-6,\ a=-4\text{ and }b=-9
\\

\displaystyle \textbf{Question 22. }\text{If }A=\begin{bmatrix}2&-3&4\end{bmatrix},\ B=\begin{bmatrix}3\\2\\2\end{bmatrix},\ X=\begin{bmatrix}1&2&3\end{bmatrix}\text{ and }
\displaystyle Y=\begin{bmatrix}2\\3\\4\end{bmatrix}, \text{then }AB+XY\text{ is equal to}
\displaystyle \text{(a) }[28]\qquad \text{(b) }[24]\qquad \text{(c) }28\qquad \text{(d) }24\hspace{2.2cm}\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given, }A=[2\ -3\ 4],\ B=\begin{bmatrix}3\\2\\2\end{bmatrix}
\displaystyle X=[1\ 2\ 3]\text{ and }Y=\begin{bmatrix}2\\3\\4\end{bmatrix}
\displaystyle \text{Now, }AB=[2\ -3\ 4]\begin{bmatrix}3\\2\\2\end{bmatrix}=[6-6+8]=[8]
\displaystyle \text{and }XY=[1\ 2\ 3]\begin{bmatrix}2\\3\\4\end{bmatrix}=[2+6+12]=[20]
\displaystyle \therefore AB+XY=[8]+[20]=[28]
\\

\displaystyle \textbf{Question 23. }\text{If }\begin{bmatrix}x+y&7\\9&x-y\end{bmatrix}=\begin{bmatrix}2&7\\9&4\end{bmatrix},\text{ then find }x\cdot y.   \hspace{2.2cm}\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }\begin{bmatrix}x+y&7\\9&x-y\end{bmatrix}=\begin{bmatrix}2&7\\9&4\end{bmatrix}
\displaystyle \Rightarrow x+y=2\text{ and }x-y=4
\displaystyle \text{Adding both equations, we get}
\displaystyle 2x=6\Rightarrow x=3
\displaystyle \text{Putting }x=3\text{ in }x+y=2,\text{ we get}
\displaystyle y=-1
\\

\displaystyle \textbf{Question 24. }\text{If }A=\begin{bmatrix}1&0\\1&1\end{bmatrix},\text{ then find }A^3.   \hspace{2.2cm}\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }A=\begin{bmatrix}1&0\\1&1\end{bmatrix}
\displaystyle \text{Now, }A^{2}=A\times A=\begin{bmatrix}1&0\\1&1\end{bmatrix}\begin{bmatrix}1&0\\1&1\end{bmatrix}
\displaystyle \Rightarrow A^{2}=\begin{bmatrix}1&0\\2&1\end{bmatrix}
\displaystyle \Rightarrow A^{3}=A^{2}\cdot A=\begin{bmatrix}1&0\\2&1\end{bmatrix}\begin{bmatrix}1&0\\1&1\end{bmatrix}
\displaystyle \Rightarrow A^{3}=\begin{bmatrix}1&0\\3&1\end{bmatrix}
\\

\displaystyle \textbf{Question 25. }\text{If }A+B=\begin{bmatrix}1&0\\1&1\end{bmatrix}\text{ and }A-2B=\begin{bmatrix}-1&1\\0&-1\end{bmatrix},\text{ then find the}
\displaystyle \text{value of }A. \hspace{2.2cm}\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }A+B=\begin{bmatrix}1&0\\1&1\end{bmatrix}
\displaystyle \Rightarrow B=\begin{bmatrix}1&0\\1&1\end{bmatrix}-A
\displaystyle \therefore A-2B=\begin{bmatrix}-1&1\\0&-1\end{bmatrix}
\displaystyle \Rightarrow A-2\left(\begin{bmatrix}1&0\\1&1\end{bmatrix}-A\right)=\begin{bmatrix}-1&1\\0&-1\end{bmatrix}
\displaystyle \Rightarrow 3A=\begin{bmatrix}-1&1\\0&-1\end{bmatrix}+2\begin{bmatrix}1&0\\1&1\end{bmatrix}=\begin{bmatrix}1&1\\2&1\end{bmatrix}
\displaystyle \Rightarrow A=\frac{1}{3}\begin{bmatrix}1&1\\2&1\end{bmatrix}=\begin{bmatrix}\frac13&\frac13\\\frac23&\frac13\end{bmatrix}
\\

\displaystyle \textbf{Question 26. }\text{Find }AB,\text{ if }A=\begin{bmatrix}0&-1\\0&2\end{bmatrix}\text{ and }B=\begin{bmatrix}3&5\\0&0\end{bmatrix}.   \hspace{2.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}0&-1\\0&2\end{bmatrix}\text{ and }B=\begin{bmatrix}3&5\\0&0\end{bmatrix}
\displaystyle \text{Now, }AB=\begin{bmatrix}0&-1\\0&2\end{bmatrix}\begin{bmatrix}3&5\\0&0\end{bmatrix}
\displaystyle =\begin{bmatrix}0+0&0+0\\0+0&0+0\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}
\displaystyle \therefore AB=0
\\

\displaystyle \textbf{Question 27. }\text{If }3A-B=\begin{bmatrix}5&0\\1&1\end{bmatrix}\text{ and }B=\begin{bmatrix}4&3\\2&5\end{bmatrix},\text{ then find the value}
\displaystyle \text{of matrix }A. \hspace{2.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }3A-B=\begin{bmatrix}4&3\\2&5\end{bmatrix}\text{ and }B=\begin{bmatrix}5&0\\1&1\end{bmatrix}
\displaystyle \Rightarrow 3A=\begin{bmatrix}5&0\\1&1\end{bmatrix}+\begin{bmatrix}4&3\\2&5\end{bmatrix}
\displaystyle \Rightarrow 3A=\begin{bmatrix}5+4&0+3\\1+2&1+5\end{bmatrix}
\displaystyle \Rightarrow 3A=\begin{bmatrix}9&3\\3&6\end{bmatrix}
\displaystyle \Rightarrow A=\frac{1}{3}\begin{bmatrix}9&3\\3&6\end{bmatrix}=\begin{bmatrix}3&1\\1&2\end{bmatrix}
\\

\displaystyle \textbf{Question 28. }\text{Find the value of }x-y,\text{ if}
\displaystyle 2\begin{bmatrix}1&3\\0&x\end{bmatrix}+\begin{bmatrix}y&0\\1&2\end{bmatrix}=\begin{bmatrix}5&6\\1&8\end{bmatrix}. \hspace{2.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }2\begin{bmatrix}1&3\\0&x\end{bmatrix}+\begin{bmatrix}y&0\\1&2\end{bmatrix}=\begin{bmatrix}5&6\\1&8\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2&6\\0&2x\end{bmatrix}+\begin{bmatrix}y&0\\1&2\end{bmatrix}=\begin{bmatrix}5&6\\1&8\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2+y&6+0\\0+1&2x+2\end{bmatrix}=\begin{bmatrix}5&6\\1&8\end{bmatrix}
\displaystyle \text{Here, both matrices are equal, so we equate the corresponding elements,}
\displaystyle \Rightarrow 2+y=5\text{ and }2x+2=8
\displaystyle \Rightarrow y=3\text{ and }2x=6\Rightarrow x=3
\displaystyle \therefore x-y=3-3=0
\\

\displaystyle \textbf{Question 29. }\text{If }A\text{ is a square matrix such that }A^2=I,\text{ then find the}
\displaystyle \text{simplified value of }(A-I)^3+(A+I)^3-7A. \hspace{2.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A^{2}=I
\displaystyle \text{Now, }(A-I)^{3}+(A+I)^{3}-7A
\displaystyle =(A^{3}-3A^{2}I+3AI^{2}-I^{3})+(A^{3}+3A^{2}I+3AI^{2}+I^{3})-7A
\displaystyle =A^{3}-3A^{2}+3AI-I+A^{3}+3A^{2}+3AI+I-7A
\displaystyle =2A^{3}+6AI-7A\qquad [\because A^{2}=I\text{ and }I^{3}=I^{2}=I]
\displaystyle =2A^{2}A+6A-7A\qquad [\because AI=A]
\displaystyle =2IA-A\qquad [\because A^{2}=I]
\displaystyle =2A-A=A
\\

\displaystyle \textbf{Question 30. }\text{Write the number of all possible matrices of order }2\times2
\displaystyle \text{with each entry }1,2\text{ or }3. \hspace{2.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{We know that a matrix of order }2\times2\text{ has }4\text{ entries.}
\displaystyle \text{Since, each entry has }3\text{ choices, namely }1,\ 2\text{ or }3,
\displaystyle \text{therefore number of required matrices}
\displaystyle 3^{4}=3\times3\times3\times3=81
\\

\displaystyle \textbf{Question 31. }\text{If }[2\ \ 1\ \ 3]\begin{bmatrix}-1&0&-1\\1&1&0\\0&1&1\end{bmatrix}\begin{bmatrix}1\\0\\-1\end{bmatrix}=A,\text{ then write the order of}
\displaystyle \text{matrix }A. \hspace{2.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }A=[2\ 1\ 3]\begin{bmatrix}-1&0&-1\\-1&1&0\\0&1&1\end{bmatrix}\begin{bmatrix}1\\0\\-1\end{bmatrix}
\displaystyle =[ -2-1\ \ 1+3]\begin{bmatrix}1\\0\\-1\end{bmatrix}=[-3\ \ 4]\begin{bmatrix}1\\0\\-1\end{bmatrix}
\displaystyle =[ -3-4]=[-4]_{1\times1}
\displaystyle \therefore \text{Order of matrix }A\text{ is }1\times1
\\

\displaystyle \textbf{Question 32. }\text{Write the element }a_{23}\text{ of a }3\times3\text{ matrix }A=[a_{ij}],
\displaystyle \text{whose elements }a_{ij}\text{ are given by }a_{ij}=\frac{i-j}{2}. \hspace{2.2cm}\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=[a_{ij}]_{3\times3}
\displaystyle \text{where, }a_{ij}=\frac{|i-j|}{2}
\displaystyle \text{Now, }a_{23}=\frac{|2-3|}{2}\qquad [\text{put }i=2\text{ and }j=3]
\displaystyle =\frac{|-1|}{2}=\frac12
\\

\displaystyle \textbf{Question 33. }\text{If }[2x\ \ 3]\begin{bmatrix}1&2\\-3&0\end{bmatrix}\begin{bmatrix}x\\3\end{bmatrix}=0,\text{ then find }x.  \hspace{2.2cm}\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, matrix equation is}
\displaystyle [2x\ \ 3]\begin{bmatrix}1&2\\-3&0\end{bmatrix}\begin{bmatrix}x\\3\end{bmatrix}=0
\displaystyle \Rightarrow [2x-9\ \ 4x]\begin{bmatrix}x\\3\end{bmatrix}=0
\displaystyle \Rightarrow [2x^{2}-9x+12x]=[0]
\displaystyle \Rightarrow 2x^{2}+3x=0
\displaystyle \Rightarrow x(2x+3)=0
\displaystyle \therefore x=0\text{ or }x=-\frac32
\\

\displaystyle \textbf{Question 34. }\text{If }2\begin{bmatrix}3&4\\5&x\end{bmatrix}+\begin{bmatrix}1&y\\0&1\end{bmatrix}=\begin{bmatrix}7&0\\10&5\end{bmatrix},\text{ then find }(x-y).   \hspace{2.2cm}\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\begin{bmatrix}3&4\\5&x\end{bmatrix}+\begin{bmatrix}1&y\\0&1\end{bmatrix}=\begin{bmatrix}7&0\\10&5\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}6&8\\10&2x\end{bmatrix}+\begin{bmatrix}1&y\\0&1\end{bmatrix}=\begin{bmatrix}7&0\\10&5\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}7&8+y\\10&2x+1\end{bmatrix}=\begin{bmatrix}7&0\\10&5\end{bmatrix}
\displaystyle \text{On equating the corresponding elements, we get}
\displaystyle 8+y=0\text{ and }2x+1=5
\displaystyle \Rightarrow y=-8\text{ and }x=\frac{5-1}{2}=2
\displaystyle \therefore x-y=2-(-8)=10
\\

\displaystyle \textbf{Question 35. }\text{Solve the following matrix equation for }x   [x\ \ 1]\begin{bmatrix}1&0\\-2&0\end{bmatrix}=O   \hspace{2.2cm}\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }[x\ \ 1]\begin{bmatrix}1&0\\-2&0\end{bmatrix}=0
\displaystyle \text{By using matrix multiplication, we get}
\displaystyle [x-2\ \ 0]=[0\ \ 0]
\displaystyle \text{On equating the corresponding elements, we get}
\displaystyle x-2=0\Rightarrow x=2
\\

\displaystyle \textbf{Question 36. }\text{If }A\text{ is a square matrix such that }A^{2}=A,\text{ then write}
\displaystyle \text{the value of }7A-(I+A)^{3},\text{ where }I\text{ is an identity matrix.}   \hspace{2.2cm}\text{[CBSE  2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A^{2}=A
\displaystyle \text{Now, }7A-(I+A)^{3}=7A-[I^{3}+A^{3}+3AI(I+A)]
\displaystyle \qquad [\because (x+y)^{3}=x^{3}+y^{3}+3xy(x+y)]
\displaystyle =7A-[I+A^{2}\cdot A+3A(I+A)]\qquad [\because I^{3}=I\text{ and }IA=A]
\displaystyle =7A-(I+A\cdot A+3AI+3A^{2})\qquad [\because A^{2}=A]
\displaystyle =7A-(I+A+3A+3A)\qquad [\because AI=A\text{ and }A^{2}=A]
\displaystyle =7A-(I+7A)=-I
\\

\displaystyle \textbf{Question 37. }\text{If }\begin{bmatrix}x-y&z\\2x-y&w\end{bmatrix}=\begin{bmatrix}-1&4\\0&5\end{bmatrix},\text{ then find the value of }x+y.
\displaystyle \hspace{2.2cm}\text{[CBSE  2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\begin{bmatrix}x-y&z\\2x-y&w\end{bmatrix}=\begin{bmatrix}-1&4\\0&5\end{bmatrix}
\displaystyle \text{On equating the corresponding elements, we get}
\displaystyle x-y=-1\qquad ...(i)
\displaystyle 2x-y=0\qquad ...(ii)
\displaystyle \text{On solving the Eqs. }(i)\text{ and }(ii),\text{ we get}
\displaystyle x=1\text{ and }y=2
\displaystyle \therefore x+y=1+2=3
\\

\displaystyle \textbf{Question 38. }\text{If }\begin{bmatrix}a+4&3b\\8&-6\end{bmatrix}=\begin{bmatrix}2a+2&b+2\\8&a-8b\end{bmatrix},\text{ then write the value}
\displaystyle \text{of }a-2b.   \hspace{2.2cm}\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\begin{bmatrix}a+4&3b\\8&-6\end{bmatrix}=\begin{bmatrix}2a+2&b+2\\8&a-8b\end{bmatrix}
\displaystyle \text{On equating the corresponding elements, we get}
\displaystyle a+4=2a+2\qquad ...(i)
\displaystyle 3b=b+2\qquad ...(ii)
\displaystyle -6=a-8b\qquad ...(iii)
\displaystyle \text{On solving the Eqs. }(i),\ (ii)\text{ and }(iii),\text{ we get}
\displaystyle a=2\text{ and }b=1
\displaystyle \text{Now, }a-2b=2-2(1)=0
\\

\displaystyle \textbf{Question 39. }\text{If }\begin{bmatrix}xy&4\\z+6&x+y\end{bmatrix}=\begin{bmatrix}8&w\\0&6\end{bmatrix},\text{ then write the value of}
\displaystyle (x+y+z).   \hspace{2.2cm}\text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\begin{bmatrix}x+y&4\\z+6&x+y\end{bmatrix}=\begin{bmatrix}8&w\\0&6\end{bmatrix}
\displaystyle \text{On equating the corresponding elements, we get}
\displaystyle x+y=8\qquad ...(i)
\displaystyle z+6=0\Rightarrow z=-6\qquad ...(ii)
\displaystyle x+y=6\qquad ...(iii)
\displaystyle \text{Now, on adding Eqs. }(ii)\text{ and }(iii),\text{ we get}
\displaystyle x+y+z=6+(-6)=0
\\

\displaystyle \textbf{Question 40. }\text{The elements }a_{ij}\text{ of a }3\times3\text{ matrix are given by}
\displaystyle a_{ij}=\frac{1}{2}|-3i+j|.\text{ Write the value of element }a_{32}. \hspace{2.2cm}\text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle a_{ij}=\frac{1}{2}|-3i+j|
\displaystyle \therefore a_{32}=\frac{1}{2}|-3(3)+2|
\displaystyle =\frac{1}{2}|-9+2|
\displaystyle =\frac{1}{2}|-7|
\displaystyle =\frac{7}{2}
\displaystyle \\

\displaystyle \textbf{Question 41. }\text{If }[2x\ \ 4]\begin{bmatrix}x\\-8\end{bmatrix}=O,\text{ then find the positive value of }x.   \hspace{1.2cm}\text{[CBSE  2014C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }[2x\ \ 4]\begin{bmatrix}x\\-8\end{bmatrix}=0\Rightarrow [2x^{2}-32]=[0]
\displaystyle \text{On equating the corresponding elements, we get}
\displaystyle 2x^{2}-32=0\Rightarrow 2x^{2}=32\Rightarrow x^{2}=16\Rightarrow x=\pm4
\displaystyle \therefore \text{Positive value of }x\text{ is }4
\\

\displaystyle \textbf{Question 42. }\text{If }2\begin{bmatrix}1&3\\0&x\end{bmatrix}+\begin{bmatrix}y&0\\1&2\end{bmatrix}=\begin{bmatrix}5&6\\1&8\end{bmatrix},
\displaystyle \text{then find the value of }(x+y).\hspace{0.2cm}\text{[CBSE 2013C; CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle 2\begin{bmatrix}1&3\\0&x\end{bmatrix}=\begin{bmatrix}2&6\\0&2x\end{bmatrix}
\displaystyle \therefore \begin{bmatrix}2&6\\0&2x\end{bmatrix}+\begin{bmatrix}y&0\\1&2\end{bmatrix}=\begin{bmatrix}5&6\\1&8\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2+y&6\\1&2x+2\end{bmatrix}=\begin{bmatrix}5&6\\1&8\end{bmatrix}
\displaystyle \therefore 2+y=5\text{ and }2x+2=8
\displaystyle \Rightarrow y=3\text{ and }2x=6
\displaystyle \Rightarrow x=3
\displaystyle \therefore x+y=3+3=6
\displaystyle \\

\displaystyle \textbf{Question 43. }\text{Find the value of }a,\text{ if}   \begin{bmatrix}a-b&2a+c\\2a-b&3c+d\end{bmatrix}=\begin{bmatrix}-1&5\\0&13\end{bmatrix}\hspace{0.2cm}\text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle a-b=-1\qquad \text{and}\qquad 2a-b=0
\displaystyle \text{Subtracting the first equation from the second,}
\displaystyle (2a-b)-(a-b)=0-(-1)
\displaystyle \Rightarrow a=1
\displaystyle \\

\displaystyle \textbf{Question 44. }\text{If }\begin{bmatrix}9&-1&4\\-2&1&3\end{bmatrix}=A+\begin{bmatrix}1&2&-1\\0&4&9\end{bmatrix},\text{ then find the matrix }A.
\displaystyle \hspace{2.2cm}\text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, matrix equation can be rewritten as}
\displaystyle A=\begin{bmatrix}9&-1&4\\-2&1&3\end{bmatrix}-\begin{bmatrix}1&2&-1\\0&4&9\end{bmatrix}
\displaystyle \therefore A=\begin{bmatrix}9-1&-1-2&4+1\\-2-0&1-4&3-9\end{bmatrix}=\begin{bmatrix}8&-3&5\\-2&-3&-6\end{bmatrix}
\\

\displaystyle \textbf{Question 45. }\text{If matrix }A=\begin{bmatrix}1&-1\\-1&1\end{bmatrix}\text{ and }A^{2}=kA,\text{ then write the value of }k.
\displaystyle \hspace{2.2cm}\text{[CBSE  2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}1&-1\\-1&1\end{bmatrix}\qquad ...(i)
\displaystyle \text{and }A^{2}=kA\qquad ...(ii)
\displaystyle \text{Now, }A^{2}=A\cdot A
\displaystyle =\begin{bmatrix}1&-1\\-1&1\end{bmatrix}\begin{bmatrix}1&-1\\-1&1\end{bmatrix}
\displaystyle =\begin{bmatrix}1+1&-1-1\\-1-1&1+1\end{bmatrix}\qquad [\text{multiplying row by column}]
\displaystyle =\begin{bmatrix}2&-2\\-2&2\end{bmatrix}=2\begin{bmatrix}1&-1\\-1&1\end{bmatrix}
\displaystyle \Rightarrow A^{2}=2A\qquad [\text{from Eq. }(i)]
\displaystyle \text{On comparing with Eq. }(ii),\text{ we get }k=2
\\

\displaystyle \textbf{Question 46. }\text{If matrix }A=\begin{bmatrix}2&-2\\-2&2\end{bmatrix}\text{ and }A^2=pA,\text{ then write the}
\displaystyle \text{value of }p.\hspace{0.2cm}\text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}2&-2\\-2&2\end{bmatrix}
\displaystyle \therefore A^2=\begin{bmatrix}2&-2\\-2&2\end{bmatrix}\begin{bmatrix}2&-2\\-2&2\end{bmatrix}
\displaystyle =\begin{bmatrix}8&-8\\-8&8\end{bmatrix}
\displaystyle =4\begin{bmatrix}2&-2\\-2&2\end{bmatrix}
\displaystyle =4A
\displaystyle \therefore p=4
\displaystyle \\

\displaystyle \textbf{Question 47. }\text{If matrix }A=\begin{bmatrix}3&-3\\-3&3\end{bmatrix}\text{ and }A^2=\lambda A,\text{ then write the}
\displaystyle \text{value of }\lambda.\hspace{0.2cm}\text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}3&-3\\-3&3\end{bmatrix}
\displaystyle \therefore A^2=\begin{bmatrix}3&-3\\-3&3\end{bmatrix}\begin{bmatrix}3&-3\\-3&3\end{bmatrix}
\displaystyle =\begin{bmatrix}18&-18\\-18&18\end{bmatrix}
\displaystyle =6\begin{bmatrix}3&-3\\-3&3\end{bmatrix}
\displaystyle =6A
\displaystyle \therefore \lambda=6
\displaystyle \\

\displaystyle \textbf{Question 48. }\text{Simplify }\cos\theta\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}+\sin\theta\begin{bmatrix}\sin\theta&-\cos\theta\\\cos\theta&\sin\theta\end{bmatrix}.
\displaystyle \hspace{2.2cm}\text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }\cos\theta\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}+\sin\theta\begin{bmatrix}\sin\theta&-\cos\theta\\\cos\theta&\sin\theta\end{bmatrix}
\displaystyle =\begin{bmatrix}\cos^{2}\theta&\sin\theta\cos\theta\\-\sin\theta\cos\theta&\cos^{2}\theta\end{bmatrix}+\begin{bmatrix}\sin^{2}\theta&-\sin\theta\cos\theta\\\sin\theta\cos\theta&\sin^{2}\theta\end{bmatrix}
\displaystyle =\begin{bmatrix}\cos^{2}\theta+\sin^{2}\theta&\sin\theta\cos\theta-\sin\theta\cos\theta\\-\sin\theta\cos\theta+\sin\theta\cos\theta&\cos^{2}\theta+\sin^{2}\theta\end{bmatrix}
\displaystyle =\begin{bmatrix}1&0\\0&1\end{bmatrix}\qquad [\because \sin^{2}\theta+\cos^{2}\theta=1]
\displaystyle =I=\text{Identity matrix}
\\

\displaystyle \textbf{Question 49. }\text{If }\begin{bmatrix}2&3\\5&7\end{bmatrix}\begin{bmatrix}1&-3\\-2&4\end{bmatrix}=\begin{bmatrix}-4&6\\-9&x\end{bmatrix},\text{ then write the value of }x.
\displaystyle \hspace{2.2cm}\text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Given matrix equation is}
\displaystyle \begin{bmatrix}2&3\\5&7\end{bmatrix}\begin{bmatrix}1&-3\\-2&4\end{bmatrix}=\begin{bmatrix}-4&6\\-9&x\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2-6&-6+12\\5-14&-15+28\end{bmatrix}=\begin{bmatrix}-4&6\\-9&x\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}-4&6\\-9&13\end{bmatrix}=\begin{bmatrix}-4&6\\-9&x\end{bmatrix}
\displaystyle \text{On equating the corresponding elements, we get}
\displaystyle x=13
\\

\displaystyle \textbf{Question 50. }\text{Find the value of }y-x\text{ from the following equation}
\displaystyle 2\begin{bmatrix}x&5\\7&y-3\end{bmatrix}+\begin{bmatrix}3&-4\\1&2\end{bmatrix}=\begin{bmatrix}7&6\\15&14\end{bmatrix}\hspace{0.2cm}\text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle 2\begin{bmatrix}x&5\\7&y-3\end{bmatrix}=\begin{bmatrix}2x&10\\14&2y-6\end{bmatrix}
\displaystyle \therefore \begin{bmatrix}2x&10\\14&2y-6\end{bmatrix}+\begin{bmatrix}3&-4\\1&2\end{bmatrix}=\begin{bmatrix}7&6\\15&14\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2x+3&6\\15&2y-4\end{bmatrix}=\begin{bmatrix}7&6\\15&14\end{bmatrix}
\displaystyle \therefore 2x+3=7\text{ and }2y-4=14
\displaystyle \Rightarrow 2x=4\text{ and }2y=18
\displaystyle \Rightarrow x=2\text{ and }y=9
\displaystyle \therefore y-x=9-2=7
\displaystyle \\

\displaystyle \textbf{Question 51. }\text{If }x\begin{bmatrix}2\\3\end{bmatrix}+y\begin{bmatrix}-1\\1\end{bmatrix}=\begin{bmatrix}10\\5\end{bmatrix},\text{ then write the value of }x.   \hspace{1.2cm}\text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }x\begin{bmatrix}2\\3\end{bmatrix}+y\begin{bmatrix}-1\\1\end{bmatrix}=\begin{bmatrix}10\\5\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2x\\3x\end{bmatrix}+\begin{bmatrix}-y\\y\end{bmatrix}=\begin{bmatrix}10\\5\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2x-y\\3x+y\end{bmatrix}=\begin{bmatrix}10\\5\end{bmatrix}
\displaystyle \text{On equating the corresponding elements, we get}
\displaystyle 2x-y=10\qquad ...(i)
\displaystyle 3x+y=5\qquad ...(ii)
\displaystyle \text{On adding Eqs. }(i)\text{ and }(ii),\text{ we get}
\displaystyle 5x=15
\displaystyle \therefore x=3
\\

\displaystyle \textbf{Question 52. }\text{If }3A-B=\begin{bmatrix}5&0\\1&1\end{bmatrix}\text{ and }B=\begin{bmatrix}4&3\\2&5\end{bmatrix},\text{ then find the matrix }   A.  \hspace{0.2cm}\text{[CBSE 2012C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }B=\begin{bmatrix}4&3\\2&5\end{bmatrix}\text{ and }3A-B=\begin{bmatrix}5&0\\1&1\end{bmatrix}
\displaystyle \Rightarrow 3A=\begin{bmatrix}5&0\\1&1\end{bmatrix}+B=\begin{bmatrix}5&0\\1&1\end{bmatrix}+\begin{bmatrix}4&3\\2&5\end{bmatrix}
\displaystyle =\begin{bmatrix}5+4&0+3\\1+2&1+5\end{bmatrix}=\begin{bmatrix}9&3\\3&6\end{bmatrix}
\displaystyle \Rightarrow 3A=3\begin{bmatrix}3&1\\1&2\end{bmatrix}
\displaystyle \therefore A=\begin{bmatrix}3&1\\1&2\end{bmatrix}
\\

\displaystyle \textbf{Question 53. }\text{Write the value of }x-y+z\text{ from following equation}
\displaystyle \begin{bmatrix}x+y+z\\x+z\\y+z\end{bmatrix}=\begin{bmatrix}9\\5\\7\end{bmatrix}.   \hspace{2.2cm}\text{[CBSE  2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Given matrix equation is}
\displaystyle \begin{bmatrix}x+y+z\\x+z\\y+z\end{bmatrix}=\begin{bmatrix}9\\5\\7\end{bmatrix}
\displaystyle \text{On equating the corresponding elements, we get}
\displaystyle x+y+z=9\qquad ...(i)
\displaystyle x+z=5\qquad ...(ii)
\displaystyle y+z=7\qquad ...(iii)
\displaystyle \text{On putting the value of }x+z\text{ from Eq. }(ii)\text{ in Eq. }(i),\text{ we get}
\displaystyle y+5=9\Rightarrow y=4
\displaystyle \text{On putting }y=4\text{ in Eq. }(iii),\text{ we get}
\displaystyle z=3
\displaystyle \text{Again, putting }z=3\text{ in Eq. }(ii),\text{ we get }x=2
\displaystyle \therefore x-y+z=2-4+3=1
\\

\displaystyle \textbf{Question 54. }\text{Write the order of product matrix }\begin{bmatrix}1\\2\\3\end{bmatrix}[2\ \ 3\ \ 4].   \hspace{2.2cm}\text{[CBSE  2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=\begin{bmatrix}1\\2\\3\end{bmatrix}\text{ and }B=[2\ 3\ 4]
\displaystyle \text{Here, order of matrix }A=3\times1
\displaystyle \text{and order of matrix }B=1\times3
\displaystyle \therefore \text{Order of product matrix }AB=3\times3
\\

\displaystyle \textbf{Question 55. }\text{If a matrix has }5\text{ elements, then write all possible }   \text{orders it can have.}\hspace{0.2cm}\text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Number of elements in a matrix of order }m\times n=mn
\displaystyle \therefore mn=5
\displaystyle \text{Since }5\text{ is prime, its factor pairs are }1\times5\text{ and }5\times1.
\displaystyle \therefore \text{Possible orders are }1\times5\text{ and }5\times1.
\displaystyle \\

\displaystyle \textbf{Question 56. }\text{For a }2\times2\text{ matrix, }A=[a_{ij}]\text{ whose elements are given}
\displaystyle \text{by }a_{ij}=\frac{i}{j},\text{ write the value of }a_{12}.\hspace{0.2cm}\text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle a_{ij}=\frac{i}{j}
\displaystyle \therefore a_{12}=\frac{1}{2}
\displaystyle \\

\displaystyle \textbf{Question 57. }\text{If }\begin{bmatrix}x&x-y\\2x+y&7\end{bmatrix}=\begin{bmatrix}3&1\\8&7\end{bmatrix},\text{ then find the value of }y.  \hspace{1.2cm}\text{[CBSE 2011C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\begin{bmatrix}x&x-y\\2x+y&7\end{bmatrix}=\begin{bmatrix}3&1\\8&7\end{bmatrix}
\displaystyle \text{On equating the corresponding elements, we get}
\displaystyle x=3\text{ and }x-y=1
\displaystyle \Rightarrow y=x-1=3-1=2
\\

\displaystyle \textbf{Question 58. }\text{If }A=\begin{bmatrix}4&2\\-1&1\end{bmatrix},\text{ then show that }   (A-2I)(A-3I)=O.   \hspace{1.2cm}\text{[CBSE  2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}4&2\\-1&1\end{bmatrix}
\displaystyle \text{LHS }=(A-2I)(A-3I)
\displaystyle =\left[\begin{bmatrix}4&2\\-1&1\end{bmatrix}-\begin{bmatrix}2&0\\0&2\end{bmatrix}\right]\left[\begin{bmatrix}4&2\\-1&1\end{bmatrix}-\begin{bmatrix}3&0\\0&3\end{bmatrix}\right]
\displaystyle =\begin{bmatrix}2&2\\-1&-1\end{bmatrix}\begin{bmatrix}1&2\\-1&-2\end{bmatrix}
\displaystyle =\begin{bmatrix}2-2&4-4\\-1+1&-2+2\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}=O=\text{RHS}
\displaystyle \text{Hence proved.}
\\

\displaystyle \textbf{Question 59. }\text{Find a matrix }A\text{ such that }2A-3B+5C=O,\text{ where}
\displaystyle B=\begin{bmatrix}-2&2&0\\3&1&4\end{bmatrix}\text{ and }C=\begin{bmatrix}2&0&-2\\7&1&6\end{bmatrix}.   \hspace{2.2cm}\text{[CBSE  2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }2A-3B+5C=O
\displaystyle \Rightarrow 2A=3B-5C
\displaystyle \Rightarrow 2A=3\begin{bmatrix}-2&2&0\\3&1&4\end{bmatrix}-5\begin{bmatrix}2&0&-2\\7&1&6\end{bmatrix}
\displaystyle \Rightarrow 2A=\begin{bmatrix}-6&6&0\\9&3&12\end{bmatrix}-\begin{bmatrix}10&0&-10\\35&5&30\end{bmatrix}
\displaystyle \Rightarrow 2A=\begin{bmatrix}-6-10&6-0&0-(-10)\\9-35&3-5&12-30\end{bmatrix}
\displaystyle \Rightarrow 2A=\begin{bmatrix}-16&6&10\\-26&-2&-18\end{bmatrix}
\displaystyle \Rightarrow A=\begin{bmatrix}-8&3&5\\-13&-1&-9\end{bmatrix}
\\

\displaystyle \textbf{Question 60. }\text{If }A=\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix},\text{ then find the value of }(A^{2}-5A).   \hspace{1.2cm}\text{[CBSE   2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}
\displaystyle \Rightarrow A^{2}=A\times A=\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}
\displaystyle =\begin{bmatrix}4+0+1&0+0-1&2+0+0\\4+2+3&0+1-3&2+3+0\\2-2+0&0-1+0&1-3-0\end{bmatrix}
\displaystyle =\begin{bmatrix}5&-1&2\\9&-2&5\\0&-1&-2\end{bmatrix}
\displaystyle \text{Now, }A^{2}-5A=\begin{bmatrix}5&-1&2\\9&-2&5\\0&-1&-2\end{bmatrix}-5\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}
\displaystyle =\begin{bmatrix}5&-1&2\\9&-2&5\\0&-1&-2\end{bmatrix}-\begin{bmatrix}10&0&5\\10&5&15\\5&-5&0\end{bmatrix}
\displaystyle =\begin{bmatrix}5-10&-1-0&2-5\\9-10&-2-5&5-15\\0-5&-1+5&-2-0\end{bmatrix}
\displaystyle =\begin{bmatrix}-5&-1&-3\\-1&-7&-10\\-5&4&-2\end{bmatrix}
\\

\displaystyle \textbf{Question 61. }\text{Find matrix }A\text{ such that }   \begin{bmatrix}2&-1\\1&0\\-3&4\end{bmatrix}A=\begin{bmatrix}-1&-8\\1&-2\\9&22\end{bmatrix}.   \hspace{2.2cm}\text{[CBSE  2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the order of }A\text{ is }m\times n.
\displaystyle \therefore m=2,\ n=2
\displaystyle \therefore \text{Order of matrix }A\text{ is }2\times2.
\displaystyle \text{Let }A=\begin{bmatrix}x&y\\s&t\end{bmatrix}
\displaystyle \begin{bmatrix}2&-1\\1&0\\-3&4\end{bmatrix}A=\begin{bmatrix}-1&-8\\1&-2\\9&22\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2&-1\\1&0\\-3&4\end{bmatrix}\begin{bmatrix}x&y\\s&t\end{bmatrix}=\begin{bmatrix}-1&-8\\1&-2\\9&22\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2x-s&2y-t\\x&y\\-3x+4s&-3y+4t\end{bmatrix}=\begin{bmatrix}-1&-8\\1&-2\\9&22\end{bmatrix}
\displaystyle \text{On equating the corresponding elements both sides, we get}
\displaystyle 2x-s=-1,\ x=1,\ y=-2\text{ and }2y-t=-8
\displaystyle \text{At }x=1,\ 2x-s=-1\Rightarrow 2\times1-s=-1
\displaystyle \Rightarrow -s=-3\Rightarrow s=3\text{ and at }y=-2,\ 2y-t=-8
\displaystyle \Rightarrow 2\times(-2)-t=-8
\displaystyle \Rightarrow -4-t=-8\Rightarrow t=4
\displaystyle \text{On putting }x=1,\ y=-2,\ s=3\text{ and }t=4\text{ in Eq. }(i),\text{ we get}
\displaystyle A=\begin{bmatrix}1&-2\\3&4\end{bmatrix}
\\

\displaystyle \textbf{Question 62. }\text{Let }A=\begin{bmatrix}2&-1\\3&4\end{bmatrix},\ B=\begin{bmatrix}5&2\\7&4\end{bmatrix},\ C=\begin{bmatrix}2&5\\3&8\end{bmatrix},\text{ find a matrix}
\displaystyle D\text{ such that }CD-AB=O.   \hspace{2.2cm}\text{[CBSE   2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}2&-1\\3&4\end{bmatrix},\ B=\begin{bmatrix}5&2\\7&4\end{bmatrix},\ C=\begin{bmatrix}2&5\\3&8\end{bmatrix}
\displaystyle \text{Let matrix }D=\begin{bmatrix}x&y\\z&w\end{bmatrix}
\displaystyle \text{According to the question, }CD-AB=O
\displaystyle \Rightarrow \begin{bmatrix}2&5\\3&8\end{bmatrix}\begin{bmatrix}x&y\\z&w\end{bmatrix}-\begin{bmatrix}2&-1\\3&4\end{bmatrix}\begin{bmatrix}5&2\\7&4\end{bmatrix}=O
\displaystyle \Rightarrow \begin{bmatrix}2x+5z&2y+5w\\3x+8z&3y+8w\end{bmatrix}-\begin{bmatrix}10-7&4-4\\15+28&6+16\end{bmatrix}=O
\displaystyle \Rightarrow \begin{bmatrix}2x+5z&2y+5w\\3x+8z&3y+8w\end{bmatrix}=\begin{bmatrix}3&0\\43&22\end{bmatrix}
\displaystyle \text{On equating corresponding elements both sides,}
\displaystyle 2x+5z=3,\ 3x+8z=43
\displaystyle \text{and }2y+5w=0,\ 3y+8w=22
\displaystyle \text{After solving, we get}
\displaystyle x=-191,\ y=-110,\ z=77\text{ and }w=44
\displaystyle \therefore D=\begin{bmatrix}x&y\\z&w\end{bmatrix}=\begin{bmatrix}-191&-110\\77&44\end{bmatrix}
\\

\displaystyle \textbf{Question 63. }\text{If }A=\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix},\text{ then find }A^{2}-5A+4I\text{ and hence}
\displaystyle \text{find a matrix }X\text{ such that }A^{2}-5A+4I+X=O.   \hspace{2.2cm}\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}
\displaystyle \text{Clearly, }A^{2}=\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}
\displaystyle =\begin{bmatrix}4+0+1&0+0-1&2+0+0\\4+2+3&0+1-3&2+3+0\\2-2+0&0-1+0&1-3+0\end{bmatrix}
\displaystyle =\begin{bmatrix}5&-1&2\\9&-2&5\\0&-1&-2\end{bmatrix}\qquad [\text{multiplying row by column}]
\displaystyle \text{Now, consider }A^{2}-5A+4I
\displaystyle =\begin{bmatrix}5&-1&2\\9&-2&5\\0&-1&-2\end{bmatrix}-5\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}+4\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}5&-1&2\\9&-2&5\\0&-1&-2\end{bmatrix}-\begin{bmatrix}10&0&5\\10&5&15\\5&-5&0\end{bmatrix}+\begin{bmatrix}4&0&0\\0&4&0\\0&0&4\end{bmatrix}
\displaystyle =\begin{bmatrix}-5&-1&-3\\-1&-7&-10\\-5&4&-2\end{bmatrix}+\begin{bmatrix}4&0&0\\0&4&0\\0&0&4\end{bmatrix}
\displaystyle =\begin{bmatrix}-1&-1&-3\\-1&-3&-10\\-5&4&2\end{bmatrix}
\displaystyle \text{Given, }A^{2}-5A+4I+X=O
\displaystyle \Rightarrow A^{2}-5A+4I+X-X=O-X\qquad [\text{subtracting matrix }X\text{ from both sides}]
\displaystyle \therefore X=-(A^{2}-5A+4I)
\displaystyle =-\begin{bmatrix}-1&-1&-3\\-1&-3&-10\\-5&4&2\end{bmatrix}
\displaystyle =\begin{bmatrix}1&1&3\\1&3&10\\5&-4&-2\end{bmatrix}
\\

\displaystyle \textbf{Question 64. }\text{If }A=\begin{bmatrix}1&-1\\2&-1\end{bmatrix},\ B=\begin{bmatrix}a&1\\b&-1\end{bmatrix}\text{ and }(A+B)^{2}=A^{2}+B^{2},   \text{then find the value of }a\text{ and }b.   \hspace{2.2cm}\text{[CBSE  2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}1&-1\\2&-1\end{bmatrix}\text{ and }B=\begin{bmatrix}a&1\\b&-1\end{bmatrix}
\displaystyle \text{Now, }A+B=\begin{bmatrix}1&-1\\2&-1\end{bmatrix}+\begin{bmatrix}a&1\\b&-1\end{bmatrix}
\displaystyle =\begin{bmatrix}1+a&0\\2+b&-2\end{bmatrix}
\displaystyle \Rightarrow (A+B)^{2}=\begin{bmatrix}1+a&0\\2+b&-2\end{bmatrix}\begin{bmatrix}1+a&0\\2+b&-2\end{bmatrix}
\displaystyle =\begin{bmatrix}1+a^{2}+2a&0\\2+2a+b+ab-4-2b&4\end{bmatrix}
\displaystyle =\begin{bmatrix}a^{2}+2a+1&0\\2a-ab-b-2&4\end{bmatrix}
\displaystyle \text{Also, }A^{2}+B^{2}=\begin{bmatrix}1&-1\\2&-1\end{bmatrix}\begin{bmatrix}1&-1\\2&-1\end{bmatrix}+\begin{bmatrix}a&1\\b&-1\end{bmatrix}\begin{bmatrix}a&1\\b&-1\end{bmatrix}
\displaystyle =\begin{bmatrix}-1&0\\0&-1\end{bmatrix}+\begin{bmatrix}a^{2}+b&a-1\\ab-b&b+1\end{bmatrix}
\displaystyle =\begin{bmatrix}a^{2}+b-1&a-1\\ab-b&b\end{bmatrix}
\displaystyle \text{Given, }(A+B)^{2}=A^{2}+B^{2}
\displaystyle \Rightarrow \begin{bmatrix}a^{2}+2a+1&0\\2a-ab-b-2&4\end{bmatrix}=\begin{bmatrix}a^{2}+b-1&a-1\\ab-b&b\end{bmatrix}
\displaystyle \text{On equating the corresponding elements, we get}
\displaystyle a^{2}+2a+1=a^{2}+b-1
\displaystyle \Rightarrow 2a-b=-2\qquad ...(i)
\displaystyle a-1=0\qquad ...(ii)
\displaystyle 2a-ab-b-2=ab-b
\displaystyle \Rightarrow 2a-2=0\Rightarrow a=1
\displaystyle \text{and }b=4
\displaystyle \text{Since, }a=1\text{ and }b=4\text{ satisfy Eq. }(i),\text{ therefore }a=1\text{ and }b=4
\\

\displaystyle \textbf{Question 65. }\text{If }f(\alpha)=\begin{bmatrix}\cos\alpha&-\sin\alpha&0\\\sin\alpha&\cos\alpha&0\\0&0&1\end{bmatrix},\text{ prove that}
\displaystyle f(\alpha)\cdot f(-\beta)=f(\alpha-\beta).   \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }f(\alpha)=\begin{bmatrix}\cos\alpha&-\sin\alpha&0\\\sin\alpha&\cos\alpha&0\\0&0&1\end{bmatrix}\qquad ...(i)
\displaystyle \text{For }f(-\beta),\text{ replace }\alpha\text{ in }f(\alpha)\text{ by }-\beta
\displaystyle \therefore f(-\beta)=\begin{bmatrix}\cos(-\beta)&-\sin(-\beta)&0\\\sin(-\beta)&\cos(-\beta)&0\\0&0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}\cos\beta&\sin\beta&0\\-\sin\beta&\cos\beta&0\\0&0&1\end{bmatrix}
\displaystyle \text{By LHS }=f(\alpha)f(-\beta)
\displaystyle =\begin{bmatrix}\cos\alpha&-\sin\alpha&0\\\sin\alpha&\cos\alpha&0\\0&0&1\end{bmatrix}\begin{bmatrix}\cos\beta&\sin\beta&0\\-\sin\beta&\cos\beta&0\\0&0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}\cos\alpha\cos\beta+\sin\alpha\sin\beta&\cos\alpha\sin\beta-\sin\alpha\cos\beta&0\\\sin\alpha\cos\beta-\cos\alpha\sin\beta&\sin\alpha\sin\beta+\cos\alpha\cos\beta&0\\0&0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}\cos(\alpha-\beta)&-\sin(\alpha-\beta)&0\\\sin(\alpha-\beta)&\cos(\alpha-\beta)&0\\0&0&1\end{bmatrix}\qquad ...(ii)
\displaystyle \text{Now, for }f(\alpha-\beta),\text{ replace }\alpha\text{ in }f(\alpha)\text{ by }(\alpha-\beta)
\displaystyle \text{RHS }=f(\alpha-\beta)
\displaystyle =\begin{bmatrix}\cos(\alpha-\beta)&-\sin(\alpha-\beta)&0\\\sin(\alpha-\beta)&\cos(\alpha-\beta)&0\\0&0&1\end{bmatrix}\qquad ...(iii)
\displaystyle \text{From Eqs. }(ii)\text{ and }(iii),\text{ we get}
\displaystyle \text{LHS}=\text{RHS}
\displaystyle \text{Hence proved.}
\\

\displaystyle \textbf{Question 66. }\text{If }A=\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix},\text{ then show that }  \\ A^{3}-6A^{2}+7A+2I=O.   \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}
\displaystyle \therefore A^{2}=A\cdot A=\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}
\displaystyle =\begin{bmatrix}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{bmatrix}
\displaystyle =\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}
\displaystyle \text{and }A^{3}=A\cdot A^{2}=\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}
\displaystyle =\begin{bmatrix}5+0+16&0+0+0&8+0+26\\0+4+8&0+8+0&0+10+13\\10+0+24&0+0+0&16+0+39\end{bmatrix}
\displaystyle =\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}
\displaystyle \text{Now, }A^{3}-6A^{2}+7A+2I
\displaystyle =\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}-6\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}+7\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}+2\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}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{bmatrix}
\displaystyle =\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=O
\displaystyle \text{Hence proved.}
\\

\displaystyle \textbf{Question 67. }\text{If }A=\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\text{ and }A^{3}-6A^{2}+7A+kI_{3}=O,\text{ then}
\displaystyle \text{find the value of }k.   \hspace{2.2cm}\text{[CBSE 2023;  2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}
\displaystyle \text{Now, }A^{2}=\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}
\displaystyle =\begin{bmatrix}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{bmatrix}
\displaystyle =\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}
\displaystyle \text{and }A^{3}=A\cdot A^{2}=\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}
\displaystyle =\begin{bmatrix}5+0+16&0+0+0&8+0+26\\0+4+8&0+8+0&0+10+13\\10+0+24&0+0+0&16+0+39\end{bmatrix}
\displaystyle =\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}
\displaystyle \text{Also, given }A^{3}-6A^{2}+7A+kI_{3}=O
\displaystyle \Rightarrow \begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}-6\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}+7\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}+k\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=O
\displaystyle \Rightarrow \begin{bmatrix}21-30+7+k&0&34-48+14\\12-12+0&8-24+14+k&23-30+7\\34-48+14&0&55-78+21+k\end{bmatrix}=O
\displaystyle \Rightarrow \begin{bmatrix}k-2&0&0\\0&k-2&0\\0&0&k-2\end{bmatrix}=O
\displaystyle \Rightarrow k-2=0
\displaystyle \therefore k=2
\displaystyle \begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}-6\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}+7\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}
\displaystyle +k\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}-\begin{bmatrix}30&0&48\\12&24&30\\48&0&78\end{bmatrix}
\displaystyle +\begin{bmatrix}7&0&14\\0&14&7\\14&0&21\end{bmatrix}+\begin{bmatrix}k&0&0\\0&k&0\\0&0&k\end{bmatrix}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}21-30+7+k&0-0+0+0&34-48+14+0\\12-12+0+0&8-24+14+k&23-30+7+0\\34-48+14+0&0-0+0+0&55-78+21+k\end{bmatrix}
\displaystyle =\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}-2+k&0&0\\0&-2+k&0\\0&0&-2+k\end{bmatrix}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}
\displaystyle \text{On equating the corresponding elements, we get}
\displaystyle -2+k=0
\displaystyle \therefore k=2
\\

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