[using elementary transformations (operations), find the inverse of the following matrices, if it exist]
Question 1
$\begin{bmatrix}1&3\\2&7\end{bmatrix}$
Sol :
A=$\begin{bmatrix}1&3\\2&7\end{bmatrix}$
A=IA
$\begin{bmatrix}1&3\\2&7\end{bmatrix}$=$\begin{bmatrix}1&0\\0&1\end{bmatrix}A$
R2→R2-2R1
$\begin{bmatrix}1&3\\0&1\end{bmatrix}$=$\begin{bmatrix}1&0\\-2&1\end{bmatrix}A$
R1→R1-3R2
$\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}7&-3\\-2&1\end{bmatrix}A$
$A^{-1}=\begin{bmatrix}7&-3\\-2&1\end{bmatrix}$
Question 2
$\begin{bmatrix}2&1\\7&4\end{bmatrix}$
Sol :
Let A=$\begin{bmatrix}2&1\\7&4\end{bmatrix}$
A=IA
$\begin{bmatrix}2&1\\7&4\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}A$
R1↔R2
$\begin{bmatrix}7&4\\2&1\end{bmatrix}=\begin{bmatrix}0&1\\1&0\end{bmatrix}A$
R1→R1-3R2
$\begin{bmatrix}1&1\\2&1\end{bmatrix}=\begin{bmatrix}-3&1\\1&0\end{bmatrix}A$
R2→R2-2R1
$\begin{bmatrix}1&1\\0&-1\end{bmatrix}=\begin{bmatrix}-3&1\\7&-2\end{bmatrix}A$
R1→-1R2
$\begin{bmatrix}1&1\\0&1\end{bmatrix}=\begin{bmatrix}-3&1\\-7&2\end{bmatrix}A$
$\begin{bmatrix}1&1\\0&-1\end{bmatrix}=\begin{bmatrix}-3&1\\7&-2\end{bmatrix}A$
$\begin{bmatrix}1&1\\0&1\end{bmatrix}=\begin{bmatrix}-3&1\\-7&2\end{bmatrix}A$
$\begin{bmatrix}1&1\\0&1\end{bmatrix}=\begin{bmatrix}4&-1\\-7&2\end{bmatrix}A$
$A^{-1}=\begin{bmatrix}4&-1\\-7&2\end{bmatrix}$
Question 3
$\begin{bmatrix}2&1\\1&1\end{bmatrix}$
Sol :
Let A=$\begin{bmatrix}2&1\\1&1\end{bmatrix}$
A=IA
$\begin{bmatrix}2&1\\1&1\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}A$
R1→R1-R2
$\begin{bmatrix}1&0\\1&1\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}A$
$\begin{bmatrix}1&0\\1&1\end{bmatrix}=\begin{bmatrix}1&-1\\-1&1\end{bmatrix}A$
$A^{-1}=\begin{bmatrix}1&-1\\-1&2\end{bmatrix}$
$\begin{bmatrix}\dfrac{1}{5}&\dfrac{2}{5}\\\dfrac{2}{5}&-\dfrac{1}{5}\end{bmatrix}$
Sol :
A=$\begin{bmatrix}\dfrac{1}{5}&\dfrac{2}{5}\\\dfrac{2}{5}&-\dfrac{1}{5}\end{bmatrix}$
A=IA
$\begin{bmatrix}\dfrac{1}{5}&\dfrac{2}{5}\\\dfrac{2}{5}&-\dfrac{1}{5}\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}A$
R1→5R1
$\begin{bmatrix}1&2\\ \dfrac{2}{5}&-\dfrac{1}{5}\end{bmatrix}=\begin{bmatrix}5&0\\0&1\end{bmatrix}A$
$\begin{bmatrix}1&2\\ 0&-1\end{bmatrix}=\begin{bmatrix}5&0\\-2&1\end{bmatrix}A$
$\begin{bmatrix}1&2\\ 0&1\end{bmatrix}=\begin{bmatrix}5&0\\2&-1\end{bmatrix}A$
R1→R1-2R2
$\begin{bmatrix}1&0\\ 0&1\end{bmatrix}=\begin{bmatrix}1&2\\2&-1\end{bmatrix}A$
$A^{-1}=\begin{bmatrix}1&2\\2&-1\end{bmatrix}$
$\left[\begin{array}{rr}1 & 2 \\ 2 & -1\end{array}\right]$
Sol :
$A=\left[\begin{array}{cc}1 & 2 \\ 2 & -1\end{array}\right]$
A=IA
$\left[\begin{array}{cc}1 & 2 \\ 2 & -1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]A$
R2→R2-2R1
$\left[\begin{array}{cc}1 & 2 \\ 0 & -5\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ -2 & 1\end{array}\right]A$
$\left[\begin{array}{cc}1 & 2 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ \frac{2}{5} & \frac{-1}{5}\end{array}\right]A$
R1→R1-2R2
$\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}\frac{1}{5} & \frac{2}{5} \\ \frac{2}{5} & -\frac{1}{5}\end{array}\right]A$
$A^{-1}=\left[\begin{array}{cc}\frac{1}{5} & \frac{2}{5} \\ \frac{2}{5} & -\frac{1}{5}\end{array}\right]$
Exercise 5.4 (Q6-Q10)
$\left[\begin{array}{ll}2 & 3 \\ 5 & 7\end{array}\right]$
Sol :
$A=\left[\begin{array}{ll}2 & 3 \\ 5 & 7\end{array}\right]$
A=IA
$\left[\begin{array}{ll}2 & 3 \\ 5 & 7\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] A$
R1↔R2
$\left[\begin{array}{ll}5 & 7 \\ 2 & 3\end{array}\right]=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] A$
R1→R1-2R2
$\left[\begin{array}{ll}1 & 1 \\ 2 & 3\end{array}\right]=\left[\begin{array}{cc}-2 & 1 \\ 1 & 0\end{array}\right]A$
R2→R2-2R1
$\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}-2 & 1 \\ 5 & -2\end{array}\right]A$
R1→R1-R2
$\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}-7 & 3 \\ 5 & -2\end{array}\right]A$
$A^{-1}=\left[\begin{array}{rr}-7 & 3 \\ 5 & -2\end{array}\right]$
$\left[\begin{array}{ll}4 & 5 \\ 3 & 4\end{array}\right]$
Sol :
$A=\left[\begin{array}{ll}4 & 5 \\ 3 & 4\end{array}\right]$
A=IA
$\left[\begin{array}{ll}4 & 5 \\ 3 & 4\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] A$
R1→R1-R2
$\left[\begin{array}{ll}1 & 1 \\ 3 & 4\end{array}\right]=\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right] A$
R2→R2-3R1
$\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}1 & -1 \\ -3 & 4\end{array}\right] A$
R1→R1-R2
$\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}4 & -5 \\ -3 & 4\end{array}\right]A$
$A^{-1}=\left[\begin{array}{cc}4 & -5 \\ -3 & 4\end{array}\right]$
$\left[\begin{array}{rr}3 & -1 \\ -4 & 2\end{array}\right]$
Sol :
$A=\left[\begin{array}{cc}3 & -1 \\ -4 & 2\end{array}\right]$
A=IA
$\left[\begin{array}{cc}3 & -1 \\ -4 & 2\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] A$
R1→R1-R2
$\left[\begin{array}{cc}-1 & 1 \\ -4 & 2\end{array}\right]=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right] A$
R1→-1R1
$\left[\begin{array}{cc}1 & -1 \\ -4 & 2\end{array}\right]=\left[\begin{array}{cc}-1 & -1 \\ 0 & 1\end{array}\right] A$
R2→R2+4R1
$\left[\begin{array}{cc}1 & -1 \\ 0 & -2\end{array}\right]=\left[\begin{array}{cc}-1 & -1 \\ -4 & -3\end{array}\right]A$
$R_{2} \rightarrow-\frac{1}{2} R_{2}$
$\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}-1 & -1 \\ 2 & \frac{3}{2}\end{array}\right]A$
R1→R1-R2
$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1 & \frac{1}{2} \\ 2 & \frac{3}{2}\end{array}\right]A$
$A^{-1}=\left[\begin{array}{ll}1 & \frac{1}{2} \\ 2 & \frac{3}{2}\end{array}\right]$
Question 9
$\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right]$
Sol :
$A=\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right]$
A=IA
$\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]A$
R1↔R2
$\left[\begin{array}{ll}4 & 2 \\ 2 & 1\end{array}\right]=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] A$
R1→R1-2R2
$\left[\begin{array}{cc}0 & 0 \\ 2 & 1\end{array}\right]=\left[\begin{array}{cc}-2 & 1 \\ 1 & 0\end{array}\right] A$
∴ A-1 का अस्तित्व नही हैं
Question 10
$\left[\begin{array}{rr}1 & -1 \\ 2 & 3\end{array}\right]$
Sol :
$A=\left[\begin{array}{rr}1 & -1 \\ 2 & 3\end{array}\right]$
A=IA
$\left[\begin{array}{rr}1 & -1 \\ 2 & 3\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]A$
R2→R2-2R1
$\left[\begin{array}{cc}1 & -1 \\ 0 & 5\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ -2 & 1\end{array}\right] A$
$R_{2} \rightarrow \frac{1}{5} R_{2}$
$\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ -\frac{2}{5} & \frac{1}{5}\end{array}\right]A$
R1→R1+R2
$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}\frac{3}{5} & \frac{1}{5} \\ -\frac{2}{5} & \frac{1}{5}\end{array}\right]A$
$A^{-1}=\left[\begin{array}{cc}\frac{3}{5} & \frac{1}{5} \\ -\frac{2}{5} & \frac{1}{5}\end{array}\right]$
$\left[\begin{array}{cr}2 & -3 \\ -1 & 2\end{array}\right]$
Sol :
$\left[\begin{array}{cc}10 & -2 \\ -5 & 1\end{array}\right]$
Sol :
Question 13
$\left[\begin{array}{ll}3 & 9 \\ 1 & 3\end{array}\right]$
Sol :
$A=\left[\begin{array}{ll}3 & 9 \\ 1 & 3\end{array}\right]$
A=IA
$\left[\begin{array}{ll}3 & 9 \\ 1 & 3\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] A$
R1→R1-3R2
$\left[\begin{array}{ll}0 & 0 \\ 1 & 3\end{array}\right]=\left[\begin{array}{cc}1 & -3 \\ 0 & 1\end{array}\right] A$
∴ A-1 का अस्तित्व नही हैं
Question 14
$\left[\begin{array}{rrr}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]$
Sol :
$A=\left[\begin{array}{rrr}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]$
A=IA
$\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]A$
R1→R1-R2
$\left[\begin{array}{ccc}5 & 1 & 0 \\ 2 & 0 & -1 \\ 0 & 1 & 3\end{array}\right]=\left[\begin{array}{ccc}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right]$
R1→R1-2R2
$\left[\begin{array}{ccc}1 & 1 & 2 \\ 2 & 0 & -1 \\ 0 & 1 & 3\end{array}\right]=\left[\begin{array}{ccc}-2 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right]A$
R2→R2-2R1
$\left[\begin{array}{ccc}1 & 1 & 2 \\ 0 & -2 & -5 \\ 0 & 1 & 3\end{array}\right]=\left[\begin{array}{ccc}-2 & 1 & 0 \\ 5 & -2 & 0 \\ 0 & 0 & 1\end{array}\right]a$
$R_{3} \rightarrow R_{3}+\frac{1}{2} R_{2}$
$\left[\begin{array}{ccc}1 & 1 & 2 \\ 0 & -2 & -5 \\ 0 & 0 & \frac{1}{2}\end{array}\right]=\left[\begin{array}{ccc}-2 & 1 & 0 \\ 5 & -2 & 0 \\ \frac{5}{2} & -1 & 1\end{array}\right]A$
R3→2R3
$\left[\begin{array}{ccc}1 & 1 & 2 \\ 0 & -2 & -5 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}-2 & 1 & 0 \\ 5 & -2 & 0 \\ 5 & -2 & 20\end{array}\right]A$
R2→R2+5R3
$\left[\begin{array}{ccc}1 & 1 & 2 \\ 0 & -2 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}-2 & 1 & 0 \\ 30 & -12 & 10 \\ 5 & -2 & 2\end{array}\right]A$
R1→R1-2R3
$\left[\begin{array}{ccc}1 & 1 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}-12 & 5 & -4 \\ 30 & -12 & 10 \\ 5 & -2 & 2\end{array}\right]A$
$R_{1} \rightarrow R_{1}+\frac{1}{2} R_{2}$
$\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}3 & -1 & 1 \\ 30 & -12 & 10 \\ 5 & -2 & 2\end{array}\right]A$
$R_{2} \rightarrow-\frac{1}{2} R_{2}$
$\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{cccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right] A$
$\therefore A^{-1}=\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$
Question 15
Using elementary transformations (operations) , find the inverse of the following matrices , if it exist
$\left[\begin{array}{rrr}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right]$
Sol :
Let $A=\left[\begin{array}{ccc}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right]$
A=IA
$\left[\begin{array}{ccc}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$\left[\begin{array}{ccc}3 & -2 & 2 \\ 2 & 2 & 3 \\ 2 & -3 & 3\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]A$
$\left[\begin{array}{rrr}1 & -4 & -1 \\ 2 & 2 & 3 \\ 2 & -3 & 3\end{array}\right]=\left[\begin{array}{rrr}0 & -1 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]A$
$R_{1}\leftrightarrow R_{3}$
$\left[\begin{array}{rrr}3 & -2 & 2 \\ 2 & 2 & 3 \\ 2 & -3 & 3\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]A$
$R_{1} \rightarrow R_{1}-R_{2}$
$\left[\begin{array}{rrr}1 & -4 & -1 \\ 2 & 2 & 3 \\ 2 & -3 & 3\end{array}\right]=\left[\begin{array}{rrr}0 & -1 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]A$
$R_{2} \rightarrow R_{2}-R_{3}$
$\left[\begin{array}{ccc}1 & -4 & -1 \\ 0 & 5 & 0 \\ 2 & -3 & 3\end{array}\right]=\left[\begin{array}{ccc}0 & -1 & 1 \\ -1 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]A$
$R_{3} \rightarrow R_{3}-2 R_{1}$
$\left[\begin{array}{rrr}1 & -4 & -1 \\ 0 & 5 & 0 \\ 0 & 5 & 5\end{array}\right]=\left[\begin{array}{rrr}0 & -1 & 1 \\ -1 & 1 & 0 \\ 1 & 2 & -2\end{array}\right]A$
$R_{3} \rightarrow R_{3}-R_{2}$
$\left[\begin{array}{ccc}1 & -4 & -1 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{array}\right]=\left[\begin{array}{rrr}0 & -1 & 1 \\ -1 & 1 & 0 \\ 2 & 1 & -2\end{array}\right]A$
$R_{3} \rightarrow \frac{1}{5} R_{3}$
$\left[\begin{array}{rrr}1 & -4 & -1 \\ 0 & 5 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{rrr}0 & -1 & 1 \\ -1 & 1 & 0 \\ \frac{2}{5} & \frac{1}{5} & -\frac{2}{5}\end{array}\right]A$
$R_{1} \rightarrow R_{1}+R_{3}$
$\left[\begin{array}{rrrr}1 & -4 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{rrr}\frac{2}{5} & -\frac{4}{5} & \frac{3}{5} \\ -1 & 1 & 0 \\ \frac{2}{5} & \frac{1}{5} & -\frac{2}{5}\end{array}\right]A$
$R_{2} \rightarrow \frac{1}{5} R_{2}$
$\left[\begin{array}{ccc}1 & -4 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}\frac{2}{5} & -\frac{4}{5} & \frac{3}{5} \\ -\frac{1}{5} & \frac{1}{5} & 0 \\ \frac{2}{5} & \frac{1}{5} & -\frac{2}{8}\end{array}\right]A$
$R_1 \longrightarrow R_{1}+4 R_{2}$
$\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{array}\right]=\left[\begin{array}{ccc}-\frac{2}{5} & 0 & \frac{3}{5} \\ -\frac{1}{5} & \frac{1}{5} & 0 \\ \frac{2}{5} & \frac{1}{5} & -\frac{2}{5}\end{array}\right]$
$\theta^{-1}=\left[\begin{array}{rrr}-\frac{2}{5} & 0 & \frac{3}{5} \\ -\frac{1}{5} & \frac{1}{5} & 0 \\ \frac{2}{5} & \frac{1}{5} & -\frac{2}{5}\end{array}\right]$
Question 16
Using elementary transformations (operations), find the inverse of the following matrices , if it exist
$\left[\begin{array}{rrr}2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1\end{array}\right]$
Sol :
$A=\left[\begin{array}{ccc}2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1\end{array}\right]$
A=IA
$\left[\begin{array}{ccc}2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]A$
$R_{1}\leftrightarrow R_{3}$
$\left[\begin{array}{ccc}-7 & 2 & 1 \\ 4 & -1 & 0 \\ 2 & 1 & 3\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right] A$
$R_{1} \rightarrow R_{1}+4 R_{3}$
$\left[\begin{array}{rrr}1 & 6 & 13 \\ 4 & -1 & 0 \\ 2 & 1 & 3\end{array}\right]=\left[\begin{array}{rrr}4 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]A$
$R_{2} \rightarrow R_{2}-2 R_{3}$
$\left[\begin{array}{rrr}1 & 6 & 13 \\ 0 & -3 & -6 \\ 2 & 1 & 3\end{array}\right]=\left[\begin{array}{rrr}4 & 0 & 1 \\ -2 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]A$
$R_{3} \rightarrow R_{3}-2 R_{1}$
$\left[\begin{array}{ccc}1 & 6 & 13 \\ 0 & -3 & -6 \\ 0 & -11 & -23\end{array}\right]=\left[\begin{array}{ccc}4 & 0 & 1 \\ -2 & 1 & 0 \\ -7 & 0 & -2\end{array}\right]A$
$R_{3} \rightarrow R_{3}-\frac{11}{3} R_{2}$
$\left[\begin{array}{ccc}1 & 6 & 13 \\ 0 & -3 & -6 \\ 0 & 0 & -1\end{array}\right]=\left[\begin{array}{ccc}4 & 0 & 1 \\ -2 & 1 & 0 \\ \frac{1}{3} & \frac{-11}{3} & -2\end{array}\right]A$
$R_{3} \rightarrow-1 R_{3}$
$\left[\begin{array}{ccc}1 & 6 & 13 \\ 0 & -3 & -6 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}4 & 0 & 1 \\ -2 & 1 & 0 \\ -\frac{1}{3} & \frac{11}{3} & 2\end{array}\right]A$
$R_{2} \rightarrow R_{2}+6 R_{3}$
$\left[\begin{array}{ccc}1 & 6 & 13 \\ 0 & -3 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}4 & 0 & 1 \\ -4 & 23 & 12 \\ -\frac{1}{3} & \frac{11}{3} & 2\end{array}\right]A$
$R_{1} \rightarrow R_{1}-13 R_{3}$
$\left[\begin{array}{ccc}1 & 6 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}\frac{25}{3} & -\frac{143}{3} & -25 \\ -4 & 23 & 12 \\ -\frac{1}{3} & \frac{11}{3} & 2\end{array}\right]$
$R_{1} \rightarrow R_{1}+2 R_{2}$
$\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{3} & \frac{-5}{3} & -1 \\ -4 & 23 & 12 \\ -\frac{1}{3} & \frac{11}{3} & 2\end{array}\right]A$
$R_{2} \rightarrow-\frac{1}{3} R_{2}$
$\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{3} & -\frac{5}{3} & -1 \\ \frac{4}{3} & -\frac{23}{3} & -4 \\ -\frac{1}{3} & \frac{11}{3} & 2\end{array}\right]$
$A^{-1}=\left[\begin{array}{ccc}\frac{1}{3} & \frac{-5}{3} & -1 \\ \frac{4}{3} & -\frac{23}{3} & -4 \\ -\frac{1}{3} & \frac{11}{3} & 2\end{array}\right]$
Question 17
Using elementary transformations (operations), find the inverse of the following matrices , if it exist
$\left[\begin{array}{rrr}2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3\end{array}\right]$
Sol :
$A=\left[\begin{array}{ccc}2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3\end{array}\right]$
A=IA
$\left[\begin{array}{ccc}2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A$
$R_{1}\leftrightarrow R_{2}$
$\left[\begin{array}{rrr}-5 & 3 & 1 \\ 2 & -1 & 3 \\ -3 & 2 & 3\end{array}\right]=\left[\begin{array}{rrr}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right]A$
$\left[\begin{array}{ccc}1 & 0 & 10 \\ 2 & -1 & 3 \\ -3 & 2 & 3\end{array}\right]=\left[\begin{array}{ccc}3 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right] A$
$R_{2} \rightarrow R_{2}-2 R_{1}$
$\left[\begin{array}{ccc}1 & 0 & 10 \\ 0 & -1 & -17 \\ -3 & 2 & 3\end{array}\right]=\left[\begin{array}{ccc}3 & 1 & 0 \\ -5 & -2 & 0 \\ 0 & 0 & 1\end{array}\right]A$
$R_{3} \rightarrow R_{3}+3 R_{1}$
$\left[\begin{array}{ccc}1 & 0 & 10 \\ 0 & -1 & -17 \\ 0 & 0 & 33\end{array}\right]=\left[\begin{array}{ccc}3 & 1 & 0 \\ -5 & -2 & 0 \\ 9 & 3 & 1\end{array}\right]A$
$R_{3} \rightarrow R_{3}+2 R_{2}$
$\left[\begin{array}{ccc}1 & 0 & 10 \\ 0 & -1 & -17 \\ 0 & 0 & -1\end{array}\right]=\left[\begin{array}{ccc}3 & 1 & 0 \\ -5 & -2 & 0 \\ -1 & -1 & 1\end{array}\right]A$
$R_{3} \rightarrow-1 R_{3}$
$\left[\begin{array}{ccc}1 & 0 & 10 \\ 0 & -1 & -17 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}3 & 1 & 0 \\ -5 & -2 & 0 \\ 1 & 1 & -1\end{array}\right]A$
$R_{2} \rightarrow R_{2}+17 R_{3}$
$\left[\begin{array}{ccc}1 & 0 & 10 \\ 0 & -1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}3 & 1 & 0 \\ 12 & 15 & -17 \\ 1 & 1 & -1\end{array}\right]A$
$R_{1} \rightarrow R_{1}-10 R_{3}$
$\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}-7 & -9 & 10 \\ 12 & 15 & -17 \\ 1 & 1 & -1\end{array}\right]A$
$R_{2} \rightarrow-1 R_{2}$
$\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}-7 & -9 & 10 \\ -12 & -15 & 17 \\ 2 & 1 & -1\end{array}\right]A$
$A^{-1}=\left[\begin{array}{rrr}-7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1\end{array}\right]$