\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 \Rightarrow \begin{bmatrix}6x-2\\-2x+4\end{bmatrix}+\begin{bmatrix}-8\\10\end{bmatrix}=\begin{bmatrix}8\\4y\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}6x-10\\-2x+14\end{bmatrix}=\begin{bmatrix}8\\4y\end{bmatrix}
\displaystyle 6x-10=8\Rightarrow 6x=18\Rightarrow x=3
\displaystyle -2x+14=4y\Rightarrow -6+14=4y\Rightarrow 4y=8\Rightarrow y=2
\displaystyle \therefore x=3\text{ and }y=2.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find }x\text{ and }y\text{ if }\begin{bmatrix}3x&8\end{bmatrix}\begin{bmatrix}1&4\\3&7\end{bmatrix}-3\begin{bmatrix}2&-7\end{bmatrix}=5\begin{bmatrix}3&2y\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle \begin{bmatrix}3x&8\end{bmatrix}\begin{bmatrix}1&4\\3&7\end{bmatrix}-3\begin{bmatrix}2&-7\end{bmatrix}=5\begin{bmatrix}3&2y\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}3x+24&12x+56\end{bmatrix}-\begin{bmatrix}6&-21\end{bmatrix}=\begin{bmatrix}15&10y\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}3x+18&12x+77\end{bmatrix}=\begin{bmatrix}15&10y\end{bmatrix}
\displaystyle 3x+18=15\Rightarrow 3x=-3\Rightarrow x=-1
\displaystyle 12x+77=10y\Rightarrow -12+77=10y\Rightarrow 10y=65\Rightarrow y=\frac{13}{2}
\displaystyle \therefore x=-1\text{ and }y=\frac{13}{2}.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If }\begin{bmatrix}x&y\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}25\end{bmatrix}\text{ and }\begin{bmatrix}-x&y\end{bmatrix}\begin{bmatrix}2x\\y\end{bmatrix}=\begin{bmatrix}-2\end{bmatrix},
\displaystyle \text{find the values of }x\text{ and }y\text{ if i) }x,y\text{ are whole numbers ii) }x,y\text{ are integers.}
\displaystyle \text{Answer:}
\displaystyle \begin{bmatrix}x&y\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}25\end{bmatrix}
\displaystyle \Rightarrow x^2+y^2=25\hspace{1cm}\text{...(i)}
\displaystyle \begin{bmatrix}-x&y\end{bmatrix}\begin{bmatrix}2x\\y\end{bmatrix}=\begin{bmatrix}-2\end{bmatrix}
\displaystyle \Rightarrow -2x^2+y^2=-2\hspace{1cm}\text{...(ii)}
\displaystyle \text{Multiplying (i) by }2\text{ and adding it to (ii),}
\displaystyle 3y^2=48\Rightarrow y^2=16\Rightarrow y=\pm4
\displaystyle \text{Substituting in (i), }x^2=25-16=9\Rightarrow x=\pm3
\displaystyle \text{i) If }x\text{ and }y\text{ are whole numbers, then }x=3,\ y=4.
\displaystyle \text{ii) If }x\text{ and }y\text{ are integers, then }(x,y)=(3,4),\ (-3,4),\ (3,-4),\ (-3,-4).
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Given }\begin{bmatrix}2&1\\-3&4\end{bmatrix}X=\begin{bmatrix}7\\6\end{bmatrix},\text{ find: i) the order of }X\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 A_{2\times2}X_{p\times q}=B_{2\times1}
\displaystyle \Rightarrow p=2,\ 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 \Rightarrow \begin{bmatrix}2x+y\\-3x+4y\end{bmatrix}=\begin{bmatrix}7\\6\end{bmatrix}
\displaystyle 2x+y=7
\displaystyle -3x+4y=6
\displaystyle \text{Solving the above equations, we get }x=2,\ y=3.
\displaystyle \therefore X=\begin{bmatrix}2\\3\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Evaluate }\begin{bmatrix}\cos45^\circ&\sin30^\circ\\\sqrt{2}\cos0^\circ&\sin0^\circ\end{bmatrix}\begin{bmatrix}\sin45^\circ&\cos90^\circ\\\sin90^\circ&\cot45^\circ\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle \begin{bmatrix}\cos45^\circ&\sin30^\circ\\\sqrt{2}\cos0^\circ&\sin0^\circ\end{bmatrix}\begin{bmatrix}\sin45^\circ&\cos90^\circ\\\sin90^\circ&\cot45^\circ\end{bmatrix}
\displaystyle =\begin{bmatrix}\frac{1}{\sqrt2}&\frac12\\\sqrt2&0\end{bmatrix}\begin{bmatrix}\frac{1}{\sqrt2}&0\\1&1\end{bmatrix}
\displaystyle =\begin{bmatrix}\frac12+\frac12&\frac12\\1&0\end{bmatrix}=\begin{bmatrix}1&\frac12\\1&0\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }A=\begin{bmatrix}0&-1\\4&-3\end{bmatrix},\ B=\begin{bmatrix}-5\\6\end{bmatrix}\text{ and }3AM=2B,\text{ find matrix }M.
\displaystyle \text{Answer:}
\displaystyle 3AM=2B
\displaystyle A_{2\times2}M_{p\times q}=B_{2\times1}
\displaystyle \Rightarrow p=2,\ q=1
\displaystyle \therefore \text{The order of }M\text{ is }2\times1.
\displaystyle \text{Let }M=\begin{bmatrix}x\\y\end{bmatrix}
\displaystyle 3\begin{bmatrix}0&-1\\4&-3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=2\begin{bmatrix}-5\\6\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}-3y\\12x-9y\end{bmatrix}=\begin{bmatrix}-10\\12\end{bmatrix}
\displaystyle -3y=-10\Rightarrow y=\frac{10}{3}
\displaystyle 12x-9y=12\Rightarrow 12x-30=12\Rightarrow x=\frac{7}{2}
\displaystyle \therefore M=\begin{bmatrix}\frac{7}{2}\\\frac{10}{3}\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 7: }\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 values 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 \Rightarrow \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 \Rightarrow \begin{bmatrix}a+1&b+2\\7&-c-1\end{bmatrix}=\begin{bmatrix}5&0\\7&3\end{bmatrix}
\displaystyle a+1=5\Rightarrow a=4
\displaystyle b+2=0\Rightarrow b=-2
\displaystyle -c-1=3\Rightarrow -c=4\Rightarrow c=-4
\displaystyle \therefore a=4,\ b=-2,\ c=-4.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If }A=\begin{bmatrix}1&2\\2&1\end{bmatrix}\text{ and }B=\begin{bmatrix}2&1\\1&2\end{bmatrix},\text{ find:}
\displaystyle \text{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}=\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}=\begin{bmatrix}13&14\\14&13\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 9: }\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 \Rightarrow \begin{bmatrix}2x+3x\\2y+4y\end{bmatrix}=\begin{bmatrix}5\\12\end{bmatrix}
\displaystyle 5x=5\Rightarrow x=1
\displaystyle 6y=12\Rightarrow y=2
\displaystyle \therefore x=1\text{ and }y=2.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If the matrix }X=\begin{bmatrix}-3&4\\2&-3\end{bmatrix}\begin{bmatrix}2\\-2\end{bmatrix}\text{ and }2X-3Y=\begin{bmatrix}10\\-8\end{bmatrix},
\displaystyle \text{find the matrices }X\text{ and }Y.
\displaystyle \text{Answer:}
\displaystyle X=\begin{bmatrix}-3&4\\2&-3\end{bmatrix}\begin{bmatrix}2\\-2\end{bmatrix}
\displaystyle =\begin{bmatrix}-14\\10\end{bmatrix}
\displaystyle 2X-3Y=\begin{bmatrix}10\\-8\end{bmatrix}
\displaystyle \Rightarrow 2\begin{bmatrix}-14\\10\end{bmatrix}-3Y=\begin{bmatrix}10\\-8\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}-28\\20\end{bmatrix}-3Y=\begin{bmatrix}10\\-8\end{bmatrix}
\displaystyle \Rightarrow -3Y=\begin{bmatrix}38\\-28\end{bmatrix}
\displaystyle \Rightarrow Y=\frac{1}{3}\begin{bmatrix}-38\\28\end{bmatrix}
\displaystyle \therefore X=\begin{bmatrix}-14\\10\end{bmatrix},\quad Y=\frac{1}{3}\begin{bmatrix}-38\\28\end{bmatrix}.
\displaystyle \\

\displaystyle \textbf{Question 11: }\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 \Rightarrow \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 \Rightarrow \begin{bmatrix}2&-1\\2&0\end{bmatrix}+X=\begin{bmatrix}-5&4\\8&2\end{bmatrix}
\displaystyle \Rightarrow X=\begin{bmatrix}-5&4\\8&2\end{bmatrix}-\begin{bmatrix}2&-1\\2&0\end{bmatrix}
\displaystyle \Rightarrow X=\begin{bmatrix}-7&5\\6&2\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 12: }\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 \Rightarrow \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 \Rightarrow \begin{bmatrix}4&36\\0&1\end{bmatrix}=\begin{bmatrix}4&x\\0&1\end{bmatrix}
\displaystyle \therefore x=36
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }A=\begin{bmatrix}2&5\\1&3\end{bmatrix},\ B=\begin{bmatrix}4&-2\\-1&3\end{bmatrix},\ I\text{ is the unit 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}\Rightarrow A^t=\begin{bmatrix}2&1\\5&3\end{bmatrix}
\displaystyle I=\begin{bmatrix}1&0\\0&1\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 \\

\displaystyle \textbf{Question 14: }\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 \Rightarrow \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 \Rightarrow \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 \Rightarrow \begin{bmatrix}2&-6\\2&0\end{bmatrix}+2X=\begin{bmatrix}-2&4\\8&2\end{bmatrix}
\displaystyle \Rightarrow 2X=\begin{bmatrix}-2&4\\8&2\end{bmatrix}-\begin{bmatrix}2&-6\\2&0\end{bmatrix}
\displaystyle \Rightarrow 2X=\begin{bmatrix}-4&10\\6&2\end{bmatrix}
\displaystyle \Rightarrow X=\frac{1}{2}\begin{bmatrix}-4&10\\6&2\end{bmatrix}=\begin{bmatrix}-2&5\\3&1\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 15: }\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-A+BC
\displaystyle =\begin{bmatrix}4&-2\\6&-3\end{bmatrix}\begin{bmatrix}4&-2\\6&-3\end{bmatrix}-\begin{bmatrix}4&-2\\6&-3\end{bmatrix}+\begin{bmatrix}0&2\\1&-1\end{bmatrix}\begin{bmatrix}-2&3\\1&-1\end{bmatrix}
\displaystyle =\begin{bmatrix}4&-2\\6&-3\end{bmatrix}-\begin{bmatrix}4&-2\\6&-3\end{bmatrix}+\begin{bmatrix}2&-2\\-3&4\end{bmatrix}
\displaystyle =\begin{bmatrix}2&-2\\-3&4\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 16: }\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 \\

\displaystyle \textbf{Question 17: }\text{If }A=\begin{bmatrix}3&a\\-4&8\end{bmatrix},\ B=\begin{bmatrix}c&4\\-3&0\end{bmatrix},\ C=\begin{bmatrix}-1&4\\3&b\end{bmatrix}\text{ and }3A-2C=6B,
\displaystyle \text{find the values of }a,\ b\text{ and }c.
\displaystyle \text{Answer:}
\displaystyle 3A-2C=6B
\displaystyle \Rightarrow 3\begin{bmatrix}3&a\\-4&8\end{bmatrix}-2\begin{bmatrix}-1&4\\3&b\end{bmatrix}=6\begin{bmatrix}c&4\\-3&0\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}9&3a\\-12&24\end{bmatrix}-\begin{bmatrix}-2&8\\6&2b\end{bmatrix}=\begin{bmatrix}6c&24\\-18&0\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}11&3a-8\\-18&24-2b\end{bmatrix}=\begin{bmatrix}6c&24\\-18&0\end{bmatrix}
\displaystyle 6c=11\Rightarrow c=\frac{11}{6}
\displaystyle 3a-8=24\Rightarrow a=\frac{32}{3}
\displaystyle 24-2b=0\Rightarrow b=12
\displaystyle \therefore a=\frac{32}{3},\ b=12,\ c=\frac{11}{6}.
\displaystyle \\

\displaystyle \textbf{Question 18: }\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},
\displaystyle \text{and }BA=C^2,\text{ find the values of }p\text{ and }q.\hfill \text{[ICSE 2008]}
\displaystyle \text{Answer:}
\displaystyle BA=C^2
\displaystyle \Rightarrow \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 \Rightarrow \begin{bmatrix}0&-2q\\p&0\end{bmatrix}=\begin{bmatrix}0&-8\\8&0\end{bmatrix}
\displaystyle -2q=-8\Rightarrow q=4
\displaystyle p=8
\displaystyle \therefore p=8\text{ and }q=4.
\displaystyle \\

\displaystyle \textbf{Question 19: }\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},\ D=\begin{bmatrix}2\\2\end{bmatrix},
\displaystyle \text{find }AB+2C-4D.\hfill \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 \\

\displaystyle \textbf{Question 20: }\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}2&1\\1&2\end{bmatrix}\begin{bmatrix}4&5\\5&4\end{bmatrix}
\displaystyle =\begin{bmatrix}13&14\\14&13\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{If }A=\begin{bmatrix}3&1\\-1&2\end{bmatrix},\ I=\begin{bmatrix}1&0\\0&1\end{bmatrix},\text{ find }A^2-5A+7I.\hfill \text{[ICSE 2012]}
\displaystyle \text{Answer:}
\displaystyle A^2-5A+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 \\

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