Question 15 Marks
Solve the following equations by the methods of inversion : $5x – y +4z = 5, 2x + 3y + 5z = 2$ and $5x – 2y + 6z = -1$
AnswerThe given equations can be written in the matrix form as :
$\left[\begin{array}{rrr}5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{r}5 \\ 2 \\ -1\end{array}\right]$
This is of the form $A X=B$, where
$\mathrm{A}=\left[\begin{array}{rrr}5 & -1 & 4 \\2 & 3 & 5 \\5 & -2 & 6\end{array}\right], \mathrm{X}=\left[\begin{array}{l}x \\y \\z\end{array}\right] \text { and } \mathrm{B}=\left[\begin{array}{r}5 \\2 \\-1\end{array}\right]$
Let us find $\mathrm{A}^{-1}$.
$|A|=\left|\begin{array}{rrr}5 & -1 & 4 \\2 & 3 & 5 \\5 & -2 & 6\end{array}\right|$
$= 5(18 + 10) + 1 (12 – 25) + 4( -4 – 15)$
$= 140 – 13 – 76 = 51 0$
$\therefore A^{-1}$ exists.
Now, we have to find the cofactor matrix
$=\left[\mathrm{A}_{i j}\right]_{3 \times 3} \text {, where } \mathrm{A}_{i j}=(-1)^{i+j} \mathrm{M}_{i j} $
$ \mathrm{~A}_{11}=(-1)^{1+1} \mathrm{M}_{11}=\left|\begin{array}{rr}3 & 5 \\ -2 & 6 \end{array}\right|=18+10=28 $
$ \mathrm{~A}_{12}=(-1)^{1+2} \mathrm{M}_{12}=-\left|\begin{array}{ll}2 & 5 \\ 5 & 6\end{array}\right|=-(12-25)=13$
$A_{13}=(-1)^{1+3} \mathrm{M}_{13}=\left|\begin{array}{lr}2 & 3 \\ 5 & -2\end{array}\right|=-4-15=-19 $
$ A_{21}=(-1)^{2+1} \mathrm{M}_{21}=-\left|\begin{array}{ll}-1 & 4 \\ -2 & 6\end{array}\right|=-(-6+8)=-2 $
$ A_{22}=(-1)^{2+2} \mathrm{M}_{22}=\left|\begin{array}{ll}5 & 4 \\ 5 & 6\end{array}\right|=(30-20)=10 $
$ A_{23}=(-1)^{2+3} \mathrm{M}_{23}=-\left|\begin{array}{ll}5 & -1 \\ 5 & -2\end{array}\right|=-(-10+5)=5 $
$ A_{31}=(-1)^{3+1} \mathrm{M}_{31}=\left|\begin{array}{rr}-1 & 4 \\ 3 & 5\end{array}\right|=-5-12=-17 $
$ A_{32}=(-1)^{3+2} \mathrm{M}_{32}=-\left|\begin{array}{rr}5 & 4 \\ 2 & 5\end{array}\right|=-(25-8)=-17 $
$ A_{33}=(-1)^{3+3} \mathrm{M}_{33}=\left|\begin{array}{rr}5 & -1 \\ 2 & 3\end{array}\right|=15+2=17$
$\therefore \text { the cofactor matrix }= $
$ \left[\begin{array}{lll}A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \mathrm{~A}_{31} & A_{32} & A_{33}\end{array}\right]=\left[\begin{array}{rrr}28 & 13 & -19 \\ -2 & 10 & 5 \\ -17 & -17 & 17\end{array}\right] $
$ \therefore \operatorname{adj} A=\left[\begin{array}{rrr}28 & -2 & -17 \\ 13 & 10 & -17 \\ -19 & 5 & 17\end{array}\right] $
$ \therefore A^{-1}=\frac{1}{|\mathrm{~A}|}(\operatorname{adj} \mathrm{A}) \\ =\frac{1}{51}\left[\begin{array}{rrr}28 & -2 & -17 \\ 13 & 10 & -17 \\ -19 & 5 & 17\end{array}\right] $
Now, premultiply $AX = B$ by$ A^{-1},$ we get,
$A^{-1}(AX) = A^{-1}B$
$\therefore (A^{-1}A)X = A^{-1}B$
$\therefore IX = A^{-1}B$
$ \therefore X=\frac{1}{51}\left[\begin{array}{rrr}28 & -2 & -17 \\ 13 & 10 & -17 \\ -19 & 5 & 17\end{array}\right]\left[\begin{array}{r}5 \\ 2 \\ -1\end{array}\right] $
$ =\frac{1}{51}\left[\begin{array}{r}140-4+17 \\ 65+20+17 \\ -95+10-17\end{array}\right]=\frac{1}{51}\left[\begin{array}{r}153 \\ 102 \\ -102\end{array}\right] $
$\therefore\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{r}3 \\ 2 \\ -2\end{array}\right]$
By equality of matrices,
$x = 3, y = 2, z = -2$ is the required solution.
View full question & answer→Question 25 Marks
Solve the following equations by the methods of inversion : $x + y + z = 1, 2x + 3y + 2z = 2$ and $ax + ay + 2az = 4, a \neq 0.$
AnswerThe given equations can be written in the matrix form as :
$\left[\begin{array}{rrr}1 & 1 & 1 \\2 & 3 & 2 \\a & a & 2 a\end{array}\right]\left[\begin{array}{l}x \\y \\z\end{array}\right]=\left[\begin{array}{l}1 \\2 \\4\end{array}\right]$
This is of the form $\mathrm{AX}=\mathrm{B}$, where
$\mathrm{A}=\left[\begin{array}{ccc}1 & 1 & 1 \\2 & 3 & 2 \\a & a & 2 a\end{array}\right], \mathrm{X}=\left[\begin{array}{l}x \\y \\z\end{array}\right] \text { and } \mathrm{B}=\left[\begin{array}{l}1 \\2 \\4\end{array}\right]$
Let us find $\mathrm{A}^{-1}$.
$|\mathrm{A}|=\left[\begin{array}{rrr}1 & 1 & 1 \\2 & 3 & 2 \\a & a & 2 a\end{array}\right]$
$= 1(6a – 2a) – 1(4a – 2a) + 1(2a – 3a)$
$= 4a – 2a – a = a \neq 0$
$\therefore A^{-1}$ exists.
Consider $AA^{-1} = I$
$\therefore\left[\begin{array}{rrr}1 & 1 & 1 \\ 2 & 3 & 2 \\ a & a & 2 a\end{array}\right] A^{-1}=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
By $R_2-2 R_1$ and $R_3-a R_1$, we get,
$\left[\begin{array}{lll}1 & 1 & 1 \\0 & 1 & 0 \\0 & 0 & a\end{array}\right] A^{-1}=\left[\begin{array}{rrr}1 & 0 & 0 \\-2 & 1 & 0 \\-a & 0 & 1\end{array}\right]$
By $R_1-R_2$, we get,
$\left[\begin{array}{lll}1 & 0 & 1 \\0 & 1 & 0 \\0 & 0 & a\end{array}\right] A^{-1}=\left[\begin{array}{rrr}3 & -1 & 0 \\-2 & 1 & 0 \\-a & 0 & 1\end{array}\right]$
By $\left(\frac{1}{a}\right) R_3 $, we get,
$\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rrr}3 & -1 & 0 \\ -2 & 1 & 0 \\ -1 & 0 & \frac{1}{a}\end{array}\right]$
By $R_1-R_3$, we get,
$\left[\begin{array}{lll}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right] \mathrm{A}^{-1}=\left[\begin{array}{rrr}4 & -1 & -\frac{1}{a} \\-2 & 1 & 0 \\-1 & 0 & \frac{1}{a}\end{array}\right]$
$\therefore A^{-1}=\left[\begin{array}{rrr}4 & -1 & -\frac{1}{a} \\ -2 & 1 & 0 \\ -1 & 0 & \frac{1}{a}\end{array}\right]$
Now, premultiply $\mathrm{AX}=\mathrm{B}$ by $\mathrm{A}^{-1}$, we get,
$ A^{-1}(A X) =A^{-1} B$
$\therefore\left(A^{-1} A\right) X =A^{-1} B $
$\therefore I X=A^{-1} B$
$\therefore X=\left[\begin{array}{rrr}4 & -1 & -\frac{1}{a} \\ -2 & 1 & 0 \\ -1 & 0 & \frac{1}{a}\end{array}\right]\left[\begin{array}{l}1 \\ 2 \\ 4\end{array}\right] $
$ \therefore\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{r}4-2-\frac{4}{a} \\ -2+2+0 \\ -1+0+\frac{4}{a}\end{array}\right]=\left[\begin{array}{c}2-\frac{4}{a} \\ 0 \\ \frac{4}{a}-1\end{array}\right]$
By equality of matrices,
$x=2-\frac{4}{a}, y=0, z=\frac{4}{a}-1$ is the required solution.
View full question & answer→Question 35 Marks
Solve the following equations by the methods of inversion : $2x – y = -2 , 3x + 4y = 5$
AnswerThe given equations can be written in the matrix form as :
$\left[\begin{array}{rr}2 & -1 \\ 3 & 4\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{r}-2 \\ 3\end{array}\right]$
This is of the form $\mathrm{AX}=\mathrm{B}$,
where $\mathrm{A}=\left[\begin{array}{rr}2 & -1 \\ 3 & 4\end{array}\right], \mathrm{X}=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{r}-2 \\ 3\end{array}\right]$
Let us find $\mathrm{A}^{-1}$.
$|A|=\left|\begin{array}{rr}2 & -1 \\3 & 4\end{array}\right|=8+3=11 \neq 0$
$\therefore \mathrm{A}^{-1}$ exists.
Mahara Consider $\mathrm{AA}^{-1}=\mathrm{I}$
$\therefore\left[\begin{array}{rr}2 & -1 \\ 3 & 4\end{array}\right] A^{-1}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
By $R_1 \leftrightarrow R_2$, we get,
$\left[\begin{array}{rr}3 & 4 \\2 & -1\end{array}\right] A^{-1}=\left[\begin{array}{ll}0 & 1 \\1 & 0\end{array}\right]$
By $R_1-R_2$, we get,
$\left[\begin{array}{rr}1 & 5 \\2 & -1\end{array}\right] A^{-1}=\left[\begin{array}{rr}-1 & 1 \\1 & 0\end{array}\right]$
By $R_2-2 R_1$, we get,
$\left[\begin{array}{rr}1 & 5 \\0 & -11\end{array}\right] A^{-1}=\left[\begin{array}{rr}-1 & 1 \\3 & -2\end{array}\right]$
By $\left(-\frac{1}{11}\right) R_2 ,$ we get,
$\left[\begin{array}{ll}1 & 5 \\0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{cc}-1 & 1 \\-\frac{3}{11} & \frac{2}{11}\end{array}\right]$
By $R_1-5 R_2$, we get,
$\begin{array}{l}{\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \mathrm{A}^{-1}=\left[\begin{array}{cc}\frac{4}{11} & \frac{1}{11} \\ -\frac{3}{11} & \frac{2}{11}\end{array}\right]} \\ \therefore A^{-1}=\left[\begin{array}{rr}\frac{4}{11} & \frac{1}{11} \\ -\frac{3}{11} & \frac{2}{11}\end{array}\right] \\\end{array}$
Now, premultiply $\mathrm{AX}=\mathrm{B}$ by $\mathrm{A}^{-1}$, we get,
$ A^{-1}(A X)=A^{-1} B $
$\therefore \left(A^{-1} A\right) X=A^{-1} B $
$ \therefore I X=A^{-1} B$
$\therefore X=\left[\begin{array}{cc}4 / 11 & 1 / 11 \\ -3 / 11 & 2 / 11\end{array}\right]\left[\begin{array}{c}-2 \\ 3\end{array}\right] $
$ \therefore\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}-\frac{8}{11}+\frac{3}{11} \\ \frac{6}{11}+\frac{6}{11}\end{array}\right]=\left[\begin{array}{r}-\frac{5}{11} \\ \frac{12}{11}\end{array}\right]$
By equality of matrices,
$x=-\frac{5}{11}, y=\frac{12}{11}$ is the required solution.
View full question & answer→Question 45 Marks
If $A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1\end{array}\right]$ and $B=\left[\begin{array}{lll}1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7\end{array}\right]$, then find a matrix $X$ such that $X A=B$.
AnswerConsider $XA = B$ By $C_3-C_1,$ we get,
$X\left[\begin{array}{lll}1 & 0 & 0 \\0 & 2 & 3 \\1 & 2 & 0\end{array}\right]=\left[\begin{array}{lll}1 & 2 & 2 \\1 & 1 & 4 \\2 & 4 & 5\end{array}\right]$
By $\left(\frac{1}{2}\right) C_2$, we get,
$X\left[\begin{array}{lll}1 & 0 & 0 \\0 & 1 & 3 \\1 & 1 & 0\end{array}\right]=\left[\begin{array}{rrr}1 & 1 & 2 \\1 & 1 / 2 & 4 \\2 & 2 & 5\end{array}\right]$
By $\mathrm{C}_3-3 \mathrm{C}_2$, we get,
$X\left[\begin{array}{rrr}1 & 0 & 0 \\0 & 1 & 0 \\1 & 1 & -3\end{array}\right]=\left[\begin{array}{rrr}1 & 1 & -1 \\1 & 1 / 2 & 5 / 2 \\2 & 2 & -1\end{array}\right]$
By $\left(-\frac{1}{3}\right) C_3$, we get,
$X\left[\begin{array}{lll}1 & 0 & 0 \\0 & 1 & 0 \\1 & 1 & 1\end{array}\right]=\left[\begin{array}{rrr}1 & 1 & 1 / 3 \\1 & 1 / 2 & -5 / 6 \\2 & 2 & 1 / 3\end{array}\right]$
By $C_1-C_3$ and $C_2-C_3$, we get,
$X\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{rrr}2 / 3 & 2 / 3 & 1 / 3 \\ 11 / 6 & 4 / 3 & -5 / 6 \\ 5 / 3 & 5 / 3 & 1 / 3\end{array}\right] $
$ \therefore X=\frac{1}{6}\left[\begin{array}{rrr}4 & 4 & 2 \\ 11 & 8 & -5 \\ 10 & 10 & 2\end{array}\right] $
View full question & answer→Question 55 Marks
Find the inverse of $\left[\begin{array}{lll}1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7\end{array}\right]$ by elementary row transformations.
AnswerLet $ \mathrm{A}=\left[\begin{array}{lll}1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7\end{array}\right] $
$ \therefore|\mathrm{A}|=\left|\begin{array}{lll}1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7\end{array}\right|$
$= 1(7 – 20) – 2(7 – 10) + 3(4 – 2)$
$= -13 + 6 + 6 = -1 \neq 0$
$\therefore A^{-1}$ exists.
Consider $AA^{-1} = I$
$\therefore\left[\begin{array}{lll}1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7\end{array}\right] A^{-1}=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
By $R_2-R_1$ and $R_3-2 R_1$, we get,
$\left[\begin{array}{rrr}1 & 2 & 3 \\0 & -1 & 2 \\0 & 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rrr}1 & 0 & 0 \\-1 & 1 & 0 \\-2 & 0 & 1\end{array}\right]$
By $(-1) R_2$, we get,
$\left[\begin{array}{rrr}1 & 2 & 3 \\ 0 & 1 & -2 \\ 0 & 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rrr}1 & 0 & 0 \\ 1 & -1 & 0 \\ -2 & 0 & 1\end{array}\right]$
By $R_1-2 R_2$, we get,
$\left[\begin{array}{rrr}1 & 0 & 7 \\0 & 1 & -2 \\0 & 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rrr}-1 & 2 & 0 \\1 & -1 & 0 \\-2 & 0 & 1\end{array}\right]$
By $R_1-7 R_3$ and $R_2+2 R_3$, we get,
${\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rrr}13 & 2 & -7 \\ -3 & -1 & 2 \\ -2 & 0 & 1\end{array}\right]} $
$ \therefore A^{-1}=\left[\begin{array}{rrr}13 & 2 & -7 \\ -3 & -1 & 2 \\ -2 & 0 & 1\end{array}\right]$
View full question & answer→Question 65 Marks
Find the inverse of $A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1\end{array}\right]$ by elementary column transformations.
Answer$|A|=\left|\begin{array}{lll}1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1\end{array}\right|$
$= 1 (2 – 6) – 0 + 1 (0 – 2) $
$= -4 – 2= -6 \neq 0$
$\therefore A^{-1}$ exists.
Consider $A^{-1}A = I$
$\therefore A^{-1}\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
By $C_3-C_1$, we get,
$A^{-1}\left[\begin{array}{lll}1 & 0 & 0 \\0 & 2 & 3 \\1 & 2 & 0\end{array}\right]=\left[\begin{array}{rrr}1 & 0 & -1 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right]$
By $\mathrm{C}_2 \leftrightarrow \mathrm{C}_3$, we get,
$A^{-1}\left[\begin{array}{lll}1 & 0 & 0 \\0 & 3 & 2 \\1 & 0 & 2\end{array}\right]=\left[\begin{array}{rrr}1 & -1 & 0 \\0 & 0 & 1 \\0 & 1 & 0\end{array}\right]$
By $\mathrm{C}_2-\mathrm{C}_3$, we get,
$A^{-1}\left[\begin{array}{rrr}1 & 0 & 0 \\0 & 1 & 2 \\1 & -2 & 2\end{array}\right]=\left[\begin{array}{rrr}1 & -1 & 0 \\0 & -1 & 1 \\0 & 1 & 0\end{array}\right]$
By $\mathrm{C}_3-2 \mathrm{C}_2$, we get,
$A^{-1}\left[\begin{array}{rrr}1 & 0 & 0 \\0 & 1 & 0 \\1 & -2 & 6\end{array}\right]=\left[\begin{array}{rrr}1 & -1 & 2 \\0 & -1 & 3 \\0 & 1 & -2\end{array}\right]$
By $\left(\frac{1}{6}\right) C_3$, we get,
$A^{-1}\left[\begin{array}{rrr}1 & 0 & 0 \\0 & 1 & 0 \\1 & -2 & 1\end{array}\right]=\left[\begin{array}{rrr}1 & -1 & \frac{1}{3} \\0 & -1 & \frac{1}{2} \\0 & 1 & -\frac{1}{3}\end{array}\right]$
By $C_1-C_3$ and $C_2+2 C_3$, we get,
$ A^{-1}\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{rrr}\frac{2}{3} & -\frac{1}{3} & \frac{1}{3} \\ -\frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{3} & \frac{1}{3} & -\frac{1}{3}\end{array}\right] $
$ \therefore A^{-1}=\frac{1}{6}\left[\begin{array}{rrr}4 & -2 & 2 \\ -3 & 0 & 3 \\ 2 & 2 & -2\end{array}\right] $
View full question & answer→Question 75 Marks
Find $A^{-1}$ by adjoint method and by elementary transformations if $A=\left[\begin{array}{lll}1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]$
Answer$|A|=\left|\begin{array}{lll}1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4\end{array}\right|$
$= 1(4 – 4) – 2(-4 – 2) + 3(-2 – 1)$
$= 0 + 12 – 9 = 3 \neq 0$
$\therefore A^{-1}$ exists.
$A^{-1}$ by adjoint method :
We have to find the cofactor matrix
$= [A_{ij}]_{3\times 3}$, where $A_{ij} = (-1)^{i+j}M_{ij}$
Now, $\mathrm{A}_{11}=(-1)^{1+1} \mathrm{M}_{11}=\left|\begin{array}{ll}1 & 2 \\ 2 & 4\end{array}\right|=4-4=0$
$A_{12}=(-1)^{1+2} M_{12}=-\left|\begin{array}{rrr}-1 & 2 \\ 1 & 4\end{array}\right|=-(-4-2)=6 $
$ A_{13}=(-1)^{1+3} M_{13}=\left|\begin{array}{rr}-1 & 1 \\ 1 & 2\end{array}\right|=-2-1=-3 \\ A_{21}=(-1)^{2+1}$
$ M_{21}=-\left|\begin{array}{ll}2 & 3 \\ 2 & 4\end{array}\right|=-(8-6)=-2 \\ A_{22}=(-1)^{2+2} $
$M_{22}=\left|\begin{array}{ll}1 & 3 \\ 1 & 4\end{array}\right|=4-3=1$
$A_{23}=(-1)^{2+3} \mathrm{M}_{23}=-\left|\begin{array}{ll}1 & 2 \\ 1 & 2\end{array}\right|=-(2-2)=0 $
$ A_{31}=(-1)^{3+1} M_{31}=\left|\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right|=4-3=1 $
$ A_{32}=(-1)^{3+2} \mathrm{M}_{32}=-\left|\begin{array}{rr}1 & 3 \\ -1 & 2\end{array}\right|=-(2+3)=-5 $
$ A_{33}=(-1)^{3+3} \mathrm{M}_{33}=\left|\begin{array}{rr}1 & 2 \\ -1 & 1\end{array}\right|=1+2=3$
$\therefore$ the cofactor matrix $=$
$\left[\begin{array}{lll}A_{11} & A_{12} & A_{13} \\\mathrm{~A}_{21} & \mathrm{~A}_{22} & \mathrm{~A}_{23} \\\mathrm{~A}_{31} & \mathrm{~A}_{32} & \mathrm{~A}_{33}\end{array}\right]=\left[\begin{array}{rrr}0 & 6 & -3 \\-2 & 1 & 0 \\1 & -5 & 3\end{array}\right]$
$ \therefore \operatorname{adj} A =\left[\begin{array}{rrr}0 & -2 & 1 \\ 6 & 1 & -5 \\ -3 & 0 & 3\end{array}\right] $
$ \therefore A^{-1} =\frac{1}{|A|}(\operatorname{adj} A) =\frac{1}{3}\left[\begin{array}{rrr}0 & -2 & 1 \\ 6 & 1 & -5 \\ -3 & 0 & 3\end{array}\right]$
$\mathbf{A}^{-1}$ by elementary transformations :
Consider $\mathrm{AA}^{-1}=\mathrm{I}$
By $R_2+R_1$ and $R_3-R_1$, we get,
$\therefore \left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 3 & 5 \\ 0 & 0 & 1\end{array}\right] \mathrm{A}^{-1}=\left[\begin{array}{rrr}1 & 0 & 0 \\ 1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right]$
By $\left(\frac{1}{3}\right) R_2$, we get,
$\therefore\left[\begin{array}{rrr}1 & 2 & 3 \\0 & 1 & 5 / 3 \\0 & 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rrr}1 & 0 & 0 \\1 / 3 & 1 / 3 & 0 \\-1 & 0 & 1\end{array}\right]$
By $R_1-2 R_2$, we get,
$\therefore\left[\begin{array}{rrr}1 & 0 & -1 / 3 \\0 & 1 & 5 / 3 \\0 & 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rrr}1 / 3 & -2 / 3 & 0 \\1 / 3 & 1 / 3 & 0 \\-1 & 0 &1\end{array}\right]$
By $R_1+\frac{1}{3} R_3$ and $R_2-\frac{5}{3} R_3$, we get,
$\begin{array}{l}\therefore\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rrr}0 & -2 / 3 & 1 / 3 \\ 2 & 1 / 3 & -5 / 3 \\ -1 & 0 & 1\end{array}\right]\end{array} $
$ \therefore A^{-1}=\frac{1}{3}\left[\begin{array}{rrr}0 & -2 & 1 \\ 6 & 1 & -5 \\ -3 & 0 & 3.\end{array}\right]$
View full question & answer→Question 85 Marks
Find the inverse of $\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1\end{array}\right]$ by adjoint method.
Answerwhere $A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1\end{array}\right]$
$|A| = 1(2 – 6) – 0(0 – 3) + 1(0 – 2)$
$|A| = -4 – 2$
$|A| = -6 \neq 0$
$\therefore A^{-1}$ exists.
First we have to find the cofactor matrix
$= [A_{ij}]3\times 3,$ where $A_{ij} = (-1)^{i+j}M_{ij}$
Now $\mathrm{A}_{11}=(-1)^{1+1} \mathrm{M}_{11}=\left|\begin{array}{ll}2 & 3 \\ 2 & 1\end{array}\right|=2-6=-4$
$ \mathrm{A}_{12}=(-1)^{1+2} \mathrm{M}_{12}=-\left|\begin{array}{ll}0 & 3 \\ 1 & 1\end{array}\right|=0-3=-3 $
$ \mathrm{~A}_{13}=(-1)^{1+3} \mathrm{M}_{13}=\left|\begin{array}{ll}0 & 2 \\ 1 & 2\end{array}\right|=0-2=-2 $
$ \mathrm{~A}_{21}=(-1)^{2+1} \mathrm{M}_{21}=-\left|\begin{array}{ll}0 & 1 \\ 2 & 1\end{array}\right|=-0-2=-2 \\ \mathrm{~A}_{22}=(-1)^{2+2} \mathrm{M}_{22}=\left|\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right|=1-1=0 $
$ \mathrm{~A}_{23}=(-1)^{2+3} \mathrm{M}_{23}=-\left|\begin{array}{ll}1 & 0 \\ 1 & 2\end{array}\right|=-2-0=-2 \\ \mathrm{~A}_{31}=(-1)^{3+1} \mathrm{M}_{31}=\left|\begin{array}{ll}0 & 1 \\ 2 & 3\end{array}\right|=0-2=-2 $
$ \mathrm{~A}_{32}=(-1)^{3+2} \mathrm{M}_{32}=-\left|\begin{array}{ll}1 & 1 \\ 0 & 3\end{array}\right|=-3-0=-3 $
$\mathrm{~A}_{33}=(-1)^{3+3} \mathrm{M}_{33} =\left|\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right|=2-0=2$
$\operatorname{adj}(A)=\left[\begin{array}{ccc}-4 & -3 & -2 \\ -2 & 0 & -2 \\ -2 & -3 & 2\end{array}\right] $
$ A^{-1}=\frac{1}{|A|} \cdot \operatorname{adj}(A) \\ A^{-0}=\frac{1}{6}\left[\begin{array}{ccc}-4 & -3 & -2 \\ -2 & 0 & -2\end{array}\right]$
View full question & answer→Question 95 Marks
Find the inverse of $\left[\begin{array}{lll}1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7\end{array}\right]$ by adjoint method.
AnswerLet $ \mathrm{A}=\left[\begin{array}{lll}1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7\end{array}\right] $
$ \therefore|\mathrm{A}|=\left|\begin{array}{lll}1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7\end{array}\right|$
$= 1(7 – 20) – 2(7 – 10) + 3(4 – 2)$
$= -13 + 6 + 6 = -1 \neq 0$
$\therefore A^{-1} $ exists.
First we have to find the cofactor matrix
$= [A_{ij}]_{3\times 3} $ where $A_{ij} = (-1)^{i+j}M_{ij}$
Now, $\mathrm{A}_{11}=(-1)^{1+1} \mathrm{M}_{11}=\left|\begin{array}{ll}1 & 5 \\ 4 & 7\end{array}\right|=7-20=-13$
$\mathrm{A}_{12}=(-1)^{1+2} \mathrm{M}_{12}=\left|\begin{array}{ll}1 & 5 \\ 2 & 7\end{array}\right|=-(7-10)=3$
$\mathrm{A}_{13}=(-1)^{1+3} \mathrm{M}_{13}=\left|\begin{array}{ll}1 & 1 \\ 2 & 4\end{array}\right|=4-2=2$
$A_{21}=(-1)^{2+1} M_{21}=-\left|\begin{array}{ll}2 & 3 \\ 4 & 7\end{array}\right|=-(14-12)=-2$
$\mathrm{A}_{22}=(-1)^{2+2} \mathrm{M}_{22}=\left|\begin{array}{ll}1 & 3 \\ 2 & 7\end{array}\right|=7-6=1$
$A_{23}=(-1)^{2+3} \mathrm{M}_{23}=-\left|\begin{array}{ll}1 & 2 \\ 2 & 4\end{array}\right|=-(4-4)=0$
$A_{31}=(-1)^{3+1} M_{31}=\left|\begin{array}{ll}2 & 3 \\ 1 & 5\end{array}\right|=10-3=7$
$A_{32}=(-1)^{3+2} M_{32}=-\left|\begin{array}{ll}1 & 3 \\ 1 & 5\end{array}\right|=-(5-3)=-2$
$A_{33}=(-1)^{3+3} M_{33}=\left|\begin{array}{ll}1 & 2 \\ 1 & 1\end{array}\right|=1-2=-1$
$\therefore$ the co $-$ factor matrix $=$
$\left[\begin{array}{lll}A_{11} & A_{12} & A_{13} \\A_{21} & A_{22} & A_{23} \\A_{31} & A_{32} & A_{33}\end{array}\right]=\left[\begin{array}{rrr}-13 & 3 & 2 \\-2 & 1 & 0 \\7 & -2 & -1\end{array}\right]$
$\therefore \operatorname{adj} A=\left[\begin{array}{rrr}-13 & -2 & 7 \\ 3 & 1 & -2 \\ 2 & 0 & -1\end{array}\right] $
$ \therefore A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)$
$ =\frac{1}{-1}\left[\begin{array}{rrr}-13 & -2 & 7 \\ 3 & 1 & -2 \\ 2 & 0 & -1\end{array}\right]$
$ \therefore A^{-1} =\left[\begin{array}{rrr}13 & 2 & -7 \\ -3 & -1 & 2 \\ -2 & 0 & 1\end{array}\right]$
View full question & answer→Question 105 Marks
If $A=\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right], B=\left[\begin{array}{ll}4 & 1 \\ 3 & 1\end{array}\right]$ and $C=\left[\begin{array}{ll}24 & 7 \\ 31 & 9\end{array}\right]$ then find matrix $X$ such that $A X B=C$
Answer$\text{AXB}=C \\ \therefore\left(\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right)(\mathrm{XB})=\left[\begin{array}{ll}24 & 7 \\ 31 & 9\end{array}\right]$
First we perform the row transformations.
By $R_2-R_1$, we get,
$\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right](X B)=\left[\begin{array}{rr}24 & 7 \\ 7 & 2\end{array}\right]$
By $R_1-R_2$, we get,
$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right](X B)=\left[\begin{array}{rr}17 & 5 \\ 7 & 2\end{array}\right]$
$\therefore X B=\left[\begin{array}{rr}17 & 5 \\ 7 & 2\end{array}\right] $
$ \therefore X\left[\begin{array}{ll}4 & 1 \\ 3 & 1\end{array}\right]=\left[\begin{array}{rr}17 & 5 \\ 7 & 2\end{array}\right]$
Now, we perform the column transformations.
By $C_1 \leftrightarrow C_3$, we get,
$X\left[\begin{array}{ll}1 & 4 \\ 1 & 3\end{array}\right]=\left[\begin{array}{rr}5 & 17 \\ 2 & 7\end{array}\right]$
By $C_2-4 C_1$, we get,
$X\left[\begin{array}{rr}1 & 0 \\ 1 & -1\end{array}\right]=\left[\begin{array}{ll}5 & -3 \\ 2 & -1\end{array}\right]$
By $(-1) C_2$, we get,
$X\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]=\left[\begin{array}{ll}5 & 3 \\ 2 & 1\end{array}\right]$
By $C_1-C_2$, we get,
$X\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}2 & 3 \\ 1 & 1\end{array}\right]$
$\therefore X=\left[\begin{array}{ll}2 & 3 \\ 1 & 1\end{array}\right]$
View full question & answer→Question 115 Marks
Find $X$, if $A X=B$ where $A=\left[\begin{array}{lll}1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]$ and $B=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$.
Answer$\text{AX}=B $
$ \therefore\left[\begin{array}{rrr}1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4\end{array}\right] X=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$
By $R_2+R_1$ and $R_3-R_1$, we get,
$\left[\begin{array}{lll}1 & 2 & 3 \\0 & 3 & 5 \\0 & 0 & 1\end{array}\right] X=\left[\begin{array}{l}1 \\3 \\2\end{array}\right]$
By $\left(\frac{1}{3}\right) R_2$, we get,
$\left[\begin{array}{lll}1 & 2 & 3 \\0 & 1 & \frac{5}{3} \\0 & 0 & 1\end{array}\right] X=\left[\begin{array}{l}1 \\1 \\2\end{array}\right]$
By $R_1-2 R_2$, we get,
$\left[\begin{array}{rrr}1 & 0 & -\frac{1}{3} \\0 & 1 & \frac{5}{3} \\0 & 0 & 1\end{array}\right] X=\left[\begin{array}{r}-1 \\1 \\2\end{array}\right]$
By $R_1+\frac{1}{3} R_3$ and $R_2-\frac{5}{3} R_3$, we get,
${\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] X=\left[\begin{array}{r}-\frac{1}{3} \\ -\frac{7}{3} \\ 2\end{array}\right]} $
$ \therefore X=\left[\begin{array}{r}-\frac{1}{3} \\ -\frac{7}{3} \\ 2\end{array}\right]$
View full question & answer→Question 125 Marks
Find matrix $X$ such that $A X=B$, where $A=\left[\begin{array}{ll}1 & 2 \\ -1 & 3\end{array}\right]$ and $B=\left[\begin{array}{ll}0 & 1 \\ 2 & 4\end{array}\right]$
AnswerAX = B$\therefore\left[\begin{array}{rr}1 & 2 \\ -1 & 3\end{array}\right] X=\left[\begin{array}{ll}0 & 1 \\ 2 & 4\end{array}\right]$
By $R_2+R_1$, we get, $\left[\begin{array}{ll}1 & 2 \\ 0 & 5\end{array}\right] X=\left[\begin{array}{ll}0 & 1 \\ 2 & 5\end{array}\right]$
By $\left(\frac{1}{5}\right) R_2$, we get, $\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right] X=\left[\begin{array}{ll}0 & 1 \\ \frac{2}{5} & 1\end{array}\right]$
By $R_1-2 R_2$, we get, $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] X=\left[\begin{array}{rr}-\frac{4}{5} & -1 \\ \frac{2}{5} & 1\end{array}\right]$
$\therefore X=\left[\begin{array}{rr}-\frac{4}{5} & -1 \\ \frac{2}{5} & 1\end{array}\right]$
View full question & answer→Question 135 Marks
If $A=\left[\begin{array}{ll}4 & 5 \\ 2 & 1\end{array}\right]$, then show that $A^{-1}=\frac{1}{6}(A-5 I)$
Answer$|A|=\left|\begin{array}{ll}4 & 5 \\ 2 & 1\end{array}\right|=4-10=-6 \neq 0$
$\therefore A^{-1}$ exists.
Consider $AA^{-1} = I$
$\therefore\left[\begin{array}{ll}4 & 5 \\ 2 & 1\end{array}\right] \mathrm{A}^{-1}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
By $\left(\frac{1}{4}\right) R_1$, we get, $\left[\begin{array}{ll}1 & \frac{5}{4} \\ 2 & 1\end{array}\right] A^{-1}=\left[\begin{array}{ll}\frac{1}{4} & 0 \\ 0 & 1\end{array}\right]$
By $R_2-2 R_1$, we get, $\left[\begin{array}{rr}1 & \frac{5}{4} \\ 0 & -\frac{3}{2}\end{array}\right] A^{-1}=\left[\begin{array}{rr}\frac{1}{4} & 0 \\ -\frac{1}{2} & 1\end{array}\right]$
By $\left(-\frac{2}{3}\right) \mathrm{R}_2$, we get, $\left[\begin{array}{ll}1 & \frac{5}{4} \\ 0 & 1\end{array}\right] \mathrm{A}^{-1}=\left[\begin{array}{rr}\frac{1}{4} & 0 \\ \frac{1}{3} & -\frac{2}{3}\end{array}\right]$
By $R_1-\frac{5}{4} R_2$, we get, $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rr}-\frac{1}{6} & \frac{5}{6} \\ \frac{1}{3} & -\frac{2}{3}\end{array}\right]$
$\begin{aligned} \therefore A^{-1} & =\frac{1}{6}\left[\begin{array}{rr}-1 & 5 \\ 2 & -4\end{array}\right] \\ \frac{1}{6}(A-5 I) & =\frac{1}{6}\left\{\left[\begin{array}{ll}4 & 5 \\ 2 & 1\end{array}\right]-5\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\right\} \\ & =\frac{1}{6}\left\{\left[\begin{array}{ll}4 & 5 \\ 2 & 1\end{array}\right]-\left[\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right]\right\} \\ & =\frac{1}{6}\left[\begin{array}{rr}-1 & 5 \\ 2 & -4\end{array}\right]\end{aligned}$
From (1) and (2), $A^{-1}=\frac{1}{6}(A-5 I)$.
View full question & answer→Question 145 Marks
If $A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right], B=\left[\begin{array}{ll}1 & 0 \\ 3 & 1\end{array}\right]$ find $A B$ and $(A B)^{-1}$. Verify that $(A B)^{-1}=B^{-1} A^{-1}$
Answer$\begin{aligned} A B=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right] & {\left[\begin{array}{ll}1 & 0 \\ 3 & 1\end{array}\right] } =\left[\begin{array}{ll}2+9 & 0+3 \\ 1+6 & 0+2\end{array}\right]=\left[\begin{array}{rr}11 & 3 \\ 7 & 2\end{array}\right]\end{aligned}$
$\therefore|A B|=\left|\begin{array}{rr}11 & 3 \\ 7 & 2\end{array}\right|=22-21=1 \neq 0 \\ \therefore(A B)^{-1} $ exists.
Now $,(A B)(A B)^{-1}=I $
$ \therefore\left[\begin{array}{rr}11 & 3 \\ 7 & 2\end{array}\right](A B)^{-1}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right].$
By $2 R_1$, we get, $\left[\begin{array}{rr}22 & 6 \\ 7 & 2\end{array}\right](AB)^{-1}=\left[\begin{array}{ll}2 & 0 \\ 0 & 1\end{array}\right]$
By $R_1-3 R_2$, we get, $\left[\begin{array}{ll}1 & 0 \\ 7 & 2\end{array}\right](A B)^{-1}=\left[\begin{array}{rr}2 & -3 \\ 0 & 1\end{array}\right]$
By $R_2-7 R_1$, we get, $\left(\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right)(A B)^{-1}=\left[\begin{array}{rr}2 & -3 \\ -14 & 22\end{array}\right]$
By $\left(\frac{1}{2}\right) R_2$, we get,
$\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)(A B)^{-1}=\left(\begin{array}{rr}2 & -3 \\ -7 & 11\end{array}\right)$
$\therefore(A B)^{-1}=\left(\begin{array}{rr}2 & -3 \\ -7 & 11\end{array}\right)$
$|A|=\left|\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right|=4-3=1 \neq 0$
$\therefore \mathrm{A}^{-1}$ exists.
Consider, $\mathrm{AA}^{-1}=\mathrm{I}$
$\therefore\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right] \mathrm{A}^{-1}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
By $R_1 \leftrightarrow R_2$, we get, $\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right] A^{-1}=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
By $R_2-2 R_1$, we get, $\left[\begin{array}{rr}1 & 2 \\ 0 & -1\end{array}\right] A^{-1}=\left[\begin{array}{rr}0 & 1 \\ 1 & -2\end{array}\right]$
By $(-1) R_2$, we get, $\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rr}0 & 1 \\ -1 & 2\end{array}\right]$
By $R_1-2 R_2$, we get, $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rr}2 & -3 \\ -1 & 2\end{array}\right]$
$\therefore A^{-1}=\left[\begin{array}{rr}2 & -3 \\ -1 & 2\end{array}\right]$
$|B|=\left|\begin{array}{ll}1 & 0 \\ 3 & 1\end{array}\right|=1-0=1 \neq 0$
$\therefore \mathrm{B}^{-1} $ exists.
Consider, $\mathrm{BB}^{-1}=\mathrm{I}$
$\therefore\left[\begin{array}{ll}1 & 0 \\ 3 & 1\end{array}\right] \mathrm{B}^{-1}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
By $R_2-3 R_1$, we get, $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] B^{-1}=\left[\begin{array}{rr}1 & 0 \\ -3 & 1\end{array}\right]$
$ \therefore B^{-1}=\left[\begin{array}{rr}1 & 0 \\ -3 & 1\end{array}\right] $
$ \therefore B^{-1} \cdot A^{-1}=\left[\begin{array}{rr}1 & 0 \\ -3 & 1\end{array}\right]\left[\begin{array}{rr}2 & -3 \\ -1 & 2\end{array}\right] $
$ =\left[\begin{array}{rr}2-0 & -3+0 \\ -6-1 & 9+2\end{array}\right] $
$ =\left[\begin{array}{rr}2 & -3 \\ -7 & 11\end{array}\right]$
From $(1)$ and $(2), (AB)^{-1} = B^{-1} ∙ A^{-1}.$
View full question & answer→Question 155 Marks
Find the inverse of $A=\left[\begin{array}{ccc}\cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right]$
$(i)$ elementary row transformations
$(ii)$ elementary column transformations
AnswerConsider $A^{-1}A = I $
$\therefore A^{-1}\left[\begin{array}{ccc}\cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\0 & 0 & 1\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right]$
By $(\cos \theta) \times \mathrm{C}_1$, we get,
$\mathrm{A}^{-1}\left[\begin{array}{ccc}\cos ^2 \theta & -\sin \theta & 0 \\\sin \theta \cos \theta & \cos \theta & 0 \\0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}\cos \theta & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right]$
By $C_1-\sin \theta \times C_2$, we get,
$A^{-1}\left[\begin{array}{ccc}1 & -\sin \theta & 0 \\0 & \cos \theta & 0 \\0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}\cos \theta & 0 & 0 \\-\sin \theta & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
By $\mathrm{C}_2+\sin \theta \times \mathrm{C}_1$, we get,
$\mathrm{A}^{-1}\left[\begin{array}{ccc}1 & 0 & 0 \\0 & \cos \theta & 0 \\0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}\cos \theta & \sin \theta \cos \theta & 0 \\-\sin \theta & \cos ^2 \theta & 0 \\0 & 0 & 1\end{array}\right]$
$\operatorname{By}\left(\frac{1}{\cos \theta}\right) C_2$, we get,
$A^{-1}\left[\begin{array}{lll}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right)=\left[\begin{array}{ccc}\cos \theta & \sin \theta & 0 \\-\sin \theta & \cos \theta & 0 \\0 & 0 & 1\end{array}\right] $
$\therefore A^{-1}=\left[\begin{array}{ccc}\cos \theta & \sin \theta & 0 \\-\sin \theta & \cos \theta & 0 \\0 & 0 & 1\end{array}\right]$
View full question & answer→Question 165 Marks
If $A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$, verify that $A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A|$.
Answer$\text { For } A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right] $
$A _{11}=(-1)^{1+1}(4)=4$
$A_{12}=(-1)^{1+2}(3)=-3$
$A_{21}=(-1)^{2+1}(2)=-2$
$A_{22}=(-1)^{2+2}(1)=1$
$ \operatorname{adj} A=\left[\begin{array}{ll} A_{11} & A_{21} \\ A_{12} & A_{22}\end{array}\right] \quad=\left[\begin{array}{cc} 4 & -2 \\ -3 & 1 \end{array}\right] $
$ \therefore \quad A (\operatorname{adj} A )=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]\left[\begin{array}{cc} 4 & -2 \\ -3 & 1 \end{array}\right]=\left[\begin{array}{cc} -2 & 0 \\ 0 & -2 \end{array}\right] $
$ (\operatorname{adj} A) \cdot A=\left[\begin{array}{cc} 4 & -2 \\ -3 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] $
$ =\left[\begin{array}{cc} 4-6 & 8-8 \\ -3+3 & -6+4 \end{array}\right] $
$ =\left[\begin{array}{cc} -2 & 0 \\ 0 & -2 \end{array}\right] $ and $|A| 1=\left|\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right|\left[\begin{array}{ll} 1 & 0 \\ 0 & 1
\end{array}\right] $
$ =(-2)\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ =\left[\begin{array}{cc} -2 & 0 \\
0 & -2 \end{array}\right] $
From $(i), (ii)$ and $(iii)$ we get, $A (\operatorname{adj} A )=(\operatorname{adj} A ) A =| A | I$
$($Note that this equation is valid for every nonsingular square matrix $A)$
View full question & answer→Question 175 Marks
If $A=\left[\begin{array}{ccc}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right],$ find $A ^{-1}$ by the adjoint method.
AnswerFor the given matrix $A$
$ A_{11}=(-1)^{1+1}\left|\begin{array}{cc} 2 & -1 \\ -1 & 2 \end{array}\right|=3 $
$A_{12}=(-1)^{1+2}\left|\begin{array}{cc} -1 & -1 \\ 1 & 2 \end{array}\right|=1 $
$ A_{13}=(-1)^{1+3}\left|\begin{array}{cc} -1 & 2 \\ 1 & -1 \end{array}\right|=-1 $
$ A_{21}=(-1)^{2+1}\left|\begin{array}{cc} -1 & 1 \\ -1 & 2 \end{array}\right|=1 $
$ A_{22}=(-1)^{2+2}\left|\begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array}\right|=3 $
$ A_{23}=(-1)^{2+3}\left|\begin{array}{cc} 2 & -1 \\ 1 & -1 \end{array}\right|=1$
$ A_{s 1}=(-1)^{3+1}\left|\begin{array}{cc} -1 & 1 \\ 2 & -1 \end{array}\right|=-1 $
$ A_{s 2}=(-1)^{3+2}\left|\begin{array}{cc} 2 & 1 \\ -1 & -1 \end{array}\right|=1 $
$A_{s s}=(-1)^{3+1}\left|\begin{array}{cc} 2 & -1 \\ -1 & 2 \end{array}\right|=3 $
$ \therefore \text { adj } A=\left[\begin{array}{ccc} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{array}\right] $
Now $| A |=\left|\begin{array}{ccc} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{array}\right| $
$ =2(4-1)+1(-2+1)+1(1-2) \\ =6-1-1 \\ =4 $
Therefore by using the formula for $A ^{-1}$
$ A ^{-1} =\frac{1}{| A |}(\text { adj } A ) $
$\therefore A ^{-1} =\frac{1}{4}\left[\begin{array}{ccc} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{array}\right]$
View full question & answer→Question 185 Marks
The total cost of $3 \ T.V$. and $2\ \text{V.C.R}$. is $₹\ 35,000$. The shopkeeper wants profit of $₹\ 1000$ per television and $₹ \ 500$ per $\text{V.C.R}.$ He can sell $2\ T.V$. and $1\ \text{ V.C.R}$. and get the total revenue as $₹\ 21,500$. Find the cost price and the selling price of a $T.V$. and a $\text{V.C.R}$.
Answer Let the cost of a $T.V$. be $₹\ x$ and the cost of a $\text{V.C.R}$. be $₹\ y$.
According to the first condition,
$3x + 2y = 35000 ......(i)$
The required profit per $T.V$. is $₹ 1000$ and per $\text{V.C.R}$. is $₹ 500$.
$\therefore$ Selling price of a $T.V$. is $₹ (x + 1000)$ and selling price of a $\text{V.C.R}$. is $₹\ (y + 500)$.
According to the second condition,
$2(x + 1000) + 1(y + 500) = 21500$
$\therefore 2x + 2000 + y + 500 = 21500$
$\therefore 2x + y = 21500 – 2500$
$\therefore 2x + y = 19000 ......(ii)$
Matrix form of equations $(i)$ and $(ii)$ is
$\left[\begin{array}{ll} 3 & 2 \\ 2 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 35000 \\ 19000 \end{array}\right]$
Applying $R_2 \rightarrow 2 R_2-R_1$, we get
$\left[\begin{array}{ll} 3 & 2 \\ 1 & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} 35000 \\ 3000 \end{array}\right]$
Applying $R_1 \leftrightarrow R_2-R_1$, we get
$ \left[\begin{array}{ll} 1 & 0 \\ 3 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} 300 \\ 35000 \end{array}\right]$
Hence, the original matrix is reduced to a lower triangular matrix.
$\therefore\left[\begin{array}{c} x+0 \\ 3 x+2 y \end{array}\right]=\left[\begin{array}{c} 3000 \\ 35000 \end{array}\right]$
$\therefore$ By equality of matrices, we get
$ \begin{array}{l} x=3000 \ldots(\text{iii})\\ 3 x+2 y=35000 \ldots(\text{iv}) \end{array}$
Substituting $x=3000$ in equation $(iv),$ we get
$ \begin{array}{l} 3(3000)+2 y=35000 \\ \therefore 2 y=35000-9000 \\ \therefore y=\frac{35000-9000}{2} \\
=\frac{26000}{2} \\ =13000 . \end{array}$
$\therefore$ The cost price of a $T.V$. is $\text{₹} 3,000$ and the cost price of a $\text{V.C.R}$. is $₹ 13,000$ .
Hence, the selling price of a $T.V.$
$= ₹ (3,000 + 1,000)$
$= ₹ 4,000$
and the selling price of a $\text{V.C.R}.$
$= ₹ (13,000 + 500)$
$= ₹ 13,500.$
View full question & answer→Question 195 Marks
If three numbers are added, their sum is $2$. If $2$ times the second number is subtracted from the sum of first and third numbers, we get $8$ and if three times the first number is added to the sum of second and third numbers, we get $4$. Find the numbers using matrices.
AnswerLet the three numbers be $x, y$ and $z$. According to the given conditions,
$x + y + z = 2$
$x + z – 2y = 8,$
i.e., $x – 2y + 2 = 8$
and $y + z + 3x = 4,$
i.e., $3x + y + z = 4$
Hence, the system of linear equations is
$x + y + z = 2$
$x – 2y + z = 8$
$3x + y + z = 4$
These equations can be written in the matrix form as :
$\left(\begin{array}{rrr}1 & 1 & 1 \\ 1 & -2 & 1 \\ 3 & 1 & 1\end{array}\right)\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{l}2 \\ 8 \\ 4\end{array}\right)$
By $R_2-R_1$ and $R_3-3 R_1$, we get,
${\left[\begin{array}{rrr}1 & 1 & 1 \\ 0 & -3 & 0 \\ 0 & -2 & -2\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{r}2 \\ 6 \\ -2\end{array}\right]}$
$ \therefore\left[\begin{array}{l}x+y+z \\ 0-3 y+0 \\ 0-2 y-2 z\end{array}\right]=\left[\begin{array}{r}2 \\ 6 \\ -2\end{array}\right]$
By equality of matrices,
$x + y + z = 2 ……(1)$
$-3y = 6 ……(2)$
$– 2y – 2z = -2 ……..(3)$
From $(2), y = -2$
Substituting $y = -2$ in $(3),$ we get,
$-2(-2) – 2z = -2$
$\therefore -2z = -6$
$ \therefore z = 3$
Substituting $y = -2, z = 3$ in $(1),$ we get,
$x – 2 + 3 = 2 $
$\therefore x = 1$
Hence, the required numbers are $1, -2 $ and $3$.
View full question & answer→Question 205 Marks
The cost of 4 pencils, 3 pens and 2 erasers is ₹ 60. The cost of 2 pencils, 4 pens and 6 erasers is ₹ 90, whereas the cost of 6 pencils, 2 pens and 3 erasers is ₹ 70. Find the cost of each item by using matrices.
Question 215 Marks
If $A=\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right]$, verify that $A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| \cdot \mid$
Answer$A=\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right] $
$ \therefore|A|=\left|\begin{array}{rrr}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right| $
$ =1(0+0)+1(9+2)+2(0-0)$
$=0+11+0=11 $
First we have to find the co $-$ factor matrix $=\left[\mathrm{A}_{i j}\right]_{3 \times 3}$
where $\mathrm{A}_{i j}=(-1)^{i+j} \mathrm{M}_{i j}$
Now $, A_{11}=(-1)^{1+1} M_{11}=\left|\begin{array}{rr}0 & -2 \\ 0 & 3\end{array}\right|=0+0=0 $
$A_{12}=(-1)^{1+2} M_{12}=-\left|\begin{array}{rr}3 & -2 \\ 1 & 3\end{array}\right|=-(9+2)=-11 $
$A_{13}=(-1)^{1+3} M_{13}=\left|\begin{array}{ll}3 & 0 \\ 1 & 0\end{array}\right|=0-0=0 $
$A_{21}=(-1)^{2+1} M_{21}=-\left|\begin{array}{rr}-1 & 2 \\ 0 & 3\end{array}\right|=-(-3-0)=3 $
$ A_{22}=(-1)^{2+2} \mathrm{M}_{22}=\left|\begin{array}{rrr}1 & 2 \\ 1 & 3\end{array}\right|=3-2=1 $
$ A_{23}=(-1)^{2+3} M_{23}=-\left|\begin{array}{rr}1 & -1 \\ 1 & 0\end{array}\right|=-(0+1)=-1 $
$ A_{31}=(-1)^{3+1} M_{31}=\left|\begin{array}{rr}-1 & 2 \\ 0 & -2\end{array}\right|=2-0=2 $
$ A_{32}=(-1)^{3+2} M_{32}=-\left|\begin{array}{rr}1 & 2 \\ 3 & -2\end{array}\right|=-(-2-6)=8 $
$ A_{33}=(-1)^{3+3} M_{33}=\left|\begin{array}{rr}1 & -1 \\ 3 & 0\end{array}\right|=0+3=3$
Hence the co $-$ factor matrix
$=\left(\begin{array}{lll}A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33}\end{array}\right)=\left(\begin{array}{rrr}0 & -11 & 0 \\ 3 & 1 & -1 \\ 2 & 8 & 3\end{array}\right)$
$\therefore \operatorname{adj} A=\left[\begin{array}{rrr}0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3\end{array}\right]$
$=\left[\begin{array}{rrr}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right]\left[\begin{array}{rrr}0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3\end{array}\right] $
$ =\left[\begin{array}{rrr}0+11+0 & 3-1-2 & 2-8+6 \\ 0+0-0 & 9+0+2 & 6+0-6 \\ 0+0+0 & 3+0-3 & 2+0+9\end{array}\right] $
$ =\left[\begin{array}{rrr}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{array}\right]$
$=\left[\begin{array}{rrr}0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3\end{array}\right]\left[\begin{array}{rrr}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right] $
$ =\left[\begin{array}{rrr}0+9+2 & 0+0+0 & 0-6+6 \\ -11+3+8 & 11+0+0 & -22-2+24 \\ 0-3+3 & 0-0+0 & 0+2+9\end{array}\right] $
$ =\left[\begin{array}{rrr}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{array}\right] $
$ |\mathrm{A}| \cdot \mathrm{I}=11\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{rrr}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{array}\right]$
From $(1), (2)$ and $(3),$ we get,
$A\text{(adj} A) = \text{adj} A)A = |A|∙I.$
Note: This relation is valid for any non $-$ singular matrix $A$.
View full question & answer→Question 225 Marks
Find the co-factors of the elements of the following matrices: $\left[\begin{array}{ccc}1 & -1 & 2 \\ -2 & 3 & 5 \\ -2 & 0 & -1\end{array}\right]$
AnswerNow, $\mathrm{M}_{11}=\left|\begin{array}{rr}3 & 5 \\ 0 & -1\end{array}\right|=-3-0=-3$
$\therefore A_{11}=(-1)^{1+1}(-3)=-3 $
$M_{12}=\left|\begin{array}{rr}-2 & 5 \\-2 & -1\end{array}\right|=2+10=12 $
$\therefore A_{12}=(-1)^{1+2}(12)=-12 $
$M_{13}=\left|\begin{array}{lr}-2 & 3 \\-2 & 0\end{array}\right|=0+6=6 $
$\therefore A_{13}=(-1)^{1+3}(6)=6$
$M_{21}=\left|\begin{array}{rr}-1 & 2 & 0 & -1\end{array}\right|=1-0=1 $
$ \therefore A_{21}=(-1)^{2+1}(1)=-1 $
$M_{22}=\left|\begin{array}{rr}1 & 2 \\ -2 & -1\end{array}\right|=-1+4=3 $
$ \therefore A_{22}=(-1)^{2+2}(3)=3 $
$ M_{23}=\left|\begin{array}{rr}1 & -1 \\ -2 & 0\end{array}\right|=0-2=-2 $
$ \therefore A_{23}=(-1)^{2+3}(-2)=2 $
$ M_{31}=\left|\begin{array}{rr}-1 & 2 \\ 3 & 5\end{array}\right|=-5-6=-11 $
$ \therefore A_{31}=(-1)^{3+1}(-11)=-11$
$ M_{32}=\left|\begin{array}{rr}1 & 2 \\ -2 & 5\end{array}\right|=5+4=9 $
$ \therefore A_{32}=(-1)^{3+2}(9)=-9 $
$ M_{33}=\left|\begin{array}{rr}1 & -1 \\ -2 & 3\end{array}\right|=3-2=1 $
$ \therefore A_{33}=(-1)^{3+3}(1)=1 \text {. } $
View full question & answer→Question 235 Marks
Find the adjoint of the following matrices. : If $A=\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right]$, verify that $A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| \cdot \mid$
Answer$ A= {\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right] } $
$ \therefore|A| =\left|\begin{array}{rrr}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right| $
$ =1(0+0)+1(9+2)+2(0-0)$
$ =0+11+0=11$
First we have to find the co-factor matrix $=\left[A_{i j}\right]_{3 \times 3}$ where $\mathrm{A}_{i j}=(-1)^{i+j} \mathrm{M}_{i j}$
Now $,A_{11}=(-1)^{1+1} \mathrm{M}_{11}=\left|\begin{array}{rr}0 & -2 \\ 0 & 3\end{array}\right|=0+0=0 $
$ A_{12}=(-1)^{1+2} M_{12}=-\left|\begin{array}{rr}3 & -2 \\ 1 & 3\end{array}\right|=-(9+2)=-11 $
$A_{13}=(-1)^{1+3} M_{13}=\left|\begin{array}{ll}3 & 0 \\ 1 & 0\end{array}\right|=0-0=0 $
$ A_{21}=(-1)^{2+1} \mathrm{M}_{21}=-\left|\begin{array}{rr}-1 & 2 \\ 0 & 3\end{array}\right|=-(-3-0)=3 $
$ A_{22}=(-1)^{2+2} \mathrm{M}_{22}=\left|\begin{array}{rrr}1 & 2 \\ 1 & 3\end{array}\right|=3-2=1 $
$ \mathrm{~A}_{23}=(-1)^{2+3} \mathrm{M}_{23}=-\left|\begin{array}{rr}1 & -1 \\ 1 & 0\end{array}\right|=-(0+1)=-1 $
$\mathrm{~A}_{31}=(-1)^{3+1} \mathrm{M}_{31}=\left|\begin{array}{rr}-1 & 2 \\ 0 & -2\end{array}\right|=2-0=2 $
$ \mathrm{~A}_{32}=(-1)^{3+2} \mathrm{M}_{32}=-\left|\begin{array}{rr}1 & 2 \\ 3 & -2\end{array}\right|=-(-2-6)=8 $
$ \mathrm{~A}_{33}=(-1)^{3+3} \mathrm{M}_{33}=\left|\begin{array}{rr}1 & -1 \\ 3 & 0\end{array}\right|=0+3=3$
Hence the co$-$factor matrix
$=\left(\begin{array}{lll}A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33}\end{array}\right)=\left(\begin{array}{rrr}0 & -11 & 0 \\ 3 & 1 & -1 \\ 2 & 8 & 3\end{array}\right)$
$\therefore \text{adj } A=\left(\begin{array}{rrr}0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3\end{array}\right)$
$=\left[\begin{array}{rrr}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right]\left[\begin{array}{rrr}0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3\end{array}\right] $
$ =\left[\begin{array}{rrr}0+11+0 & 3-1-2 & 2-8+6 \\ 0+0-0 & 9+0+2 & 6+0-6 \\ 0+0+0 & 3+0-3 & 2+0+9\end{array}\right] $
$ =\left[\begin{array}{rrr}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{array}\right]$
$(\operatorname{adj} A) \mathrm{A}$
$=\left[\begin{array}{rrr}0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3\end{array}\right]\left[\begin{array}{rrr}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right] $
$ =\left[\begin{array}{rrr}0+9+2 & 0+0+0 & 0-6+6 \\ -11+3+8 & 11+0+0 & -22-2+24 \\ 0-3+3 & 0-0+0 & 0+2+9\end{array}\right] $
$ =\left(\begin{array}{rrr}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{array}\right)$
$|A| \cdot I=11\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)=\left(\begin{array}{rrr}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{array}\right)$
From $(1), (2)$ and $(3),$ we get,
$A(\text{adj} A) = (\text{adj} A)A = |A|∙I.$
Note: This relation is valid for any non $-$ singular matrix $A$.
View full question & answer→