\displaystyle \textbf{Question 1. }\text{A furniture factory uses three types of wood namely, teakwood, rosewood}
\displaystyle \text{and satinwood for manufacturing three types of furniture, that are, table, chair and cot.}
\displaystyle \text{The wood requirements (in tonnes) for each type of furniture are given below} \ \
\displaystyle \begin{array}{|c|c|c|c|}\hline & \text{Table} & \text{Chair} & \text{Cot} \\\hline \text{Teakwood} & 2 & 3 & 4 \\\hline \text{Rosewood} & 1 & 1 & 2 \\\hline \text{Satinwood} & 3 & 2 & 1 \\\hline \end{array} \ \
\displaystyle \text{It is found that }29\text{ tonnes of teakwood, }13\text{ tonnes of rosewood and }16\text{ tonnes of }
\displaystyle \text{satinwood are available to make all three types of furniture.} \ \
\displaystyle \text{Using the above information, answer the following questions:} \ \
\displaystyle \text{(i) Express the data given in the table above in the form of a set of simultaneous equations.} \ \
\displaystyle \text{(ii) Solve the set of simultaneous equations formed in subpart (i) by matrix method.} \ \
\displaystyle \text{(iii) Hence, find the number of table(s), chair(s) and cot(s) produced. \hspace{0.2cm} ISC 2024} \ \
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let } x,\ y \text{ and } z \text{ be the number of tables, chairs and cots respectively.}
\displaystyle \text{The wood requirements (in tonnes) are:}
\displaystyle \text{Teakwood: } 2x+3y+4z=29
\displaystyle \text{Rosewood: } x+y+2z=13
\displaystyle \text{Satinwood: } 3x+2y+z=16
\displaystyle \text{Thus, the system of equations is}
\displaystyle 2x+3y+4z=29,\ x+y+2z=13,\ 3x+2y+z=16
\displaystyle \text{(ii) Writing in matrix form } AX=B
\displaystyle \begin{bmatrix}2&3&4\\1&1&2\\3&2&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}29\\13\\16\end{bmatrix}
\displaystyle \text{where } A=\begin{bmatrix}2&3&4\\1&1&2\\3&2&1\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix},\ B=\begin{bmatrix}29\\13\\16\end{bmatrix}
\displaystyle \text{Now, } |A|=2(-1-4)-3(1-6)+4(2-3)
\displaystyle =2(-5)-3(-5)+4(-1)=-10+15-4=1\neq 0
\displaystyle \text{So, } A^{-1} \text{ exists}
\displaystyle \text{Cofactors are } C_{11}=-3,\ C_{12}=5,\ C_{13}=-1,
\displaystyle C_{21}=5,\ C_{22}=-10,\ C_{23}=5,\ C_{31}=2,\ C_{32}=0,\ C_{33}=-1
\displaystyle \therefore \mathrm{adj}(A)=\begin{bmatrix}-3&5&2\\5&-10&0\\-1&5&-1\end{bmatrix}
\displaystyle \therefore A^{-1}=\frac{1}{|A|}\mathrm{adj}(A)=\begin{bmatrix}-3&5&2\\5&-10&0\\-1&5&-1\end{bmatrix}
\displaystyle \text{Now, } X=A^{-1}B
\displaystyle \Rightarrow \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-3&5&2\\5&-10&0\\-1&5&-1\end{bmatrix}\begin{bmatrix}29\\13\\16\end{bmatrix}
\displaystyle =\begin{bmatrix}-87+65+32\\145-130+0\\-29+65-16\end{bmatrix}=\begin{bmatrix}10\\15\\20\end{bmatrix}
\displaystyle \Rightarrow x=10,\ y=15,\ z=20
\displaystyle \text{(iii) Hence, number of tables }=10,\ \text{chairs }=15,\ \text{cots }=20
\\

\displaystyle \textbf{Question 2. }\text{Use matrix method to solve the following system of} \ \
\displaystyle \text{equations } \frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4,\ \frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1,\ \frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2  \text{\hspace{1.2cm} ISC 2023, 11} \ \
\displaystyle \textbf{Answer:}
\displaystyle \text{Given equation, } \frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4,\ \frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1,\ \frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2
\displaystyle \text{Writing in matrix form}
\displaystyle \begin{bmatrix}2&3&10\\4&-6&5\\6&9&-20\end{bmatrix}\begin{bmatrix}\frac{1}{x}\\\frac{1}{y}\\\frac{1}{z}\end{bmatrix}=\begin{bmatrix}4\\1\\2\end{bmatrix}
\displaystyle \text{Let } A=\begin{bmatrix}2&3&10\\4&-6&5\\6&9&-20\end{bmatrix},\ X=\begin{bmatrix}\frac{1}{x}\\\frac{1}{y}\\\frac{1}{z}\end{bmatrix},\ B=\begin{bmatrix}4\\1\\2\end{bmatrix}
\displaystyle \text{Now, } |A|=2(120-45)-3(-80-30)+10(36+36)
\displaystyle =2(75)-3(-110)+10(72)=150+330+720=1200\neq 0
\displaystyle \text{So, } A^{-1} \text{ exists}
\displaystyle \text{Cofactors are } A_{11}=75,\ A_{12}=110,\ A_{13}=72,
\displaystyle A_{21}=150,\ A_{22}=-100,\ A_{23}=0,\ A_{31}=75,\ A_{32}=30,\ A_{33}=-24
\displaystyle \therefore \mathrm{adj}(A)=\begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix}
\displaystyle \therefore A^{-1}=\frac{1}{1200}\begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix}
\displaystyle \text{Now, } X=A^{-1}B
\displaystyle \Rightarrow \begin{bmatrix}\frac{1}{x}\\\frac{1}{y}\\\frac{1}{z}\end{bmatrix}=\frac{1}{1200}\begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix}\begin{bmatrix}4\\1\\2\end{bmatrix}
\displaystyle =\frac{1}{1200}\begin{bmatrix}300+150+150\\440-100+60\\288+0-48\end{bmatrix}=\frac{1}{1200}\begin{bmatrix}600\\400\\240\end{bmatrix}
\displaystyle =\begin{bmatrix}\frac{1}{2}\\\frac{1}{3}\\\frac{1}{5}\end{bmatrix}
\displaystyle \text{On comparing, we get } x=2,\ y=3,\ z=5
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\displaystyle \textbf{Question 3. }\text{If }A\text{ is a square matrix of order }3,\text{ then }|2A|\text{ is equal to \hspace{0.2cm} ISC 2023} \ \
\displaystyle \text{(a) }2|A| \ \           \displaystyle \text{(b) }4|A| \ \           \displaystyle \text{(c) }8|A| \ \           \displaystyle \text{(d) }6|A| \ \
\displaystyle \text{Answer:}
\displaystyle \text{(c) We know that } |kA|=k^{n}|A|, \text{ where } n \text{ is the order of } A.
\displaystyle \therefore |2A|=2^{3}|A| \text{ since the order of } A \text{ is } 3
\displaystyle =8|A|
\\

\displaystyle \textbf{Question 4. }\text{Solve the following system of linear equations using} \ \
\displaystyle \text{matrices}   x-2y=10,\ 2x-y-z=8,\ -2y+z=7   \text{\hspace{3.2cm} ISC 2020} \ \
\displaystyle \textbf{Answer:}
\displaystyle \text{Given system of equations}
\displaystyle x-2y=10,\ 2x-y-z=8,\ -2y+z=7
\displaystyle \text{The given system can be written in matrix form as}
\displaystyle \begin{bmatrix}1&-2&0\\2&-1&-1\\0&-2&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}10\\8\\7\end{bmatrix}
\displaystyle \text{i.e. } AX=B \Rightarrow X=A^{-1}B
\displaystyle \text{Now, } |A|=1(-1-2)-(-2)(2-0)+0=-3+4=1\neq 0
\displaystyle \text{So, } A \text{ is non-singular and inverse exists}
\displaystyle \text{Cofactors are } A_{11}=-3,\ A_{12}=-2,\ A_{13}=-4,
\displaystyle A_{21}=2,\ A_{22}=1,\ A_{23}=2,\ A_{31}=2,\ A_{32}=1,\ A_{33}=3
\displaystyle \therefore \mathrm{adj}(A)=\begin{bmatrix}-3&2&2\\-2&1&1\\-4&2&3\end{bmatrix}
\displaystyle \therefore A^{-1}=\begin{bmatrix}-3&2&2\\-2&1&1\\-4&2&3\end{bmatrix}
\displaystyle \text{Now, } X=A^{-1}B
\displaystyle \Rightarrow \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-3&2&2\\-2&1&1\\-4&2&3\end{bmatrix}\begin{bmatrix}10\\8\\7\end{bmatrix}
\displaystyle =\begin{bmatrix}-30+16+14\\-20+8+7\\-40+16+21\end{bmatrix}
\displaystyle =\begin{bmatrix}0\\-5\\-3\end{bmatrix}
\displaystyle \therefore x=0,\ y=-5,\ z=-3
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\displaystyle \textbf{Question 5. }\text{Using determinants, find the values of }k,\text{ if the area} \ \
\displaystyle \text{of triangle with vertices }(-2,0),\ (0,4)\text{ and }(0,k)\text{ is }4\text{ sq units. \hspace{2.2cm} ISC 2019} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, area of a triangle with vertices } (-2,0),\ (0,4) \text{ and } (0,k) \text{ is } 4 \text{ sq units.}
\displaystyle \therefore \frac{1}{2}\begin{vmatrix}-2&0&1\\0&4&1\\0&k&1\end{vmatrix}=\pm 4 \Rightarrow \begin{vmatrix}-2&0&1\\0&4&1\\0&k&1\end{vmatrix}=\pm 8
\displaystyle \text{Expanding along } C_{1}, \text{ we get}
\displaystyle -2(4-k)=\pm 8 \Rightarrow -8+2k=\pm 8
\displaystyle \text{Taking } + \text{ sign, } -8+2k=8 \Rightarrow 2k=16 \Rightarrow k=8
\displaystyle \text{Taking } - \text{ sign, } -8+2k=-8 \Rightarrow 2k=0 \Rightarrow k=0
\displaystyle \text{Hence, the values of } k \text{ are } 0 \text{ and } 8.
\\

\displaystyle \textbf{Question 6. }\text{Solve the following system of linear equations using} \ \
\displaystyle \text{matrix method }  \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=9,\ \frac{2}{x}+\frac{5}{y}+\frac{7}{z}=52,\ \frac{2}{x}+\frac{1}{y}-\frac{1}{z}=0  \text{\hspace{0.2cm} ISC 2019} \ \
\displaystyle \textbf{Answer:}
\displaystyle \text{Given system of equations}
\displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=9,\ \frac{2}{x}+\frac{5}{y}+\frac{7}{z}=52,\ \frac{2}{x}+\frac{1}{y}-\frac{1}{z}=0
\displaystyle \text{Let } \frac{1}{x}=u,\ \frac{1}{y}=v,\ \frac{1}{z}=w
\displaystyle \text{Then, the system becomes}
\displaystyle u+v+w=9,\ 2u+5v+7w=52,\ 2u+v-w=0
\displaystyle \text{This system can be written in matrix form as}
\displaystyle \begin{bmatrix}1&1&1\\2&5&7\\2&1&-1\end{bmatrix}\begin{bmatrix}u\\v\\w\end{bmatrix}=\begin{bmatrix}9\\52\\0\end{bmatrix}
\displaystyle \text{i.e. } AX=B \Rightarrow X=A^{-1}B
\displaystyle \text{where, } A=\begin{bmatrix}1&1&1\\2&5&7\\2&1&-1\end{bmatrix},\ X=\begin{bmatrix}u\\v\\w\end{bmatrix},\ B=\begin{bmatrix}9\\52\\0\end{bmatrix}
\displaystyle \text{Now, } |A|=1(5(-1)-7\cdot 1)-1(2(-1)-7\cdot 2)+1(2\cdot 1-5\cdot 2)
\displaystyle =1(-5-7)-1(-2-14)+1(2-10)
\displaystyle =-12+16-8=-4\neq 0
\displaystyle \text{Since, } |A|\neq 0,\ \text{unique solution exists}
\displaystyle \text{Cofactors are } C_{11}=-12,\ C_{12}=16,\ C_{13}=-8,
\displaystyle C_{21}=2,\ C_{22}=-3,\ C_{23}=1,\ C_{31}=2,\ C_{32}=-5,\ C_{33}=3
\displaystyle \therefore \mathrm{adj}(A)=\begin{bmatrix}-12&2&2\\16&-3&-5\\-8&1&3\end{bmatrix}
\displaystyle \therefore A^{-1}=\frac{1}{-4}\begin{bmatrix}-12&2&2\\16&-3&-5\\-8&1&3\end{bmatrix}
\displaystyle \text{Now, } X=A^{-1}B
\displaystyle \Rightarrow \begin{bmatrix}u\\v\\w\end{bmatrix}=\frac{-1}{4}\begin{bmatrix}-12&2&2\\16&-3&-5\\-8&1&3\end{bmatrix}\begin{bmatrix}9\\52\\0\end{bmatrix}
\displaystyle =\frac{-1}{4}\begin{bmatrix}-108+104+0\\144-156+0\\-72+52+0\end{bmatrix}
\displaystyle =\frac{-1}{4}\begin{bmatrix}-4\\-12\\-20\end{bmatrix}=\begin{bmatrix}1\\3\\5\end{bmatrix}
\displaystyle \text{On comparing, we get } u=1,\ v=3,\ w=5
\displaystyle \therefore \frac{1}{x}=1,\ \frac{1}{y}=3,\ \frac{1}{z}=5
\displaystyle \Rightarrow x=1,\ y=\frac{1}{3},\ z=\frac{1}{5}
\\

\displaystyle \textbf{Question 7. }\text{Given that }A=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix}\text{ and }B=\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}, \ \
\displaystyle \text{find }AB.\text{ Using this result, solve the following system } \text{of equation }
\displaystyle x-y=3,\ 2x+3y+4z=17\text{ and }y+2z=7.\text{ \hspace{3.2cm} ISC 2017} \ \
\displaystyle \textbf{Answer:}
\displaystyle \text{Given, } A=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix} \text{ and } B=\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}
\displaystyle \text{Now, } AB=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix}\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}
\displaystyle =\begin{bmatrix}2+4+0&2-2+0&-4+4+0\\4-12+8&4+6-4&-8-12+20\\0-4+4&0+2-2&0-4+10\end{bmatrix}
\displaystyle =\begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix}=6I_{3}
\displaystyle \text{Now, the given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3\\17\\7\end{bmatrix}
\displaystyle \text{i.e. } AX=C \Rightarrow X=A^{-1}C
\displaystyle \text{Since } AB=6I_{3},\ \therefore A^{-1}=\frac{1}{6}B
\displaystyle \Rightarrow X=\frac{1}{6}B\begin{bmatrix}3\\17\\7\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{6}\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}\begin{bmatrix}3\\17\\7\end{bmatrix}
\displaystyle =\frac{1}{6}\begin{bmatrix}6+34-28\\-12+34-28\\6-17+35\end{bmatrix}
\displaystyle =\frac{1}{6}\begin{bmatrix}12\\-6\\24\end{bmatrix}=\begin{bmatrix}2\\-1\\4\end{bmatrix}
\displaystyle \therefore x=2,\ y=-1,\ z=4
\displaystyle \text{From Eq. (i), we have } AB=6I_{3}
\displaystyle \Rightarrow A\left(\frac{B}{6}\right)=I_{3} \Rightarrow A^{-1}=\frac{1}{6}B
\displaystyle \Rightarrow A^{-1}=\frac{1}{6}\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}
\displaystyle \text{On putting the value of } A^{-1} \text{ in Eq. (ii), we get}
\displaystyle X=\frac{1}{6}\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}\begin{bmatrix}3\\17\\7\end{bmatrix}
\displaystyle =\frac{1}{6}\begin{bmatrix}6+34-28\\-12+34-28\\6-17+35\end{bmatrix}
\displaystyle =\frac{1}{6}\begin{bmatrix}12\\-6\\24\end{bmatrix}=\begin{bmatrix}2\\-1\\4\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2\\-1\\4\end{bmatrix}
\displaystyle \therefore x=2,\ y=-1 \text{ and } z=4
\\

\displaystyle \textbf{Question 8. }\text{Solve the following system of linear equations by} \ \
\displaystyle \text{using matrix method }  3x+y+z=1,\ 2x+2z=0,\ 5x+y+2z=2   \text{\hspace{0.2cm} ISC 2016} \ \
\displaystyle \textbf{Answer:}
\displaystyle \text{Given system of linear equations is}
\displaystyle 3x+y+z=1
\displaystyle 2x+2z=0
\displaystyle 5x+y+2z=2
\displaystyle \text{So, we can rewrite it as}
\displaystyle \begin{bmatrix}3&1&1\\2&0&2\\5&1&2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\0\\2\end{bmatrix}
\displaystyle \text{where, } A=\begin{bmatrix}3&1&1\\2&0&2\\5&1&2\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix} \text{ and } B=\begin{bmatrix}1\\0\\2\end{bmatrix}
\displaystyle \text{Now, } |A|=\begin{vmatrix}3&1&1\\2&0&2\\5&1&2\end{vmatrix}
\displaystyle =3(0\cdot 2-1\cdot 2)-1(2\cdot 2-5\cdot 2)+1(2\cdot 1-5\cdot 0)
\displaystyle =3(0-2)-(4-10)+(2-0)
\displaystyle =3(-2)-(-6)+2=-6+6+2=2\neq 0
\displaystyle \text{So, } A^{-1} \text{ exists and the system has a unique solution}
\displaystyle \therefore X=A^{-1}B
\displaystyle \text{To find } A^{-1}, \text{ we have cofactors of the elements of matrix } A \text{ as}
\displaystyle A_{11}=\begin{vmatrix}0&2\\1&2\end{vmatrix}=0-2=-2
\displaystyle A_{12}=-\begin{vmatrix}2&2\\5&2\end{vmatrix}=-(4-10)=6
\displaystyle A_{13}=\begin{vmatrix}2&0\\5&1\end{vmatrix}=2-0=2
\displaystyle A_{21}=-\begin{vmatrix}1&1\\1&2\end{vmatrix}=-(2-1)=-1
\displaystyle A_{22}=\begin{vmatrix}3&1\\5&2\end{vmatrix}=6-5=1
\displaystyle A_{23}=-\begin{vmatrix}3&1\\5&1\end{vmatrix}=-(3-5)=2
\displaystyle A_{31}=\begin{vmatrix}1&1\\0&2\end{vmatrix}=2-0=2
\displaystyle A_{32}=-\begin{vmatrix}3&1\\2&2\end{vmatrix}=-(6-2)=-4
\displaystyle A_{33}=\begin{vmatrix}3&1\\2&0\end{vmatrix}=0-2=-2
\displaystyle \text{Now, } \mathrm{adj}(A)=\begin{bmatrix}-2&-1&2\\6&1&-4\\2&2&-2\end{bmatrix}
\displaystyle \therefore A^{-1}=\frac{1}{|A|}\mathrm{adj}(A)=\frac{1}{2}\begin{bmatrix}-2&-1&2\\6&1&-4\\2&2&-2\end{bmatrix}
\displaystyle =\begin{bmatrix}-1&-\frac{1}{2}&1\\3&\frac{1}{2}&-2\\1&1&-1\end{bmatrix}
\displaystyle \therefore X=A^{-1}B=\begin{bmatrix}-1&-\frac{1}{2}&1\\3&\frac{1}{2}&-2\\1&1&-1\end{bmatrix}\begin{bmatrix}1\\0\\2\end{bmatrix}
\displaystyle =\begin{bmatrix}-1\cdot 1+\left(-\frac{1}{2}\right)\cdot 0+1\cdot 2\\3\cdot 1+\frac{1}{2}\cdot 0+(-2)\cdot 2\\1\cdot 1+1\cdot 0+(-1)\cdot 2\end{bmatrix}
\displaystyle =\begin{bmatrix}-1+0+2\\3+0-4\\1+0-2\end{bmatrix}=\begin{bmatrix}1\\-1\\-1\end{bmatrix}
\displaystyle \Rightarrow x=1,\ y=-1,\ z=-1
\\

\displaystyle \textbf{Question 9. }\text{Find the matrix }X\text{ for which }\begin{bmatrix}5&4\\1&1\end{bmatrix}X=\begin{bmatrix}1&-2\\1&3\end{bmatrix}.\text{ \hspace{0.2cm} ISC 2016} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given } \begin{bmatrix}5&4\\1&1\end{bmatrix}X=\begin{bmatrix}1&-2\\1&3\end{bmatrix}
\displaystyle \text{Let } A=\begin{bmatrix}5&4\\1&1\end{bmatrix},\ B=\begin{bmatrix}1&-2\\1&3\end{bmatrix}
\displaystyle \text{Then, } |A|=5\cdot 1-1\cdot 4=5-4=1\neq 0
\displaystyle \text{So, } A^{-1} \text{ exists}
\displaystyle \text{Now, cofactors of elements of matrix } A \text{ are } A_{11}=1,\ A_{12}=-1,\ A_{21}=-4,\ A_{22}=5
\displaystyle \text{Now, } \mathrm{adj}(A)=\begin{bmatrix}1&-4\\-1&5\end{bmatrix}
\displaystyle \therefore A^{-1}=\frac{1}{|A|}\mathrm{adj}(A)=\begin{bmatrix}1&-4\\-1&5\end{bmatrix}
\displaystyle \text{So, } AX=B \Rightarrow X=A^{-1}B
\displaystyle \Rightarrow X=\begin{bmatrix}1&-4\\-1&5\end{bmatrix}\begin{bmatrix}1&-2\\1&3\end{bmatrix}
\displaystyle \Rightarrow X=\begin{bmatrix}1\cdot 1+(-4)\cdot 1&1\cdot(-2)+(-4)\cdot 3\\-1\cdot 1+5\cdot 1&-1\cdot(-2)+5\cdot 3\end{bmatrix}
\displaystyle \Rightarrow X=\begin{bmatrix}-3&-14\\4&17\end{bmatrix}
\\

\displaystyle \textbf{Question 10. }\text{Given two matrices }A\text{ and }B, \ \
\displaystyle A=\begin{bmatrix}1&-2&3\\1&4&1\\1&-3&2\end{bmatrix}\text{ and }B=\begin{bmatrix}11&-5&-14\\-1&-1&2\\-7&1&6\end{bmatrix}; \ \
\displaystyle AB\text{ and use this result to solve the following system of equations} \ \
\displaystyle  x-2y+3z=6,\ x+4y+z=12,\ x-3y+2z=1.\text{ \hspace{0.2cm} ISC 2015} \ \
\displaystyle \textbf{Answer:}
\displaystyle A=\begin{bmatrix}1&-2&3\\1&4&1\\1&-3&2\end{bmatrix},\quad B=\begin{bmatrix}11&-5&-14\\-1&-1&2\\-7&1&6\end{bmatrix}
\displaystyle \text{First, find }AB
\displaystyle AB=\begin{bmatrix}1&-2&3\\1&4&1\\1&-3&2\end{bmatrix}\begin{bmatrix}11&-5&-14\\-1&-1&2\\-7&1&6\end{bmatrix}
\displaystyle =\begin{bmatrix}1(11)+(-2)(-1)+3(-7)&1(-5)+(-2)(-1)+3(1)&1(-14)+(-2)(2)+3(6)\\1(11)+4(-1)+1(-7)&1(-5)+4(-1)+1(1)&1(-14)+4(2)+1(6)\\1(11)+(-3)(-1)+2(-7)&1(-5)+(-3)(-1)+2(1)&1(-14)+(-3)(2)+2(6)\end{bmatrix}
\displaystyle =\begin{bmatrix}-8&0&0\\0&-8&0\\0&0&-8\end{bmatrix}
\displaystyle =-8I
\displaystyle \therefore\ A^{-1}=-\frac{1}{8}B
\displaystyle \text{The given system is}
\displaystyle x-2y+3z=6,\quad x+4y+z=12,\quad x-3y+2z=1
\displaystyle \text{In matrix form, }AX=C,\ \text{where }X=\begin{bmatrix}x\\y\\z\end{bmatrix},\ C=\begin{bmatrix}6\\12\\1\end{bmatrix}
\displaystyle \therefore\ X=A^{-1}C=-\frac{1}{8}BC
\displaystyle BC=\begin{bmatrix}11&-5&-14\\-1&-1&2\\-7&1&6\end{bmatrix}\begin{bmatrix}6\\12\\1\end{bmatrix}
\displaystyle =\begin{bmatrix}11(6)+(-5)(12)+(-14)(1)\\(-1)(6)+(-1)(12)+2(1)\\(-7)(6)+1(12)+6(1)\end{bmatrix}
\displaystyle =\begin{bmatrix}66-60-14\\-6-12+2\\-42+12+6\end{bmatrix}
\displaystyle =\begin{bmatrix}-8\\-16\\-24\end{bmatrix}
\displaystyle \therefore\ X=-\frac{1}{8}\begin{bmatrix}-8\\-16\\-24\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix}
\displaystyle \therefore\ x=1,\quad y=2,\quad z=3
\\

\displaystyle \textbf{Question 11. }\text{Using the matrix method, solve the following system} \ \
\displaystyle \text{of equations }  x-2y=10,\ 2x+y+3z=8\text{ and }-2y+z=7.\text{ \hspace{0.2cm} ISC 2014} \ \
\displaystyle \textbf{Answer:}
\displaystyle \text{Given system of equations:}
\displaystyle x-2y=10
\displaystyle 2x+y+3z=8
\displaystyle -2y+z=7
\displaystyle \text{Writing in matrix form, }AX=B
\displaystyle \begin{bmatrix}1&-2&0\\2&1&3\\0&-2&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}10\\8\\7\end{bmatrix}
\displaystyle \text{Let }A=\begin{bmatrix}1&-2&0\\2&1&3\\0&-2&1\end{bmatrix}
\displaystyle \text{Now, }|A|=1\begin{vmatrix}1&3\\-2&1\end{vmatrix}-(-2)\begin{vmatrix}2&3\\0&1\end{vmatrix}+0
\displaystyle =1(1+6)+2(2)=7+4=11
\displaystyle \text{Cofactors of }A\text{ are:}
\displaystyle C_{11}=\begin{vmatrix}1&3\\-2&1\end{vmatrix}=7,\quad C_{12}=-\begin{vmatrix}2&3\\0&1\end{vmatrix}=-2,\quad C_{13}=\begin{vmatrix}2&1\\0&-2\end{vmatrix}=-4
\displaystyle C_{21}=-\begin{vmatrix}-2&0\\-2&1\end{vmatrix}=2,\quad C_{22}=\begin{vmatrix}1&0\\0&1\end{vmatrix}=1,\quad C_{23}=-\begin{vmatrix}1&-2\\0&-2\end{vmatrix}=2
\displaystyle C_{31}=\begin{vmatrix}-2&0\\1&3\end{vmatrix}=-6,\quad C_{32}=-\begin{vmatrix}1&0\\2&3\end{vmatrix}=-3,\quad C_{33}=\begin{vmatrix}1&-2\\2&1\end{vmatrix}=5
\displaystyle \therefore\ \mathrm{adj}(A)=\begin{bmatrix}7&2&-6\\-2&1&-3\\-4&2&5\end{bmatrix}
\displaystyle \therefore\ A^{-1}=\frac{1}{11}\begin{bmatrix}7&2&-6\\-2&1&-3\\-4&2&5\end{bmatrix}
\displaystyle \text{Hence }X=A^{-1}B
\displaystyle =\frac{1}{11}\begin{bmatrix}7&2&-6\\-2&1&-3\\-4&2&5\end{bmatrix}\begin{bmatrix}10\\8\\7\end{bmatrix}
\displaystyle =\frac{1}{11}\begin{bmatrix}70+16-42\\-20+8-21\\-40+16+35\end{bmatrix}
\displaystyle =\frac{1}{11}\begin{bmatrix}44\\-33\\11\end{bmatrix}
\displaystyle =\begin{bmatrix}4\\-3\\1\end{bmatrix}
\displaystyle \therefore\ x=4,\quad y=-3,\quad z=1
\\

\displaystyle \textbf{Question 12. }\text{Find }A^{-1},\text{ where }A=\begin{bmatrix}4&2&3\\1&1&1\\3&1&-2\end{bmatrix}. \ \
\displaystyle \text{Then, solve the following system of linear equations} \ \
\displaystyle 4x+2y+3z=2,\ x+y+z=1\text{ and }3x+y-2z=5.\text{ \hspace{2.2cm} ISC 2013} \ \
\displaystyle \textbf{Answer:}
\displaystyle \text{Given, system of equations } 4x+2y+3z=2,\ x+y+z=1,\ 3x+y-2z=5
\displaystyle \text{The given system can be written in matrix form as}
\displaystyle \begin{bmatrix}4&2&3\\1&1&1\\3&1&-2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2\\1\\5\end{bmatrix}
\displaystyle \text{i.e. } AX=B \Rightarrow X=A^{-1}B
\displaystyle \text{where, } A=\begin{bmatrix}4&2&3\\1&1&1\\3&1&-2\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix},\ B=\begin{bmatrix}2\\1\\5\end{bmatrix}
\displaystyle \text{Now, } |A|=4(-2-1)-2(-2-3)+3(1-3)
\displaystyle =4(-3)-2(-5)+3(-2)=-12+10-6=-8\neq 0
\displaystyle \text{So, } A \text{ is non-singular matrix and inverse exists}
\displaystyle \text{Cofactors are } A_{11}=-3,\ A_{12}=5,\ A_{13}=-2,
\displaystyle A_{21}=7,\ A_{22}=-17,\ A_{23}=2,\ A_{31}=-1,\ A_{32}=-1,\ A_{33}=2
\displaystyle \therefore \mathrm{adj}(A)=\begin{bmatrix}-3&7&-1\\5&-17&-1\\-2&2&2\end{bmatrix}
\displaystyle \therefore A^{-1}=\frac{1}{|A|}\mathrm{adj}(A)=\frac{1}{-8}\begin{bmatrix}-3&7&-1\\5&-17&-1\\-2&2&2\end{bmatrix}
\displaystyle =\frac{1}{8}\begin{bmatrix}3&-7&1\\-5&17&1\\2&-2&-2\end{bmatrix}
\displaystyle \text{Now, } X=A^{-1}B
\displaystyle \Rightarrow \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{8}\begin{bmatrix}3&-7&1\\-5&17&1\\2&-2&-2\end{bmatrix}\begin{bmatrix}2\\1\\5\end{bmatrix}
\displaystyle =\frac{1}{8}\begin{bmatrix}6-7+5\\-10+17+5\\4-2-10\end{bmatrix}
\displaystyle =\frac{1}{8}\begin{bmatrix}4\\12\\-8\end{bmatrix}=\begin{bmatrix}\frac{1}{2}\\\frac{3}{2}\\-1\end{bmatrix}
\displaystyle \therefore x=\frac{1}{2},\ y=\frac{3}{2},\ z=-1

 

\displaystyle \textbf{Question 13. }\text{If }A=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\text{, then find }x\text{ such that }A^{2}=xA-2I. \ \
\displaystyle \text{Hence, find }A^{-1}.\text{ \hspace{7.2cm} ISC 2011} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } A=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}
\displaystyle \text{Now, } A^{2}=A\cdot A=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\begin{bmatrix}3&-2\\4&-2\end{bmatrix}
\displaystyle =\begin{bmatrix}9-8&-6+4\\12-8&-8+4\end{bmatrix}=\begin{bmatrix}1&-2\\4&-4\end{bmatrix}
\displaystyle \text{Now, } xA=x\begin{bmatrix}3&-2\\4&-2\end{bmatrix}=\begin{bmatrix}3x&-2x\\4x&-2x\end{bmatrix}
\displaystyle \text{Also, } -2I=-2\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}-2&0\\0&-2\end{bmatrix}
\displaystyle \text{Given } A^{2}=xA-2I
\displaystyle \Rightarrow \begin{bmatrix}1&-2\\4&-4\end{bmatrix}=\begin{bmatrix}3x&-2x\\4x&-2x\end{bmatrix}-\begin{bmatrix}2&0\\0&2\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}1&-2\\4&-4\end{bmatrix}=\begin{bmatrix}3x-2&-2x\\4x&-2x-2\end{bmatrix}
\displaystyle \text{On equating the corresponding elements, we get } 3x-2=1
\displaystyle \Rightarrow 3x=3 \Rightarrow x=1
\displaystyle \text{Hence, } A^{2}=A-2I \Rightarrow A^{2}-A+2I=0
\displaystyle \text{On pre-multiplying both sides by } A^{-1} \text{ in Eq. (iv), we get}
\displaystyle A^{-1}A^{2}-A^{-1}A+2A^{-1}I=A^{-1}O
\displaystyle \Rightarrow (A^{-1}A)A-I+2A^{-1}=0
\displaystyle \text{[since } A^{-1}A=I \text{ and } A^{-1}I=A^{-1}]
\displaystyle \Rightarrow IA-I+2A^{-1}=0
\displaystyle \Rightarrow A-I+2A^{-1}=0
\displaystyle \Rightarrow 2A^{-1}=I-A
\displaystyle \Rightarrow 2A^{-1}=\begin{bmatrix}1&0\\0&1\end{bmatrix}-\begin{bmatrix}3&-2\\4&-2\end{bmatrix}
\displaystyle \Rightarrow 2A^{-1}=\begin{bmatrix}1-3&0+2\\0-4&1+2\end{bmatrix}=\begin{bmatrix}-2&2\\-4&3\end{bmatrix}
\displaystyle \therefore A^{-1}=\frac{1}{2}\begin{bmatrix}-2&2\\-4&3\end{bmatrix}=\begin{bmatrix}-1&1\\-2&\frac{3}{2}\end{bmatrix}
\\

\displaystyle \textbf{Question 14. }\text{If the matrix }A=\begin{bmatrix}6&x&2\\2&-1&2\\-10&5&2\end{bmatrix}\text{ is a singular matrix,} \ \
\displaystyle \text{then find the value of }x.\text{ \hspace{5.2cm} ISC 2010} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } A=\begin{bmatrix}6&x&2\\2&-1&2\\-10&5&2\end{bmatrix}
\displaystyle \text{Since, } A \text{ is a singular matrix, } |A|=0
\displaystyle \therefore \begin{vmatrix}6&x&2\\2&-1&2\\-10&5&2\end{vmatrix}=0
\displaystyle \text{Expanding along } R_{1}, \text{ we get}
\displaystyle 6\begin{vmatrix}-1&2\\5&2\end{vmatrix}-x\begin{vmatrix}2&2\\-10&2\end{vmatrix}+2\begin{vmatrix}2&-1\\-10&5\end{vmatrix}=0
\displaystyle \Rightarrow 6(-2-10)-x(4+20)+2(10-10)=0
\displaystyle \Rightarrow 6(-12)-24x+0=0
\displaystyle \Rightarrow -72-24x=0 \Rightarrow 24x=-72 \Rightarrow x=-3
\\

\displaystyle \textbf{Question 15. }\text{If }A=\begin{bmatrix}\cos x&\sin x\\-\sin x&\cos x\end{bmatrix}\text{ and} \ \
\displaystyle A\cdot (\mathrm{adj}\ A)=k\begin{bmatrix}1&0\\0&1\end{bmatrix}\text{, then find the value of }k.\text{ \hspace{3.2cm} ISC 2005} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } A=\begin{bmatrix}\cos x&\sin x\\-\sin x&\cos x\end{bmatrix}
\displaystyle \text{Now, cofactors of } A \text{ are }
\displaystyle A_{11}=\cos x,\ A_{12}=\sin x,\ A_{21}=-\sin x,\ A_{22}=\cos x
\displaystyle \therefore \mathrm{adj}(A)=\begin{bmatrix}\cos x&-\sin x\\\sin x&\cos x\end{bmatrix}
\displaystyle \text{Now, } A\cdot \mathrm{adj}(A)=\begin{bmatrix}\cos x&\sin x\\-\sin x&\cos x\end{bmatrix}\begin{bmatrix}\cos x&-\sin x\\\sin x&\cos x\end{bmatrix}
\displaystyle =\begin{bmatrix}\cos^{2}x+\sin^{2}x&-\sin x\cos x+\cos x\sin x\\-\sin x\cos x+\cos x\sin x&\sin^{2}x+\cos^{2}x\end{bmatrix}
\displaystyle =\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle \text{[since } \sin^{2}x+\cos^{2}x=1]
\displaystyle \text{Also given, } A\cdot \mathrm{adj}(A)=kI
\displaystyle \Rightarrow \begin{bmatrix}1&0\\0&1\end{bmatrix}=k\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle \Rightarrow k=1
\\

\displaystyle \textbf{Question 16. }\text{If the matrix }A=\begin{bmatrix}2&3\\5&-2\end{bmatrix}\text{, then show that }A^{-1}=\frac{1}{19}A.\text{ \hspace{0.2cm} ISC 2004} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } A=\begin{bmatrix}2&3\\5&-2\end{bmatrix}
\displaystyle \text{Now, } |A|=2(-2)-5\cdot 3=-4-15=-19\neq 0
\displaystyle \text{Since, } A \text{ is non-singular, } A^{-1} \text{ exists}
\displaystyle \text{Cofactors are } A_{11}=-2,\ A_{12}=-5,\ A_{21}=-3,\ A_{22}=2
\displaystyle \therefore \mathrm{adj}(A)=\begin{bmatrix}-2&-3\\-5&2\end{bmatrix}
\displaystyle \therefore A^{-1}=\frac{1}{|A|}\mathrm{adj}(A)=\frac{1}{-19}\begin{bmatrix}-2&-3\\-5&2\end{bmatrix}
\displaystyle =\frac{1}{19}\begin{bmatrix}2&3\\5&-2\end{bmatrix}=\frac{1}{19}A
\\

\displaystyle \textbf{Question 3. }\text{Using properties of determinants, find the value of the} \ \
\displaystyle \text{determinant }\left|\begin{matrix}x^{3} & x^{2} & x\\y^{3} & y^{2} & y\\z^{3} & z^{2} & z\end{matrix}\right|.\text{ \hspace{5.2cm} ISC 2003} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } \Delta=\begin{vmatrix}x^{3}&x^{2}&x\\y^{3}&y^{2}&y\\z^{3}&z^{2}&z\end{vmatrix}
\displaystyle \text{On taking } x,\ y \text{ and } z \text{ common from } R_{1},\ R_{2} \text{ and } R_{3} \text{ respectively,}
\displaystyle \Delta=xyz\begin{vmatrix}x^{2}&x&1\\y^{2}&y&1\\z^{2}&z&1\end{vmatrix}
\displaystyle \text{On expanding along } R_{1}, \text{ we get}
\displaystyle \Delta=(xyz)\left[x^{2}(y-z)-x(y^{2}-z^{2})+1(yz^{2}-zy^{2})\right]
\displaystyle \Delta=(xyz)\left[x^{2}(y-z)-x(y+z)(y-z)+yz(z-y)\right]
\displaystyle \Delta=(xyz)(y-z)\left[x^{2}-xy-xz-yz\right]
\displaystyle \Delta=(xyz)(y-z)\left[x(x-y)-z(x+y)\right]
\displaystyle \Delta=(xyz)(y-z)(x-y)(x-z)
\displaystyle \Delta=(xyz)(x-y)(y-z)(z-x)
\\

\displaystyle \textbf{Question 17. }\text{Find the inverse of the matrix }A=\begin{bmatrix}3&1\\4&2\end{bmatrix}.\text{ \hspace{2.2cm} ISC 2002} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } A=\begin{bmatrix}3&1\\4&2\end{bmatrix}
\displaystyle \text{Now, } |A|=3\cdot 2-4\cdot 1=6-4=2\neq 0
\displaystyle \text{So, } A^{-1} \text{ exists}
\displaystyle \text{Cofactors are } A_{11}=2,\ A_{12}=-4,\ A_{21}=-1,\ A_{22}=3
\displaystyle \therefore \mathrm{adj}(A)=\begin{bmatrix}2&-1\\-4&3\end{bmatrix}
\displaystyle \therefore A^{-1}=\frac{1}{2}\begin{bmatrix}2&-1\\-4&3\end{bmatrix}
\displaystyle \text{Now, } \mathrm{adj}(A)=\begin{bmatrix}A_{11}&A_{21}\\A_{12}&A_{22}\end{bmatrix}=\begin{bmatrix}2&-1\\-4&3\end{bmatrix}
\displaystyle \therefore A^{-1}=\frac{1}{|A|}\mathrm{adj}(A)=\frac{1}{2}\begin{bmatrix}2&-1\\-4&3\end{bmatrix}=\begin{bmatrix}1&-\frac{1}{2}\\-2&\frac{3}{2}\end{bmatrix}
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