\displaystyle \textbf{Question 1: }\text{Find }x\text{ and }y,\text{ if} \begin{bmatrix}3&-2\\-1&4\end{bmatrix}\begin{bmatrix}2x\\1\end{bmatrix}+2\begin{bmatrix}-4\\5\end{bmatrix}=4\begin{bmatrix}2\\y\end{bmatrix}.\hfill \text{[ICSE 2003]}
\displaystyle \text{Answer:}
\displaystyle \begin{bmatrix}3&-2\\-1&4\end{bmatrix}\begin{bmatrix}2x\\1\end{bmatrix}+2\begin{bmatrix}-4\\5\end{bmatrix}=4\begin{bmatrix}2\\y\end{bmatrix}
\displaystyle \begin{bmatrix}6x-2\\-2x+4\end{bmatrix}+\begin{bmatrix}-8\\10\end{bmatrix}=\begin{bmatrix}8\\4y\end{bmatrix}
\displaystyle \begin{bmatrix}6x-10\\-2x+14\end{bmatrix}=\begin{bmatrix}8\\4y\end{bmatrix}
\displaystyle \therefore 6x-10=8\text{ and }-2x+14=4y
\displaystyle 6x=18
\displaystyle x=3
\displaystyle -2(3)+14=4y
\displaystyle 8=4y
\displaystyle y=2
\displaystyle \therefore x=3\text{ and }y=2.
\\

\displaystyle \textbf{Question 2: }\text{Given }\begin{bmatrix}2&1\\-3&4\end{bmatrix}X=\begin{bmatrix}7\\6\end{bmatrix}
\displaystyle \text{Find: (i) the order of matrix }X\qquad\text{(ii) the matrix }X.\hfill \text{[ICSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \begin{bmatrix}2&1\\-3&4\end{bmatrix}X=\begin{bmatrix}7\\6\end{bmatrix}
\displaystyle \text{Let the order of }X\text{ be }p\times q.
\displaystyle (2\times2)\times(p\times q)=(2\times1)
\displaystyle \therefore p=2\text{ and }q=1
\displaystyle \therefore \text{The order of }X\text{ is }2\times1.
\displaystyle \text{Let }X=\begin{bmatrix}x\\y\end{bmatrix}
\displaystyle \begin{bmatrix}2&1\\-3&4\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}7\\6\end{bmatrix}
\displaystyle \begin{bmatrix}2x+y\\-3x+4y\end{bmatrix}=\begin{bmatrix}7\\6\end{bmatrix}
\displaystyle \therefore 2x+y=7\text{ and }-3x+4y=6
\displaystyle \text{Solving, we get }x=2\text{ and }y=3
\displaystyle \therefore X=\begin{bmatrix}2\\3\end{bmatrix}.
\\

\displaystyle \textbf{Question 3: }\text{If }\begin{bmatrix}a&3\\4&1\end{bmatrix}+\begin{bmatrix}2&b\\1&-2\end{bmatrix}-\begin{bmatrix}1&1\\-2&c\end{bmatrix}=\begin{bmatrix}5&0\\7&3\end{bmatrix}
\displaystyle \text{Find the value of }a,\ b\text{ and }c.\hfill \text{[ICSE 1981]}
\displaystyle \text{Answer:}
\displaystyle \begin{bmatrix}a&3\\4&1\end{bmatrix}+\begin{bmatrix}2&b\\1&-2\end{bmatrix}-\begin{bmatrix}1&1\\-2&c\end{bmatrix}=\begin{bmatrix}5&0\\7&3\end{bmatrix}
\displaystyle \begin{bmatrix}a+2-1&3+b-1\\4+1+2&1-2-c\end{bmatrix}=\begin{bmatrix}5&0\\7&3\end{bmatrix}
\displaystyle \begin{bmatrix}a+1&b+2\\7&-c-1\end{bmatrix}=\begin{bmatrix}5&0\\7&3\end{bmatrix}
\displaystyle \therefore a+1=5,\quad b+2=0,\quad -c-1=3
\displaystyle a=4,\quad b=-2,\quad c=-4
\displaystyle \therefore a=4,\ b=-2\text{ and }c=-4.
\\

\displaystyle \textbf{Question 4: }\text{If }A=\begin{bmatrix}1&2\\2&1\end{bmatrix}\text{ and }B=\begin{bmatrix}2&1\\1&2\end{bmatrix}
\displaystyle \text{Find: (i) }A(BA)\qquad\text{(ii) }(AB)B.\hfill \text{[ICSE 1991]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) }A(BA)=\begin{bmatrix}1&2\\2&1\end{bmatrix}\left(\begin{bmatrix}2&1\\1&2\end{bmatrix}\begin{bmatrix}1&2\\2&1\end{bmatrix}\right)
\displaystyle =\begin{bmatrix}1&2\\2&1\end{bmatrix}\begin{bmatrix}4&5\\5&4\end{bmatrix}
\displaystyle =\begin{bmatrix}14&13\\13&14\end{bmatrix}
\displaystyle \text{(ii) }(AB)B=\left(\begin{bmatrix}1&2\\2&1\end{bmatrix}\begin{bmatrix}2&1\\1&2\end{bmatrix}\right)\begin{bmatrix}2&1\\1&2\end{bmatrix}
\displaystyle =\begin{bmatrix}4&5\\5&4\end{bmatrix}\begin{bmatrix}2&1\\1&2\end{bmatrix}
\displaystyle =\begin{bmatrix}13&14\\14&13\end{bmatrix}
\\

\displaystyle \textbf{Question 5: }\text{Find }x\text{ and }y,\text{ if }\begin{bmatrix}x&3x\\y&4y\end{bmatrix}\begin{bmatrix}2\\1\end{bmatrix}=\begin{bmatrix}5\\12\end{bmatrix}.\hfill \text{[ICSE 1992, 2013]}
\displaystyle \text{Answer:}
\displaystyle \begin{bmatrix}x&3x\\y&4y\end{bmatrix}\begin{bmatrix}2\\1\end{bmatrix}=\begin{bmatrix}5\\12\end{bmatrix}
\displaystyle \begin{bmatrix}2x+3x\\2y+4y\end{bmatrix}=\begin{bmatrix}5\\12\end{bmatrix}
\displaystyle \therefore 5x=5\text{ and }6y=12
\displaystyle x=1\text{ and }y=2
\displaystyle \therefore x=1\text{ and }y=2.
\\

\displaystyle \textbf{Question 6: }\text{Given }A=\begin{bmatrix}2&-1\\2&0\end{bmatrix},\ B=\begin{bmatrix}-3&2\\4&0\end{bmatrix}\text{ and }C=\begin{bmatrix}1&0\\0&2\end{bmatrix}
\displaystyle \text{Find }X\text{ such that }A+X=2B+C.\hfill \text{[ICSE 2005]}
\displaystyle \text{Answer:}
\displaystyle A+X=2B+C
\displaystyle \begin{bmatrix}2&-1\\2&0\end{bmatrix}+X=2\begin{bmatrix}-3&2\\4&0\end{bmatrix}+\begin{bmatrix}1&0\\0&2\end{bmatrix}
\displaystyle \begin{bmatrix}2&-1\\2&0\end{bmatrix}+X=\begin{bmatrix}-5&4\\8&2\end{bmatrix}
\displaystyle X=\begin{bmatrix}-5&4\\8&2\end{bmatrix}-\begin{bmatrix}2&-1\\2&0\end{bmatrix}
\displaystyle X=\begin{bmatrix}-7&5\\6&2\end{bmatrix}
\displaystyle \therefore X=\begin{bmatrix}-7&5\\6&2\end{bmatrix}.
\\

\displaystyle \textbf{Question 7: }\text{Find the value of }x\text{ given that }A^2=B,\ A=\begin{bmatrix}2&12\\0&1\end{bmatrix} \\ \text{ and }B=\begin{bmatrix}4&x\\0&1\end{bmatrix}.\hfill \text{[ICSE 2005]}
\displaystyle \text{Answer:}
\displaystyle A^2=B
\displaystyle \begin{bmatrix}2&12\\0&1\end{bmatrix}\begin{bmatrix}2&12\\0&1\end{bmatrix}=\begin{bmatrix}4&x\\0&1\end{bmatrix}
\displaystyle \begin{bmatrix}4&36\\0&1\end{bmatrix}=\begin{bmatrix}4&x\\0&1\end{bmatrix}
\displaystyle \therefore x=36
\\

\displaystyle \textbf{Question 8: }\text{If }A=\begin{bmatrix}2&5\\1&3\end{bmatrix},\ B=\begin{bmatrix}4&-2\\-1&3\end{bmatrix}\text{ and }I\text{ is the identity matrix}
\displaystyle \text{of the same order and }A^t\text{ is the transpose of }A,\text{ find }A^tB+BI.\hfill \text{[ICSE 2011]}
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}2&5\\1&3\end{bmatrix}
\displaystyle A^t=\begin{bmatrix}2&1\\5&3\end{bmatrix}
\displaystyle A^tB+BI
\displaystyle =\begin{bmatrix}2&1\\5&3\end{bmatrix}\begin{bmatrix}4&-2\\-1&3\end{bmatrix}+\begin{bmatrix}4&-2\\-1&3\end{bmatrix}\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}7&-1\\17&-1\end{bmatrix}+\begin{bmatrix}4&-2\\-1&3\end{bmatrix}
\displaystyle =\begin{bmatrix}11&-3\\16&2\end{bmatrix}
\displaystyle \therefore A^tB+BI=\begin{bmatrix}11&-3\\16&2\end{bmatrix}.
\\

\displaystyle \textbf{Question 9: }\text{Given }A=\begin{bmatrix}2&-6\\2&0\end{bmatrix},\ B=\begin{bmatrix}-3&2\\4&0\end{bmatrix}\text{ and }C=\begin{bmatrix}4&0\\0&2\end{bmatrix}
\displaystyle \text{Find the matrix }X\text{ such that }A+2X=2B+C.\hfill \text{[ICSE 2013]}
\displaystyle \text{Answer:}
\displaystyle A+2X=2B+C
\displaystyle \begin{bmatrix}2&-6\\2&0\end{bmatrix}+2X=2\begin{bmatrix}-3&2\\4&0\end{bmatrix}+\begin{bmatrix}4&0\\0&2\end{bmatrix}
\displaystyle \begin{bmatrix}2&-6\\2&0\end{bmatrix}+2X=\begin{bmatrix}-6&4\\8&0\end{bmatrix}+\begin{bmatrix}4&0\\0&2\end{bmatrix}
\displaystyle \begin{bmatrix}2&-6\\2&0\end{bmatrix}+2X=\begin{bmatrix}-2&4\\8&2\end{bmatrix}
\displaystyle 2X=\begin{bmatrix}-2&4\\8&2\end{bmatrix}-\begin{bmatrix}2&-6\\2&0\end{bmatrix}
\displaystyle 2X=\begin{bmatrix}-4&10\\6&2\end{bmatrix}
\displaystyle X=\frac{1}{2}\begin{bmatrix}-4&10\\6&2\end{bmatrix}
\displaystyle X=\begin{bmatrix}-2&5\\3&1\end{bmatrix}
\\

\displaystyle \textbf{Question 10: }\text{Let }A=\begin{bmatrix}4&-2\\6&-3\end{bmatrix},\ B=\begin{bmatrix}0&2\\1&-1\end{bmatrix}\text{ and }C=\begin{bmatrix}-2&3\\1&-1\end{bmatrix}
\displaystyle \text{Find }A^2-A+BC.\hfill \text{[ICSE 2006]}
\displaystyle \text{Answer:}
\displaystyle A^2=\begin{bmatrix}4&-2\\6&-3\end{bmatrix}\begin{bmatrix}4&-2\\6&-3\end{bmatrix}
\displaystyle =\begin{bmatrix}16-12&-8+6\\24-18&-12+9\end{bmatrix}
\displaystyle =\begin{bmatrix}4&-2\\6&-3\end{bmatrix}=A
\displaystyle BC=\begin{bmatrix}0&2\\1&-1\end{bmatrix}\begin{bmatrix}-2&3\\1&-1\end{bmatrix}
\displaystyle =\begin{bmatrix}2&-2\\-3&4\end{bmatrix}
\displaystyle \therefore A^2-A+BC=A-A+BC
\displaystyle =\begin{bmatrix}2&-2\\-3&4\end{bmatrix}
\\

\displaystyle \textbf{Question 11: }\text{Let }A=\begin{bmatrix}1&0\\2&1\end{bmatrix},\ B=\begin{bmatrix}2&3\\-1&0\end{bmatrix}
\displaystyle \text{Find }A^2+AB+B^2.\hfill \text{[ICSE 2007]}
\displaystyle \text{Answer:}
\displaystyle A^2+AB+B^2
\displaystyle =\begin{bmatrix}1&0\\2&1\end{bmatrix}\begin{bmatrix}1&0\\2&1\end{bmatrix}+\begin{bmatrix}1&0\\2&1\end{bmatrix}\begin{bmatrix}2&3\\-1&0\end{bmatrix}+\begin{bmatrix}2&3\\-1&0\end{bmatrix}\begin{bmatrix}2&3\\-1&0\end{bmatrix}
\displaystyle =\begin{bmatrix}1&0\\4&1\end{bmatrix}+\begin{bmatrix}2&3\\3&6\end{bmatrix}+\begin{bmatrix}1&6\\-2&-3\end{bmatrix}
\displaystyle =\begin{bmatrix}4&9\\5&4\end{bmatrix}
\displaystyle \therefore A^2+AB+B^2=\begin{bmatrix}4&9\\5&4\end{bmatrix}.
\\

\displaystyle \textbf{Question 12: }\text{Given }A=\begin{bmatrix}p&0\\0&2\end{bmatrix},\ B=\begin{bmatrix}0&-q\\1&0\end{bmatrix}\text{ and }C=\begin{bmatrix}2&-2\\2&2\end{bmatrix}
\displaystyle \text{and }BA=C^2,\text{ find the value of }p\text{ and }q.\hfill \text{[ICSE 2008]}
\displaystyle \text{Answer:}
\displaystyle BA=C^2
\displaystyle \begin{bmatrix}0&-q\\1&0\end{bmatrix}\begin{bmatrix}p&0\\0&2\end{bmatrix}=\begin{bmatrix}2&-2\\2&2\end{bmatrix}\begin{bmatrix}2&-2\\2&2\end{bmatrix}
\displaystyle \begin{bmatrix}0&-2q\\p&0\end{bmatrix}=\begin{bmatrix}0&-8\\8&0\end{bmatrix}
\displaystyle \therefore -2q=-8\text{ and }p=8
\displaystyle q=4\text{ and }p=8
\displaystyle \therefore p=8\text{ and }q=4.
\\

\displaystyle \textbf{Question 13: } \text{Given } A = \begin{bmatrix} 3 & -2 \\ -1 & 4 \end{bmatrix} \text{, } B = \begin{bmatrix} 6 \\ 1 \end{bmatrix} \text{, } C = \begin{bmatrix} -4 \\ 5 \end{bmatrix} \text{ and } D = \begin{bmatrix} 2 \\ 2 \end{bmatrix}
\displaystyle \text{Find } AB+2C-4D \hspace{1.0cm} \text{[ICSE 2010] }
\displaystyle \text{Answer:}
\displaystyle AB+2C-4D
\displaystyle = \begin{bmatrix} 3 & -2 \\ -1 & 4 \end{bmatrix}. \begin{bmatrix} 6 \\ 1 \end{bmatrix} + 2 \begin{bmatrix} -4 \\ 5 \end{bmatrix} - 4 \begin{bmatrix} 2 \\ 2 \end{bmatrix}
\displaystyle = \begin{bmatrix} 16 \\ -2 \end{bmatrix}+\begin{bmatrix} -8 \\ 10 \end{bmatrix}- \begin{bmatrix} 8 \\ 8 \end{bmatrix}
\displaystyle = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

\displaystyle \textbf{Question 14: } \text{Evaluate } \begin{bmatrix} 4\sin{30^o} & 2\cos{60^o} \\ \sin{90^o} & 2\cos{0^o} \end{bmatrix} . \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix} \hspace{1.0cm} \text{[ICSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \begin{bmatrix} 4\sin{30^o} & 2\cos{60^o} \\ \sin{90^o} & 2\cos{0^o} \end{bmatrix} . \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix}
\displaystyle = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}. \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix}
\displaystyle = \begin{bmatrix} 13 & 14 \\ 14 & 13 \end{bmatrix}

\displaystyle \textbf{Question 15: } \text{If } A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \text{, } I= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
\displaystyle \text{Find } A^2-5A+7I \hspace{1.0cm} \text{[ICSE 2012]}
\displaystyle \text{Answer:}
\displaystyle A^2-5A+7I
\displaystyle = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}.\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}-5\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}+7\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
\displaystyle = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}-\begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}+\begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}
\displaystyle = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}

\displaystyle \textbf{Question 16: } \text{If } A = \begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix}, B = \begin{bmatrix} 4 & 1 \\ -3 & -2 \end{bmatrix} \text{ and } C = \begin{bmatrix} -3 & 2 \\ -1 & 4 \end{bmatrix}
\displaystyle \text{Find } A^2 + AC-5B \hspace{1.0cm} \text{[ICSE 2014]}
\displaystyle \text{Answer:}
\displaystyle A^2 + AC-5B
\displaystyle = \begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix} + \begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix} \begin{bmatrix} -3 & 2 \\ -1 & 4 \end{bmatrix} - 5 \begin{bmatrix} 4 & 1 \\ -3 & -2 \end{bmatrix}
\displaystyle = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} + \begin{bmatrix} -7 & 8 \\ 2 & -8 \end{bmatrix} - \begin{bmatrix} 20 & 5 \\ -15 & -10 \end{bmatrix} = \begin{bmatrix} -23 & 3 \\ 17 & 14 \end{bmatrix}

\displaystyle \textbf{Question 17: }\text{Solve for }x\text{ and }y
\displaystyle \text{(i) }\begin{bmatrix}2&5\\5&2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-7\\14\end{bmatrix}
\displaystyle \text{(ii) }\begin{bmatrix}x+y&x-4\end{bmatrix}\begin{bmatrix}-1&-2\\2&2\end{bmatrix}=\begin{bmatrix}-7&14\end{bmatrix}
\displaystyle \text{(iii) }\begin{bmatrix}-2&0\\3&1\end{bmatrix}\begin{bmatrix}-1\\2x\end{bmatrix}+3\begin{bmatrix}-2\\1\end{bmatrix}=2\begin{bmatrix}y\\3\end{bmatrix}\hfill \text{[ICSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) }\begin{bmatrix}2&5\\5&2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-7\\14\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2x+5y\\5x+2y\end{bmatrix}=\begin{bmatrix}-7\\14\end{bmatrix}
\displaystyle \therefore 2x+5y=-7\qquad\cdots(1)
\displaystyle 5x+2y=14\qquad\cdots(2)
\displaystyle \text{Solving }(1)\text{ and }(2),\text{ we get }x=4\text{ and }y=-3.
\displaystyle \text{(ii) }\begin{bmatrix}x+y&x-4\end{bmatrix}\begin{bmatrix}-1&-2\\2&2\end{bmatrix}=\begin{bmatrix}-7&14\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}-(x+y)+2(x-4)&-2(x+y)+2(x-4)\end{bmatrix}=\begin{bmatrix}-7&14\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}x-y-8&-2y-8\end{bmatrix}=\begin{bmatrix}-7&14\end{bmatrix}
\displaystyle \therefore x-y-8=-7\qquad\cdots(3)
\displaystyle -2y-8=14\qquad\cdots(4)
\displaystyle \Rightarrow x-y=1\qquad\cdots(5)
\displaystyle -2y=22
\displaystyle y=-11
\displaystyle \text{Substituting }y=-11\text{ in }(5),
\displaystyle x+11=1
\displaystyle x=-10
\displaystyle \therefore x=-10\text{ and }y=-11.
\displaystyle \text{(iii) }\begin{bmatrix}-2&0\\3&1\end{bmatrix}\begin{bmatrix}-1\\2x\end{bmatrix}+3\begin{bmatrix}-2\\1\end{bmatrix}=2\begin{bmatrix}y\\3\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2\\-3+2x\end{bmatrix}+\begin{bmatrix}-6\\3\end{bmatrix}=\begin{bmatrix}2y\\6\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}-4\\2x\end{bmatrix}=\begin{bmatrix}2y\\6\end{bmatrix}
\displaystyle \therefore 2y=-4\text{ and }2x=6
\displaystyle y=-2\text{ and }x=3
\\

\displaystyle \textbf{Question 18: }\text{If }A=\begin{bmatrix}3&5\\4&-2\end{bmatrix}\text{ and }B=\begin{bmatrix}2\\4\end{bmatrix},\text{ is the product }AB\text{ possible?}\hfill \text{[ICSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{The order of }A\text{ is }2\times2\text{ and the order of }B\text{ is }2\times1.
\displaystyle \text{Since the number of columns of }A\text{ is equal to the number of rows of }B,\text{ the product }AB\text{ is possible.}
\displaystyle AB=\begin{bmatrix}3&5\\4&-2\end{bmatrix}\begin{bmatrix}2\\4\end{bmatrix}
\displaystyle =\begin{bmatrix}3(2)+5(4)\\4(2)-2(4)\end{bmatrix}
\displaystyle =\begin{bmatrix}26\\0\end{bmatrix}
\displaystyle \therefore AB=\begin{bmatrix}26\\0\end{bmatrix}.
\\

\displaystyle \textbf{Question 19: }\text{Find }x\text{ and }y\text{ if }\begin{bmatrix}-3&2\\0&-5\end{bmatrix}\begin{bmatrix}x\\2\end{bmatrix}=\begin{bmatrix}-5\\y\end{bmatrix}.\hfill \text{[ICSE 2001]}
\displaystyle \text{Answer:}
\displaystyle \begin{bmatrix}-3&2\\0&-5\end{bmatrix}\begin{bmatrix}x\\2\end{bmatrix}=\begin{bmatrix}-5\\y\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}-3x+2(2)\\0(x)-5(2)\end{bmatrix}=\begin{bmatrix}-5\\y\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}-3x+4\\-10\end{bmatrix}=\begin{bmatrix}-5\\y\end{bmatrix}
\displaystyle \text{Equating corresponding elements,}
\displaystyle -3x+4=-5\text{ and }y=-10
\displaystyle -3x=-9
\displaystyle x=3
\displaystyle \therefore x=3\text{ and }y=-10.
\\

\displaystyle \textbf{Question 20: }\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\text{ and }y.\hfill \text{[ICSE 2007]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }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}
\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{Equating the corresponding elements,}
\displaystyle 8+y=0\text{ and }2x+1=5
\displaystyle y=-8\text{ and }2x=4
\displaystyle x=2
\displaystyle \therefore x=2\text{ and }y=-8.
\\

\displaystyle \textbf{Question 21: }\text{If }\begin{bmatrix}1&4\\-2&3\end{bmatrix}+2M=3\begin{bmatrix}3&2\\0&-3\end{bmatrix},\text{ then find matrix }M.\hfill \text{[ICSE 2008]}
\displaystyle \text{Answer:}
\displaystyle \begin{bmatrix}1&4\\-2&3\end{bmatrix}+2M=3\begin{bmatrix}3&2\\0&-3\end{bmatrix}
\displaystyle \therefore 2M=3\begin{bmatrix}3&2\\0&-3\end{bmatrix}-\begin{bmatrix}1&4\\-2&3\end{bmatrix}
\displaystyle =\begin{bmatrix}9&6\\0&-9\end{bmatrix}-\begin{bmatrix}1&4\\-2&3\end{bmatrix}
\displaystyle =\begin{bmatrix}8&2\\2&-12\end{bmatrix}
\displaystyle \therefore M=\frac{1}{2}\begin{bmatrix}8&2\\2&-12\end{bmatrix}
\displaystyle =\begin{bmatrix}4&1\\1&-6\end{bmatrix}
\displaystyle \therefore M=\begin{bmatrix}4&1\\1&-6\end{bmatrix}.
\\

\displaystyle \textbf{Question 22: }\text{Given }A=\begin{bmatrix}p&0\\0&2\end{bmatrix},\ B=\begin{bmatrix}0&-q\\1&0\end{bmatrix},\ C=\begin{bmatrix}2&-2\\2&2\end{bmatrix}\text{ and } \\ BA=C^2,  \text{ find the values of }p\text{ and }q.\ \text{[ICSE 2008]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}p&0\\0&2\end{bmatrix},\ B=\begin{bmatrix}0&-q\\1&0\end{bmatrix}\text{ and }C=\begin{bmatrix}2&-2\\2&2\end{bmatrix}
\displaystyle \text{Also, given }BA=C^2\qquad\cdots\text{(i)}
\displaystyle \text{Here, }BA=\begin{bmatrix}0&-q\\1&0\end{bmatrix}\begin{bmatrix}p&0\\0&2\end{bmatrix}
\displaystyle =\begin{bmatrix}0\times p+(-q)\times0&0\times0+(-q)\times2\\1\times p+0\times0&1\times0+0\times2\end{bmatrix}\qquad[\text{multiplying row by column}]
\displaystyle =\begin{bmatrix}0&-2q\\p&0\end{bmatrix}
\displaystyle \text{and }C^2=\begin{bmatrix}2&-2\\2&2\end{bmatrix}\begin{bmatrix}2&-2\\2&2\end{bmatrix}
\displaystyle =\begin{bmatrix}2\times2+(-2)\times2&2\times(-2)+(-2)\times2\\2\times2+2\times2&2\times(-2)+2\times2\end{bmatrix}\qquad[\text{multiplying row by column}]
\displaystyle =\begin{bmatrix}4-4&-4-4\\4+4&-4+4\end{bmatrix}=\begin{bmatrix}0&-8\\8&0\end{bmatrix}
\displaystyle \text{According to the question, }BA=C^2
\displaystyle \Rightarrow \begin{bmatrix}0&-2q\\p&0\end{bmatrix}=\begin{bmatrix}0&-8\\8&0\end{bmatrix}
\displaystyle \text{As above two matrices are equal, their corresponding elements are also equal.}
\displaystyle -2q=-8\Rightarrow q=\frac{-8}{-2}\Rightarrow q=4
\displaystyle \text{and }p=8
\displaystyle \text{Hence, the values of }p\text{ and }q\text{ are }8\text{ and }4,\text{ respectively.}

\displaystyle \textbf{Question 23: }\text{Find }x\text{ and }y,\text{ if }\begin{bmatrix}2x&x\\y&3y\end{bmatrix}\begin{bmatrix}3\\2\end{bmatrix}=\begin{bmatrix}16\\9\end{bmatrix}.\ \text{[ICSE 2009]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\begin{bmatrix}2x&x\\y&3y\end{bmatrix}\begin{bmatrix}3\\2\end{bmatrix}=\begin{bmatrix}16\\9\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}6x+2x\\3y+6y\end{bmatrix}=\begin{bmatrix}16\\9\end{bmatrix}\Rightarrow \begin{bmatrix}8x\\9y\end{bmatrix}=\begin{bmatrix}16\\9\end{bmatrix}
\displaystyle \text{As above two matrices are equal, their corresponding elements are also equal.}
\displaystyle 8x=16\Rightarrow x=\frac{16}{8}=2\text{ and }9y=9\Rightarrow y=\frac{9}{9}=1
\displaystyle \therefore x=2\text{ and }y=1

\displaystyle \textbf{Question 24: }\text{Given }A=\begin{bmatrix}3&-2\\-1&4\end{bmatrix},\ B=\begin{bmatrix}6\\1\end{bmatrix},\ C=\begin{bmatrix}-4\\5\end{bmatrix}\text{ and }D=\begin{bmatrix}2\\2\end{bmatrix}.
\displaystyle \text{Find }AB+2C-4D.\hfill \text{[ICSE 2010]}
\displaystyle \text{Answer:}
\displaystyle AB=\begin{bmatrix}3&-2\\-1&4\end{bmatrix}\begin{bmatrix}6\\1\end{bmatrix}=\begin{bmatrix}3\times6-2\times1\\-1\times6+4\times1\end{bmatrix}
\displaystyle =\begin{bmatrix}18-2\\-6+4\end{bmatrix}=\begin{bmatrix}16\\-2\end{bmatrix}
\displaystyle AB+2C-4D=\begin{bmatrix}16\\-2\end{bmatrix}+2\begin{bmatrix}-4\\5\end{bmatrix}-4\begin{bmatrix}2\\2\end{bmatrix}
\displaystyle =\begin{bmatrix}16\\-2\end{bmatrix}+\begin{bmatrix}-8\\10\end{bmatrix}-\begin{bmatrix}8\\8\end{bmatrix}
\displaystyle =\begin{bmatrix}16-8-8\\-2+10-8\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}
\displaystyle \therefore AB+2C-4D=\begin{bmatrix}0\\0\end{bmatrix}.
\\

\displaystyle \textbf{Question 25: }\text{Evaluate }\begin{bmatrix}4\sin30^\circ&2\cos60^\circ\\\sin90^\circ&2\cos0^\circ\end{bmatrix}\begin{bmatrix}4&5\\5&4\end{bmatrix}.\hfill \text{[ICSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \begin{bmatrix}4\sin30^\circ&2\cos60^\circ\\\sin90^\circ&2\cos0^\circ\end{bmatrix}\begin{bmatrix}4&5\\5&4\end{bmatrix}
\displaystyle =\begin{bmatrix}4\times\frac{1}{2}&2\times\frac{1}{2}\\1&2\times1\end{bmatrix}\begin{bmatrix}4&5\\5&4\end{bmatrix}
\displaystyle =\begin{bmatrix}2&1\\1&2\end{bmatrix}\begin{bmatrix}4&5\\5&4\end{bmatrix}
\displaystyle =\begin{bmatrix}2\times4+1\times5&2\times5+1\times4\\1\times4+2\times5&1\times5+2\times4\end{bmatrix}
\displaystyle =\begin{bmatrix}8+5&10+4\\4+10&5+8\end{bmatrix}=\begin{bmatrix}13&14\\14&13\end{bmatrix}
\displaystyle \therefore \text{The required matrix is }\begin{bmatrix}13&14\\14&13\end{bmatrix}.
\\

\displaystyle \textbf{Question 26: }\text{If }A=\begin{bmatrix}3&5\\4&-2\end{bmatrix}\text{ and }B=\begin{bmatrix}2\\4\end{bmatrix},\text{ is the product }AB\text{ possible?}
\displaystyle \text{Give a reason. If yes, find it.}\hfill \text{[ICSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{The order of }A\text{ is }2\times2\text{ and the order of }B\text{ is }2\times1.
\displaystyle \text{Since the number of columns of }A\text{ is equal to the number of rows of }B,
\displaystyle \text{the product }AB\text{ is possible.}
\displaystyle AB=\begin{bmatrix}3&5\\4&-2\end{bmatrix}\begin{bmatrix}2\\4\end{bmatrix}
\displaystyle =\begin{bmatrix}3\times2+5\times4\\4\times2+(-2)\times4\end{bmatrix}
\displaystyle =\begin{bmatrix}6+20\\8-8\end{bmatrix}=\begin{bmatrix}26\\0\end{bmatrix}
\displaystyle \therefore AB=\begin{bmatrix}26\\0\end{bmatrix}.
\\

\displaystyle \textbf{Question 27: }\text{If }A=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\text{ and }I=\begin{bmatrix}1&0\\0&1\end{bmatrix},\text{ find }A^2-5A+7I.\hfill \text{[ICSE 2012]}
\displaystyle \text{Answer:}
\displaystyle A^2=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}
\displaystyle =\begin{bmatrix}3\times3+1\times(-1)&3\times1+1\times2\\(-1)\times3+2\times(-1)&(-1)\times1+2\times2\end{bmatrix}
\displaystyle =\begin{bmatrix}9-1&3+2\\-3-2&-1+4\end{bmatrix}=\begin{bmatrix}8&5\\-5&3\end{bmatrix}
\displaystyle \therefore A^2-5A+7I=\begin{bmatrix}8&5\\-5&3\end{bmatrix}-5\begin{bmatrix}3&1\\-1&2\end{bmatrix}+7\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}8&5\\-5&3\end{bmatrix}+\begin{bmatrix}-15&-5\\5&-10\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}
\displaystyle =\begin{bmatrix}8-15+7&5-5+0\\-5+5+0&3-10+7\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}
\displaystyle \therefore A^2-5A+7I=\begin{bmatrix}0&0\\0&0\end{bmatrix}.
\\

\displaystyle \textbf{Question 28: }\text{Given }\begin{bmatrix}2&1\\-3&4\end{bmatrix}X=\begin{bmatrix}7\\6\end{bmatrix},\text{ write (i) the order of }X
\displaystyle \text{(ii) the matrix }X.\hfill \text{[ICSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given, }\begin{bmatrix}2&1\\-3&4\end{bmatrix}X=\begin{bmatrix}7\\6\end{bmatrix}\qquad\cdots\text{(A)}
\displaystyle \text{The matrix on the right side is of order }2\times1.
\displaystyle \text{Hence the product on the left side must also be of order }2\times1.
\displaystyle \text{Since }\begin{bmatrix}2&1\\-3&4\end{bmatrix}\text{ is of order }2\times2,\text{ the order of }X\text{ is }2\times1.
\displaystyle \text{(ii) Let }X=\begin{bmatrix}x_1\\x_2\end{bmatrix}
\displaystyle \text{From Eq. (A), }\begin{bmatrix}2&1\\-3&4\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}7\\6\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2x_1+x_2\\-3x_1+4x_2\end{bmatrix}=\begin{bmatrix}7\\6\end{bmatrix}
\displaystyle \Rightarrow \begin{cases}2x_1+x_2=7\qquad\cdots\text{(B)}\\-3x_1+4x_2=6\qquad\cdots\text{(C)}\end{cases}
\displaystyle \text{Multiplying Eq. (B) by }4,\text{ we get }8x_1+4x_2=28
\displaystyle \text{Subtracting Eq. (C),}
\displaystyle 11x_1=22
\displaystyle x_1=2
\displaystyle \text{Substituting }x_1=2\text{ in Eq. (B),}
\displaystyle 2(2)+x_2=7
\displaystyle x_2=3
\displaystyle \therefore X=\begin{bmatrix}2\\3\end{bmatrix}.
\\

\displaystyle \textbf{Question 29: }\text{Given }A=\begin{bmatrix}2&-6\\2&0\end{bmatrix},\ B=\begin{bmatrix}-3&2\\4&0\end{bmatrix},\ C=\begin{bmatrix}4&0\\0&2\end{bmatrix}.
\displaystyle \text{Find matrix }X\text{ such that }A+2X=2B+C.\hfill \text{[ICSE 2013]}
\displaystyle \text{Answer:}
\displaystyle A+2X=2B+C
\displaystyle \therefore 2X=2\begin{bmatrix}-3&2\\4&0\end{bmatrix}+\begin{bmatrix}4&0\\0&2\end{bmatrix}-\begin{bmatrix}2&-6\\2&0\end{bmatrix}
\displaystyle =\begin{bmatrix}-6&4\\8&0\end{bmatrix}+\begin{bmatrix}4&0\\0&2\end{bmatrix}-\begin{bmatrix}2&-6\\2&0\end{bmatrix}
\displaystyle =\begin{bmatrix}-4&10\\6&2\end{bmatrix}
\displaystyle X=\frac{1}{2}\begin{bmatrix}-4&10\\6&2\end{bmatrix}
\displaystyle =\begin{bmatrix}-2&5\\3&1\end{bmatrix}
\displaystyle \therefore X=\begin{bmatrix}-2&5\\3&1\end{bmatrix}.
\\

\displaystyle \textbf{Question 30: }\text{Find }x\text{ and }y,\text{ if }\begin{bmatrix}x&3x\\y&4y\end{bmatrix}\begin{bmatrix}2\\1\end{bmatrix}=\begin{bmatrix}5\\12\end{bmatrix}.\hfill \text{[ICSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \begin{bmatrix}x&3x\\y&4y\end{bmatrix}\begin{bmatrix}2\\1\end{bmatrix}=\begin{bmatrix}5\\12\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}x\times2+3x\times1\\y\times2+4y\times1\end{bmatrix}=\begin{bmatrix}5\\12\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2x+3x\\2y+4y\end{bmatrix}=\begin{bmatrix}5\\12\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}5x\\6y\end{bmatrix}=\begin{bmatrix}5\\12\end{bmatrix}
\displaystyle \text{Equating the corresponding elements,}
\displaystyle 5x=5\text{ and }6y=12
\displaystyle x=1\text{ and }y=2
\displaystyle \therefore x=1\text{ and }y=2.
\\

\displaystyle \textbf{Question 31: }\text{Find }x\text{ and }y,\text{ if }\begin{bmatrix}-2&0\\3&1\end{bmatrix}\begin{bmatrix}-1\\2x\end{bmatrix}+3\begin{bmatrix}-2\\1\end{bmatrix}=2\begin{bmatrix}y\\3\end{bmatrix}.\hfill \text{[ICSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \begin{bmatrix}-2&0\\3&1\end{bmatrix}\begin{bmatrix}-1\\2x\end{bmatrix}+3\begin{bmatrix}-2\\1\end{bmatrix}=2\begin{bmatrix}y\\3\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}-2(-1)+0(2x)\\3(-1)+1(2x)\end{bmatrix}+\begin{bmatrix}-6\\3\end{bmatrix}=\begin{bmatrix}2y\\6\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2\\-3+2x\end{bmatrix}+\begin{bmatrix}-6\\3\end{bmatrix}=\begin{bmatrix}2y\\6\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}-4\\2x\end{bmatrix}=\begin{bmatrix}2y\\6\end{bmatrix}
\displaystyle \text{Equating the corresponding elements,}
\displaystyle -4=2y\text{ and }2x=6
\displaystyle y=-2\text{ and }x=3
\displaystyle \therefore x=3\text{ and }y=-2.
\\

\displaystyle \textbf{Question 32: }\text{If }A=\begin{bmatrix}2&1\\0&-2\end{bmatrix},\ B=\begin{bmatrix}4&1\\-3&-2\end{bmatrix}\text{ and }C=\begin{bmatrix}-3&2\\-1&4\end{bmatrix},
\displaystyle \text{ then find }A^2+AC-5B.\ \text{[ICSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}2&1\\0&-2\end{bmatrix},\ B=\begin{bmatrix}4&1\\-3&-2\end{bmatrix}\text{ and }C=\begin{bmatrix}-3&2\\-1&4\end{bmatrix}
\displaystyle \text{We have to find }A^2+AC-5B
\displaystyle \text{Here, }A^2=\begin{bmatrix}2&1\\0&-2\end{bmatrix}\begin{bmatrix}2&1\\0&-2\end{bmatrix}
\displaystyle =\begin{bmatrix}2\times2+1\times0&2\times1+1\times(-2)\\0\times2+(-2)\times0&0\times1+(-2)\times(-2)\end{bmatrix}\qquad[\text{multiplying row by column}]
\displaystyle =\begin{bmatrix}4+0&2-2\\0-0&0+4\end{bmatrix}=\begin{bmatrix}4&0\\0&4\end{bmatrix}\qquad\cdots\text{(i)}
\displaystyle AC=\begin{bmatrix}2&1\\0&-2\end{bmatrix}\begin{bmatrix}-3&2\\-1&4\end{bmatrix}
\displaystyle =\begin{bmatrix}2\times(-3)+1\times(-1)&2\times2+1\times4\\0\times(-3)+(-2)\times(-1)&0\times2+(-2)\times4\end{bmatrix}\qquad[\text{multiplying row by column}]
\displaystyle =\begin{bmatrix}-6-1&4+4\\0+2&0-8\end{bmatrix}=\begin{bmatrix}-7&8\\2&-8\end{bmatrix}\qquad\cdots\text{(ii)}
\displaystyle 5B=5\begin{bmatrix}4&1\\-3&-2\end{bmatrix}=\begin{bmatrix}5\times4&5\times1\\5\times(-3)&5\times(-2)\end{bmatrix}\qquad[\text{multiplying each element of matrix by }5]
\displaystyle =\begin{bmatrix}20&5\\-15&-10\end{bmatrix}\qquad\cdots\text{(iii)}
\displaystyle \text{Now, }A^2+AC-5B=\begin{bmatrix}4&0\\0&4\end{bmatrix}+\begin{bmatrix}-7&8\\2&-8\end{bmatrix}-\begin{bmatrix}20&5\\-15&-10\end{bmatrix}
\displaystyle =\begin{bmatrix}4-7-20&0+8-5\\0+2-(-15)&4-8-(-10)\end{bmatrix}=\begin{bmatrix}-23&3\\17&6\end{bmatrix}

\displaystyle \textbf{Question 33: }\text{If }A=\begin{bmatrix}3&x\\0&1\end{bmatrix}\text{ and }B=\begin{bmatrix}9&16\\0&-y\end{bmatrix},\text{ find }x\text{ and }y\text{ when } \\ A^2=B.\ \text{[ICSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}3&x\\0&1\end{bmatrix}\text{ and }B=\begin{bmatrix}9&16\\0&-y\end{bmatrix}
\displaystyle \text{Also, it is given }A^2=B\qquad\cdots\text{(i)}
\displaystyle \text{Here, }A^2=A\cdot A=\begin{bmatrix}3&x\\0&1\end{bmatrix}\begin{bmatrix}3&x\\0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}9+0&3x+x\\0+0&0+1\end{bmatrix}=\begin{bmatrix}9&4x\\0&1\end{bmatrix}
\displaystyle \text{From Eq. (i), }A^2=B
\displaystyle \Rightarrow \begin{bmatrix}9&4x\\0&1\end{bmatrix}=\begin{bmatrix}9&16\\0&-y\end{bmatrix}
\displaystyle \text{By comparing the corresponding elements, we get }4x=16\text{ and }1=-y
\displaystyle \Rightarrow x=\frac{16}{4}\text{ and }y=-1
\displaystyle \therefore x=4\text{ and }y=-1

\displaystyle \textbf{Question 34: }\text{If }A=\begin{bmatrix}3&7\\2&4\end{bmatrix},\ B=\begin{bmatrix}0&2\\5&3\end{bmatrix}\text{ and }C=\begin{bmatrix}1&-5\\-4&6\end{bmatrix},\text{ find}
\displaystyle AB-5C.\hfill \text{[ICSE 2015]}
\displaystyle \text{Answer:}
\displaystyle AB=\begin{bmatrix}3&7\\2&4\end{bmatrix}\begin{bmatrix}0&2\\5&3\end{bmatrix}
\displaystyle =\begin{bmatrix}3\times0+7\times5&3\times2+7\times3\\2\times0+4\times5&2\times2+4\times3\end{bmatrix}
\displaystyle =\begin{bmatrix}35&27\\20&16\end{bmatrix}
\displaystyle 5C=5\begin{bmatrix}1&-5\\-4&6\end{bmatrix}=\begin{bmatrix}5&-25\\-20&30\end{bmatrix}
\displaystyle \therefore AB-5C=\begin{bmatrix}35&27\\20&16\end{bmatrix}-\begin{bmatrix}5&-25\\-20&30\end{bmatrix}
\displaystyle =\begin{bmatrix}30&52\\40&-14\end{bmatrix}
\displaystyle \therefore AB-5C=\begin{bmatrix}30&52\\40&-14\end{bmatrix}.
\\

\displaystyle \textbf{Question 35: }\text{Given }A=\begin{bmatrix}4\sin30^\circ&\cos0^\circ\\\cos0^\circ&4\sin30^\circ\end{bmatrix}\text{ and }B=\begin{bmatrix}4\\5\end{bmatrix}.
\displaystyle \text{If }AX=B,\text{ then (i) write the order of matrix }X\text{ (ii) find matrix }X.\ \text{[ICSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given, }A=\begin{bmatrix}4\sin30^\circ&\cos0^\circ\\\cos0^\circ&4\sin30^\circ\end{bmatrix}_{2\times2}
\displaystyle \text{and }B=\begin{bmatrix}4\\5\end{bmatrix}_{2\times1}
\displaystyle \text{It is given that, }AX=B
\displaystyle \text{Since, product of matrices }A\text{ and }X\text{ are possible,}
\displaystyle \text{Number of columns of }A=\text{Number of rows of }X
\displaystyle \Rightarrow \text{Rows in matrix }X\text{ is }2
\displaystyle \text{Since, }AX\text{ results a matrix }B\text{ which have order }2\times1
\displaystyle \text{Matrix }X\text{ will have same number of columns as matrix }B
\displaystyle \Rightarrow \text{Number of columns in matrix }X=1
\displaystyle \therefore \text{Order of matrix }X\text{ is }2\times1
\displaystyle \text{(ii) We have, }A=\begin{bmatrix}4\sin30^\circ&\cos0^\circ\\\cos0^\circ&4\sin30^\circ\end{bmatrix}
\displaystyle =\begin{bmatrix}2&1\\1&2\end{bmatrix}\qquad\left[\because\ \sin30^\circ=\frac{1}{2}\text{ and }\cos0^\circ=1\right]
\displaystyle \text{Let }X=\begin{bmatrix}x_1\\x_2\end{bmatrix}_{2\times1}
\displaystyle \text{Now, }AX=B
\displaystyle \Rightarrow \begin{bmatrix}2&1\\1&2\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}4\\5\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2x_1+x_2\\x_1+2x_2\end{bmatrix}=\begin{bmatrix}4\\5\end{bmatrix}
\displaystyle \text{On equating the corresponding elements both sides, we get}
\displaystyle 2x_1+x_2=4\qquad\cdots\text{(i)}
\displaystyle x_1+2x_2=5\qquad\cdots\text{(ii)}
\displaystyle \text{On solving Eqs. (i) and (ii), we get }x_1=1\text{ and }x_2=2
\displaystyle \therefore X=\begin{bmatrix}1\\2\end{bmatrix}

\displaystyle \textbf{Question 36: }\text{Given }A=\begin{bmatrix}2&0\\-1&7\end{bmatrix}\text{ and }I=\begin{bmatrix}1&0\\0&1\end{bmatrix}\text{ and }A^2=9A+mI,
\displaystyle \text{find }m.\hfill \text{[ICSE 2016]}
\displaystyle \text{Answer:}
\displaystyle A^2=A\cdot A=\begin{bmatrix}2&0\\-1&7\end{bmatrix}\begin{bmatrix}2&0\\-1&7\end{bmatrix}
\displaystyle =\begin{bmatrix}2\times2+0\times(-1)&2\times0+0\times7\\(-1)\times2+7\times(-1)&(-1)\times0+7\times7\end{bmatrix}
\displaystyle =\begin{bmatrix}4&0\\-9&49\end{bmatrix}
\displaystyle \text{Since }A^2=9A+mI,
\displaystyle \begin{bmatrix}4&0\\-9&49\end{bmatrix}=9\begin{bmatrix}2&0\\-1&7\end{bmatrix}+m\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}4&0\\-9&49\end{bmatrix}=\begin{bmatrix}18&0\\-9&63\end{bmatrix}+\begin{bmatrix}m&0\\0&m\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}4&0\\-9&49\end{bmatrix}=\begin{bmatrix}18+m&0\\-9&63+m\end{bmatrix}
\displaystyle \text{Equating the corresponding elements,}
\displaystyle 18+m=4\text{ and }63+m=49
\displaystyle m=4-18=-14
\displaystyle \therefore m=-14.
\\

\displaystyle \textbf{Question 37: }\text{If }A=\begin{bmatrix}1&3\\3&4\end{bmatrix}\text{ and }B=\begin{bmatrix}-2&1\\-3&2\end{bmatrix}\text{ and }A^2-5B^2=5C,
\displaystyle \text{find matrix }C,\text{ where }C\text{ is }2\times2.\hfill \text{[ICSE 2017]}
\displaystyle \text{Answer:}
\displaystyle A^2-5B^2=5C
\displaystyle \therefore C=\frac{1}{5}A^2-B^2
\displaystyle A^2=\begin{bmatrix}1&3\\3&4\end{bmatrix}\begin{bmatrix}1&3\\3&4\end{bmatrix}
\displaystyle =\begin{bmatrix}1+9&3+12\\3+12&9+16\end{bmatrix}=\begin{bmatrix}10&15\\15&25\end{bmatrix}
\displaystyle \frac{1}{5}A^2=\begin{bmatrix}2&3\\3&5\end{bmatrix}
\displaystyle B^2=\begin{bmatrix}-2&1\\-3&2\end{bmatrix}\begin{bmatrix}-2&1\\-3&2\end{bmatrix}
\displaystyle =\begin{bmatrix}4-3&-2+2\\6-6&-3+4\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle \therefore C=\frac{1}{5}A^2-B^2
\displaystyle =\begin{bmatrix}2&3\\3&5\end{bmatrix}-\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}1&3\\3&4\end{bmatrix}
\displaystyle \therefore C=\begin{bmatrix}1&3\\3&4\end{bmatrix}.
\\

\displaystyle \textbf{Question 38: }\text{Given }B=\begin{bmatrix}1&1\\8&3\end{bmatrix},\text{ find matrix }X\text{ if }X=B^2-4B.
\displaystyle \text{Hence, solve for }a\text{ and }b\text{ given }X\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}5\\50\end{bmatrix}.\hfill \text{[ICSE 2017]}
\displaystyle \text{Answer:}
\displaystyle B^2=\begin{bmatrix}1&1\\8&3\end{bmatrix}\begin{bmatrix}1&1\\8&3\end{bmatrix}
\displaystyle =\begin{bmatrix}1+8&1+3\\8+24&8+9\end{bmatrix}=\begin{bmatrix}9&4\\32&17\end{bmatrix}
\displaystyle 4B=4\begin{bmatrix}1&1\\8&3\end{bmatrix}=\begin{bmatrix}4&4\\32&12\end{bmatrix}
\displaystyle \therefore X=B^2-4B
\displaystyle =\begin{bmatrix}9&4\\32&17\end{bmatrix}-\begin{bmatrix}4&4\\32&12\end{bmatrix}
\displaystyle =\begin{bmatrix}5&0\\0&5\end{bmatrix}
\displaystyle \text{Given, }X\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}5\\50\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}5&0\\0&5\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}5\\50\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}5a\\5b\end{bmatrix}=\begin{bmatrix}5\\50\end{bmatrix}
\displaystyle \therefore 5a=5\text{ and }5b=50
\displaystyle a=1\text{ and }b=10
\displaystyle \therefore X=\begin{bmatrix}5&0\\0&5\end{bmatrix},\ a=1\text{ and }b=10.
\\

\displaystyle \textbf{Question 39: }\text{If }A=\begin{bmatrix}4&4\\-2&6\end{bmatrix}\text{ and }B=\begin{bmatrix}2&1\\3&-2\end{bmatrix},\text{ find matrix }D\text{ such that}
\displaystyle 3A-2B+2D=0.\hfill \text{[ICSE 2017]}
\displaystyle \text{Answer:}
\displaystyle 3A-2B+2D=0
\displaystyle \Rightarrow 2D=2B-3A
\displaystyle \Rightarrow D=B-\frac{3}{2}A
\displaystyle =\begin{bmatrix}2&1\\3&-2\end{bmatrix}-\frac{3}{2}\begin{bmatrix}4&4\\-2&6\end{bmatrix}
\displaystyle =\begin{bmatrix}2&1\\3&-2\end{bmatrix}+\begin{bmatrix}-6&-6\\3&-9\end{bmatrix}
\displaystyle =\begin{bmatrix}-4&-5\\6&-11\end{bmatrix}
\displaystyle \therefore D=\begin{bmatrix}-4&-5\\6&-11\end{bmatrix}.
\\

\displaystyle \textbf{Question 40: }\text{If }\begin{bmatrix}2&4\\6&2\end{bmatrix}\begin{bmatrix}3x\\2\end{bmatrix}+2\begin{bmatrix}3\\4\end{bmatrix}=5\begin{bmatrix}4\\y\end{bmatrix},\text{ find }x\text{ and }y.\hfill \text{[ICSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \begin{bmatrix}2&4\\6&2\end{bmatrix}\begin{bmatrix}3x\\2\end{bmatrix}+2\begin{bmatrix}3\\4\end{bmatrix}=5\begin{bmatrix}4\\y\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}6x+8\\18x+4\end{bmatrix}+\begin{bmatrix}6\\8\end{bmatrix}=\begin{bmatrix}20\\5y\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}6x+14\\18x+12\end{bmatrix}=\begin{bmatrix}20\\5y\end{bmatrix}
\displaystyle \text{Equating the corresponding elements,}
\displaystyle 6x+14=20\qquad\cdots\text{(1)}
\displaystyle 18x+12=5y\qquad\cdots\text{(2)}
\displaystyle \text{From Eq. (1), }6x=6
\displaystyle x=1
\displaystyle \text{Substituting }x=1\text{ in Eq. (2),}
\displaystyle 18(1)+12=5y
\displaystyle 5y=30
\displaystyle y=6
\displaystyle \therefore x=1\text{ and }y=6.
\\

\displaystyle \textbf{Question 41: }\text{Find the values of }x\text{ and }y,\text{ if}
\displaystyle 2\begin{bmatrix}x&7\\9&y-5\end{bmatrix}+\begin{bmatrix}6&-7\\4&5\end{bmatrix}=\begin{bmatrix}10&7\\22&15\end{bmatrix}.\hfill \text{[ICSE 2018]}
\displaystyle \text{Answer:}
\displaystyle 2\begin{bmatrix}x&7\\9&y-5\end{bmatrix}+\begin{bmatrix}6&-7\\4&5\end{bmatrix}=\begin{bmatrix}10&7\\22&15\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2x&14\\18&2y-10\end{bmatrix}+\begin{bmatrix}6&-7\\4&5\end{bmatrix}=\begin{bmatrix}10&7\\22&15\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2x+6&7\\22&2y-5\end{bmatrix}=\begin{bmatrix}10&7\\22&15\end{bmatrix}
\displaystyle \text{Equating the corresponding elements,}
\displaystyle 2x+6=10\text{ and }2y-5=15
\displaystyle 2x=4\Rightarrow x=2
\displaystyle 2y=20\Rightarrow y=10
\displaystyle \therefore x=2\text{ and }y=10.
\\

\displaystyle \textbf{Question 42: }\text{If }A=\begin{bmatrix}2&3\\5&7\end{bmatrix},\ B=\begin{bmatrix}0&4\\-1&7\end{bmatrix},\ C=\begin{bmatrix}1&0\\-1&4\end{bmatrix},\text{ find}
\displaystyle AC+B^2-10C.\hfill \text{[ICSE 2018]}
\displaystyle \text{Answer:}
\displaystyle AC=\begin{bmatrix}2&3\\5&7\end{bmatrix}\begin{bmatrix}1&0\\-1&4\end{bmatrix}=\begin{bmatrix}2-3&0+12\\5-7&0+28\end{bmatrix}=\begin{bmatrix}-1&12\\-2&28\end{bmatrix}
\displaystyle B^2=\begin{bmatrix}0&4\\-1&7\end{bmatrix}\begin{bmatrix}0&4\\-1&7\end{bmatrix}=\begin{bmatrix}-4&28\\-7&45\end{bmatrix}
\displaystyle AC+B^2-10C=\begin{bmatrix}-1&12\\-2&28\end{bmatrix}+\begin{bmatrix}-4&28\\-7&45\end{bmatrix}-10\begin{bmatrix}1&0\\-1&4\end{bmatrix}
\displaystyle =\begin{bmatrix}-5&40\\-9&73\end{bmatrix}-\begin{bmatrix}10&0\\-10&40\end{bmatrix}
\displaystyle =\begin{bmatrix}-15&40\\1&33\end{bmatrix}
\displaystyle \therefore AC+B^2-10C=\begin{bmatrix}-15&40\\1&33\end{bmatrix}.
\\

\displaystyle \textbf{Question 43: }\text{Simplify }\sin A\begin{bmatrix}\sin A&-\cos A\\\cos A&\sin A\end{bmatrix}+\cos A\begin{bmatrix}\cos A&\sin A\\-\sin A&\cos A\end{bmatrix}.\hfill \text{[ICSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \sin A\begin{bmatrix}\sin A&-\cos A\\\cos A&\sin A\end{bmatrix}+\cos A\begin{bmatrix}\cos A&\sin A\\-\sin A&\cos A\end{bmatrix}
\displaystyle =\begin{bmatrix}\sin^2A&-\sin A\cos A\\\sin A\cos A&\sin^2A\end{bmatrix}+\begin{bmatrix}\cos^2A&\cos A\sin A\\-\sin A\cos A&\cos^2A\end{bmatrix}
\displaystyle =\begin{bmatrix}\sin^2A+\cos^2A&-\sin A\cos A+\cos A\sin A\\\sin A\cos A-\sin A\cos A&\sin^2A+\cos^2A\end{bmatrix}
\displaystyle =\begin{bmatrix}1&0\\0&1\end{bmatrix}\qquad\left[\because\ \sin^2A+\cos^2A=1\right]
\displaystyle \therefore \text{The simplified matrix is }\begin{bmatrix}1&0\\0&1\end{bmatrix}.
\\

\displaystyle \textbf{Question 44: }\text{Given }\begin{bmatrix}4&2\\-1&1\end{bmatrix}M=6I,\text{ where }M\text{ is a matrix and }I\text{ is the unit}
\displaystyle \text{matrix of order }2\times2,\text{ (i) state the order of matrix }M\text{ (ii) find matrix }M.\hfill \text{[ICSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Since }\begin{bmatrix}4&2\\-1&1\end{bmatrix}\text{ is of order }2\times2\text{ and }6I\text{ is also of order }2\times2,
\displaystyle \text{the order of }M\text{ is }2\times2.
\displaystyle \text{(ii) Let }M=\begin{bmatrix}a&b\\c&d\end{bmatrix}
\displaystyle \begin{bmatrix}4&2\\-1&1\end{bmatrix}\begin{bmatrix}a&b\\c&d\end{bmatrix}=6\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}4a+2c&4b+2d\\-a+c&-b+d\end{bmatrix}=\begin{bmatrix}6&0\\0&6\end{bmatrix}
\displaystyle \Rightarrow 4a+2c=6,\quad 4b+2d=0,\quad -a+c=0,\quad -b+d=6
\displaystyle \text{From }-a+c=0,\ c=a
\displaystyle \Rightarrow 4a+2a=6
\displaystyle \Rightarrow 6a=6
\displaystyle \Rightarrow a=1,\ c=1
\displaystyle \text{Also, }4b+2d=0\text{ and }-b+d=6
\displaystyle \Rightarrow d=b+6
\displaystyle \Rightarrow 4b+2(b+6)=0
\displaystyle \Rightarrow 6b=-12
\displaystyle \Rightarrow b=-2,\ d=4
\displaystyle \therefore M=\begin{bmatrix}1&-2\\1&4\end{bmatrix}.
\\

\displaystyle \textbf{Question 45: }\text{If }A=\begin{bmatrix}3&0\\5&1\end{bmatrix}\text{ and }B=\begin{bmatrix}-4&2\\1&0\end{bmatrix},\text{ find }A^2-2AB+B^2.\hfill \text{[ICSE 2020]}
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}3&0\\5&1\end{bmatrix},\qquad B=\begin{bmatrix}-4&2\\1&0\end{bmatrix}
\displaystyle A^2=\begin{bmatrix}3&0\\5&1\end{bmatrix}\begin{bmatrix}3&0\\5&1\end{bmatrix}=\begin{bmatrix}9&0\\20&1\end{bmatrix}
\displaystyle AB=\begin{bmatrix}3&0\\5&1\end{bmatrix}\begin{bmatrix}-4&2\\1&0\end{bmatrix}=\begin{bmatrix}-12&6\\-19&10\end{bmatrix}
\displaystyle 2AB=\begin{bmatrix}-24&12\\-38&20\end{bmatrix}
\displaystyle B^2=\begin{bmatrix}-4&2\\1&0\end{bmatrix}\begin{bmatrix}-4&2\\1&0\end{bmatrix}=\begin{bmatrix}18&-8\\-4&2\end{bmatrix}
\displaystyle \therefore A^2-2AB+B^2=\begin{bmatrix}9&0\\20&1\end{bmatrix}-\begin{bmatrix}-24&12\\-38&20\end{bmatrix}+\begin{bmatrix}18&-8\\-4&2\end{bmatrix}
\displaystyle =\begin{bmatrix}51&-20\\54&-17\end{bmatrix}
\displaystyle \therefore A^2-2AB+B^2=\begin{bmatrix}51&-20\\54&-17\end{bmatrix}.
\\

\displaystyle \textbf{Question 46: }\text{Given }A=\begin{bmatrix}x&3\\y&3\end{bmatrix},\text{ if }A^2=3I,\text{ where }I
\displaystyle \text{is the identity matrix of order }2,\text{ find }x\text{ and }y.\hfill \text{[ICSE 2020]}
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}x&3\\y&3\end{bmatrix}
\displaystyle A^2=A\cdot A=\begin{bmatrix}x^2+3y&3x+9\\xy+3y&3y+9\end{bmatrix}
\displaystyle 3I=\begin{bmatrix}3&0\\0&3\end{bmatrix}
\displaystyle \text{Since }A^2=3I,\text{ equating corresponding elements,}
\displaystyle 3x+9=0\Rightarrow x=-3
\displaystyle 3y+9=3\Rightarrow y=-2
\displaystyle \therefore x=-3\text{ and }y=-2.
\\

\displaystyle \textbf{Question 47: }\text{If }A=\begin{bmatrix}3&-1\\0&2\end{bmatrix},\text{ find matrix }B\text{ such that }
\displaystyle A^2-2B=3A-5I,  \ \text{where }I\text{ is the identity matrix of order }2.\hfill \text{[ICSE 2020]}
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}3&-1\\0&2\end{bmatrix},\qquad I=\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle A^2=\begin{bmatrix}3&-1\\0&2\end{bmatrix}\begin{bmatrix}3&-1\\0&2\end{bmatrix}
\displaystyle =\begin{bmatrix}9&-5\\0&4\end{bmatrix}
\displaystyle A^2-2B=3A-5I
\displaystyle \Rightarrow 2B=A^2-3A+5I
\displaystyle =\begin{bmatrix}9&-5\\0&4\end{bmatrix}-3\begin{bmatrix}3&-1\\0&2\end{bmatrix}+5\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}9&-5\\0&4\end{bmatrix}+\begin{bmatrix}-9&3\\0&-6\end{bmatrix}+\begin{bmatrix}5&0\\0&5\end{bmatrix}
\displaystyle =\begin{bmatrix}5&-2\\0&3\end{bmatrix}
\displaystyle \therefore B=\frac{1}{2}\begin{bmatrix}5&-2\\0&3\end{bmatrix}=\begin{bmatrix}\frac{5}{2}&-1\\0&\frac{3}{2}\end{bmatrix}.
\\

\displaystyle \textbf{Question 48: }\text{Find }x\text{ and }y,\text{ if }3\begin{bmatrix}5&-6\\4&x\end{bmatrix}-\begin{bmatrix}6&y\\0&6\end{bmatrix}=3\begin{bmatrix}3&-2\\4&0\end{bmatrix}.\hfill \text{[ICSE 2020]}
\displaystyle \text{Answer:}
\displaystyle 3\begin{bmatrix}5&-6\\4&x\end{bmatrix}-\begin{bmatrix}6&y\\0&6\end{bmatrix}=3\begin{bmatrix}3&-2\\4&0\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}15&-18\\12&3x\end{bmatrix}-\begin{bmatrix}6&y\\0&6\end{bmatrix}=\begin{bmatrix}9&-6\\12&0\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}9&-18-y\\12&3x-6\end{bmatrix}=\begin{bmatrix}9&-6\\12&0\end{bmatrix}
\displaystyle \text{Equating the corresponding elements,}
\displaystyle -18-y=-6\text{ and }3x-6=0
\displaystyle y=-12\text{ and }x=2
\displaystyle \therefore x=2\text{ and }y=-12.
\\

\displaystyle \textbf{Question 49: }\text{The product }AB\text{ of two matrices }A\text{ and }B\text{ is possible if }\hfill \text{[ICSE 2021]}
\displaystyle \text{(a) Same number of rows}\qquad\text{(b) Columns of }A=\text{ rows of }B
\displaystyle \text{(c) Rows of }A=\text{ columns of }B\qquad\text{(d) Same number of columns}
\displaystyle \text{Answer:}
\displaystyle \textbf{(b) }\text{The product of two matrices }A\text{ and }B\text{ is possible if}
\displaystyle \text{the number of columns of }A\text{ is equal to the number of rows of }B.
\\

\displaystyle \textbf{Question 50: }\text{If matrix }A\text{ is of order }3\times2\text{ and }B\text{ is of order }2\times2,\text{ then }AB\text{ is of order}
\displaystyle \text{(a) }3\times2\qquad\text{(b) }3\times1\qquad\text{(c) }2\times3\qquad\text{(d) }1\times3
\displaystyle \text{Answer:}
\displaystyle \textbf{(a) Given, order of matrix }A\text{ is }3\times2\text{ and order of matrix }B\text{ is }2\times2.
\displaystyle \text{Since the number of columns of }A\text{ is equal to the number of rows of }B,
\displaystyle \text{the product }AB\text{ is possible.}
\displaystyle \text{Order of }AB=\text{(Number of rows of }A)\times\text{(Number of columns of }B)
\displaystyle =3\times2
\displaystyle \therefore \text{The order of matrix }AB\text{ is }3\times2.
\\

\displaystyle \textbf{Question 51: }\text{Which of the following statement is not true?}\hfill \text{[ICSE 2022]}
\displaystyle \text{(a) All identity matrices are square.}\qquad\text{(b) All null matrices are square.}
\displaystyle \text{(c) For a square matrix, rows equal columns.}\qquad\text{(d) Diagonal matrix has zeros except leading diagonal.}
\displaystyle \text{Answer:}
\displaystyle \textbf{(b) }\text{A null matrix is a matrix of order }m\times n\text{ in which all the elements are zero.}
\displaystyle \text{A null matrix is square only when }m=n.
\displaystyle \therefore \text{The statement ``All null matrices are square'' is not true.}
\\

\displaystyle \textbf{Question 52: }\text{If }A=\begin{bmatrix}2&0\\-1&7\end{bmatrix},\text{ then }A^2\text{ is}\hfill \text{[ICSE 2022]}
\displaystyle \text{(a) }\begin{bmatrix}4&0\\1&49\end{bmatrix}\qquad\text{(b) }\begin{bmatrix}4&0\\-9&49\end{bmatrix}
\displaystyle \text{(c) }\begin{bmatrix}4&0\\9&49\end{bmatrix}\qquad\text{(d) }\begin{bmatrix}1&9\\-9&48\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \textbf{(b) Given, }A=\begin{bmatrix}2&0\\-1&7\end{bmatrix}
\displaystyle A^2=\begin{bmatrix}2&0\\-1&7\end{bmatrix}\begin{bmatrix}2&0\\-1&7\end{bmatrix}
\displaystyle =\begin{bmatrix}2\times2+0\times(-1)&2\times0+0\times7\\(-1)\times2+7\times(-1)&(-1)\times0+7\times7\end{bmatrix}
\displaystyle =\begin{bmatrix}4&0\\-9&49\end{bmatrix}
\displaystyle \therefore \text{Option (b) is correct.}
\\

\displaystyle \textbf{Question 53: }\text{If }A=\begin{bmatrix}3&5\\1&4\end{bmatrix},\ B=\begin{bmatrix}2&4\\0&3\end{bmatrix},\ C=\begin{bmatrix}1&-1\\2&1\end{bmatrix},\text{ then }
\displaystyle 5A-BC\text{ is equal to }\hfill \text{[ICSE 2022]}
\displaystyle \text{(a) }\begin{bmatrix}-5&-23\\1&17\end{bmatrix}\qquad\text{(b) }\begin{bmatrix}5&23\\1&17\end{bmatrix}
\displaystyle \text{(c) }\begin{bmatrix}-2&8\\-3&3\end{bmatrix}\qquad\text{(d) }\begin{bmatrix}5&23\\-1&17\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \textbf{(d) Given, }A=\begin{bmatrix}3&5\\1&4\end{bmatrix},\ B=\begin{bmatrix}2&4\\0&3\end{bmatrix},\ C=\begin{bmatrix}1&-1\\2&1\end{bmatrix}
\displaystyle 5A-BC=5\begin{bmatrix}3&5\\1&4\end{bmatrix}-\begin{bmatrix}2&4\\0&3\end{bmatrix}\begin{bmatrix}1&-1\\2&1\end{bmatrix}
\displaystyle =\begin{bmatrix}15&25\\5&20\end{bmatrix}-\begin{bmatrix}10&2\\6&3\end{bmatrix}
\displaystyle =\begin{bmatrix}5&23\\-1&17\end{bmatrix}
\displaystyle \therefore \text{Option (d) is correct.}
\\

\displaystyle \textbf{Question 54: }\text{If }A=\begin{bmatrix}5&10\\3&-4\end{bmatrix}\text{ and }I=\begin{bmatrix}1&0\\0&1\end{bmatrix},\text{ find }AI.\hfill \text{[ICSE 2022]}
\displaystyle \text{(a) }\begin{bmatrix}1&0\\0&1\end{bmatrix}\qquad\text{(b) }\begin{bmatrix}5&10\\-3&4\end{bmatrix}
\displaystyle \text{(c) }\begin{bmatrix}5&10\\3&-4\end{bmatrix}\qquad\text{(d) }\begin{bmatrix}15&15\\-1&-1\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \textbf{(c) Given, }A=\begin{bmatrix}5&10\\3&-4\end{bmatrix}\text{ and }I=\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle AI=\begin{bmatrix}5&10\\3&-4\end{bmatrix}\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}5&10\\3&-4\end{bmatrix}
\displaystyle \therefore \text{Option (c) is correct.}
\\

\displaystyle \textbf{Question 55: }\text{If }\begin{bmatrix}2&0\\0&4\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}2\\-8\end{bmatrix},\text{ find }x\text{ and }y.\hfill \text{[ICSE 2023]}
\displaystyle \text{(a) }1,-2\qquad\text{(b) }-2,1\qquad\text{(c) }1,2\qquad\text{(d) }-2,-1
\displaystyle \text{Answer:}
\displaystyle \textbf{(a) }\begin{bmatrix}2&0\\0&4\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}2\\-8\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2x+0y\\0x+4y\end{bmatrix}=\begin{bmatrix}2\\-8\end{bmatrix}
\displaystyle \Rightarrow 2x=2\text{ and }4y=-8
\displaystyle \Rightarrow x=1\text{ and }y=-2
\displaystyle \therefore \text{Option (a) is correct.}
\\

\displaystyle \textbf{Question 56: }\text{Given }\begin{bmatrix}a&b\\c&d\end{bmatrix}X=\begin{bmatrix}p\\q\end{bmatrix},\text{ find the order of }X.\hfill \text{[ICSE 2023]}
\displaystyle \text{(a) }2\times2\qquad\text{(b) }1\times2\qquad\text{(c) }2\times1\qquad\text{(d) }1\times1
\displaystyle \text{Answer:}
\displaystyle \textbf{(c) Let }A=\begin{bmatrix}a&b\\c&d\end{bmatrix}_{2\times2}\text{ and }B=\begin{bmatrix}p\\q\end{bmatrix}_{2\times1}
\displaystyle \text{Given }AX=B.
\displaystyle \text{Since the product }AX\text{ is defined, the number of columns of }A
\displaystyle \text{must be equal to the number of rows of }X.
\displaystyle \therefore \text{The number of rows of }X=2.
\displaystyle \text{Also, }AX\text{ is of order }2\times1,\text{ so the number of columns of }X=1.
\displaystyle \therefore \text{The order of matrix }X\text{ is }2\times1.
\displaystyle \therefore \text{Option (c) is correct.}
\\

\displaystyle \textbf{Question 57: }\text{If }\begin{bmatrix}2&x\\0&1\end{bmatrix}+3\begin{bmatrix}2&1\\4&0\end{bmatrix}=\begin{bmatrix}8&8\\12&1\end{bmatrix},\text{ find }x.\hfill \text{[ICSE 2023]}
\displaystyle \text{(a) }2\qquad\text{(b) }3\qquad\text{(c) }4\qquad\text{(d) }5
\displaystyle \text{Answer:}
\displaystyle \textbf{(d) Given, }\begin{bmatrix}2&x\\0&1\end{bmatrix}+3\begin{bmatrix}2&1\\4&0\end{bmatrix}=\begin{bmatrix}8&8\\12&1\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}2&x\\0&1\end{bmatrix}+\begin{bmatrix}6&3\\12&0\end{bmatrix}=\begin{bmatrix}8&8\\12&1\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}8&x+3\\12&1\end{bmatrix}=\begin{bmatrix}8&8\\12&1\end{bmatrix}
\displaystyle \text{Equating the corresponding elements,}
\displaystyle x+3=8
\displaystyle \Rightarrow x=5
\displaystyle \therefore \text{Option (d) is correct.}
\\

\displaystyle \textbf{Question 58: }\text{If }A=\begin{bmatrix}1&3\\2&4\end{bmatrix},\ B=\begin{bmatrix}1&2\\2&4\end{bmatrix},\ C=\begin{bmatrix}4&1\\1&5\end{bmatrix},\ I=\begin{bmatrix}1&0\\0&1\end{bmatrix},\text{ find }
\displaystyle A(B+C)-14I.\hfill \text{[ICSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}1&3\\2&4\end{bmatrix},\ B=\begin{bmatrix}1&2\\2&4\end{bmatrix},\ C=\begin{bmatrix}4&1\\1&5\end{bmatrix}
\displaystyle B+C=\begin{bmatrix}1+4&2+1\\2+1&4+5\end{bmatrix}=\begin{bmatrix}5&3\\3&9\end{bmatrix}
\displaystyle A(B+C)=\begin{bmatrix}1&3\\2&4\end{bmatrix}\begin{bmatrix}5&3\\3&9\end{bmatrix}
\displaystyle =\begin{bmatrix}1\times5+3\times3&1\times3+3\times9\\2\times5+4\times3&2\times3+4\times9\end{bmatrix}
\displaystyle =\begin{bmatrix}14&30\\22&42\end{bmatrix}
\displaystyle 14I=14\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}14&0\\0&14\end{bmatrix}
\displaystyle \therefore A(B+C)-14I=\begin{bmatrix}14&30\\22&42\end{bmatrix}-\begin{bmatrix}14&0\\0&14\end{bmatrix}
\displaystyle =\begin{bmatrix}0&30\\22&28\end{bmatrix}
\\

\displaystyle \textbf{Question 59: }\text{If }A=\begin{bmatrix}3&-2\\-1&4\end{bmatrix},\ B=\begin{bmatrix}6\\1\end{bmatrix},\ C=\begin{bmatrix}-4\\5\end{bmatrix},\text{ evaluate }
\displaystyle AB-5C.\hfill \text{[ICSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=\begin{bmatrix}3&-2\\-1&4\end{bmatrix},\ B=\begin{bmatrix}6\\1\end{bmatrix},\ C=\begin{bmatrix}-4\\5\end{bmatrix}
\displaystyle AB=\begin{bmatrix}3&-2\\-1&4\end{bmatrix}\begin{bmatrix}6\\1\end{bmatrix}=\begin{bmatrix}3\times6+(-2)\times1\\(-1)\times6+4\times1\end{bmatrix}
\displaystyle =\begin{bmatrix}16\\-2\end{bmatrix}
\displaystyle 5C=5\begin{bmatrix}-4\\5\end{bmatrix}=\begin{bmatrix}-20\\25\end{bmatrix}
\displaystyle \therefore AB-5C=\begin{bmatrix}16\\-2\end{bmatrix}-\begin{bmatrix}-20\\25\end{bmatrix}
\displaystyle =\begin{bmatrix}16-(-20)\\-2-25\end{bmatrix}=\begin{bmatrix}36\\-27\end{bmatrix}
\\

\displaystyle \textbf{Question 60: }\text{If matrix }A=\begin{bmatrix}2&2\\0&2\end{bmatrix}\text{ and }A^2=\begin{bmatrix}4&x\\0&4\end{bmatrix},\text{ then find }x.\hfill \text{[ICSE 2024]}
\displaystyle \text{(a) }2\qquad\text{(b) }4\qquad\text{(c) }8\qquad\text{(d) }10
\displaystyle \text{Answer:}
\displaystyle \textbf{(c) Given, }A=\begin{bmatrix}2&2\\0&2\end{bmatrix}\text{ and }A^2=\begin{bmatrix}4&x\\0&4\end{bmatrix}
\displaystyle A^2=A\times A
\displaystyle \Rightarrow \begin{bmatrix}2&2\\0&2\end{bmatrix}\begin{bmatrix}2&2\\0&2\end{bmatrix}=\begin{bmatrix}4&8\\0&4\end{bmatrix}
\displaystyle \text{Comparing with }\begin{bmatrix}4&x\\0&4\end{bmatrix},\text{ we get }x=8.
\displaystyle \therefore \text{Option (c) is correct.}
\\

\displaystyle \textbf{Question 61: }\text{If }A=\begin{bmatrix}x&0\\1&1\end{bmatrix},\ B=\begin{bmatrix}4&0\\y&1\end{bmatrix}\text{ and }C=\begin{bmatrix}4&0\\x&1\end{bmatrix},\text{ then}
\displaystyle \text{find the values of }x\text{ and }y,\text{ if }AB=C.\hfill \text{[ICSE 2024]}
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}x&0\\1&1\end{bmatrix},\quad B=\begin{bmatrix}4&0\\y&1\end{bmatrix},\quad C=\begin{bmatrix}4&0\\x&1\end{bmatrix}
\displaystyle \text{Since }AB=C,
\displaystyle \begin{bmatrix}x&0\\1&1\end{bmatrix}\begin{bmatrix}4&0\\y&1\end{bmatrix}=\begin{bmatrix}4&0\\x&1\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}4x&0\\4+y&1\end{bmatrix}=\begin{bmatrix}4&0\\x&1\end{bmatrix}
\displaystyle \text{Equating the corresponding elements,}
\displaystyle 4x=4\Rightarrow x=1
\displaystyle 4+y=x
\displaystyle \Rightarrow 4+y=1
\displaystyle \Rightarrow y=-3
\displaystyle \therefore x=1\text{ and }y=-3.
\\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.