KC Sinha Mathematics Solution Class 12 Chapter 5 आव्यूह ( Matrices ) Exercise 5.4 (Q16-Q17)

Ex-5.4

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Exercise 5.4

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

$\left[\begin{array}{ccc}2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]A$

$R_{1}\leftrightarrow R_{3}$

$\left[\begin{array}{ccc}-7 & 2 & 1 \\ 4 & -1 & 0 \\ 2 & 1 & 3\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right] A$

$R_{1} \rightarrow R_{1}+4 R_{3}$

$\left[\begin{array}{rrr}1 & 6 & 13 \\ 4 & -1 & 0 \\ 2 & 1 & 3\end{array}\right]=\left[\begin{array}{rrr}4 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]A$

$R_{2} \rightarrow R_{2}-2 R_{3}$

$\left[\begin{array}{rrr}1 & 6 & 13 \\ 0 & -3 & -6 \\ 2 & 1 & 3\end{array}\right]=\left[\begin{array}{rrr}4 & 0 & 1 \\ -2 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]A$

$R_{3} \rightarrow R_{3}-2 R_{1}$

$\left[\begin{array}{ccc}1 & 6 & 13 \\ 0 & -3 & -6 \\ 0 & -11 & -23\end{array}\right]=\left[\begin{array}{ccc}4 & 0 & 1 \\ -2 & 1 & 0 \\ -7 & 0 & -2\end{array}\right]A$

$R_{3} \rightarrow R_{3}-\frac{11}{3} R_{2}$

$\left[\begin{array}{ccc}1 & 6 & 13 \\ 0 & -3 & -6 \\ 0 & 0 & -1\end{array}\right]=\left[\begin{array}{ccc}4 & 0 & 1 \\ -2 & 1 & 0 \\ \frac{1}{3} & \frac{-11}{3} & -2\end{array}\right]A$

$R_{3} \rightarrow-1 R_{3}$

$\left[\begin{array}{ccc}1 & 6 & 13 \\ 0 & -3 & -6 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}4 & 0 & 1 \\ -2 & 1 & 0 \\ -\frac{1}{3} & \frac{11}{3} & 2\end{array}\right]A$

$R_{2} \rightarrow R_{2}+6 R_{3}$

$\left[\begin{array}{ccc}1 & 6 & 13 \\ 0 & -3 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}4 & 0 & 1 \\ -4 & 23 & 12 \\ -\frac{1}{3} & \frac{11}{3} & 2\end{array}\right]A$

$R_{1} \rightarrow R_{1}-13 R_{3}$

$\left[\begin{array}{ccc}1 & 6 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}\frac{25}{3} & -\frac{143}{3} & -25 \\ -4 & 23 & 12 \\ -\frac{1}{3} & \frac{11}{3} & 2\end{array}\right]$

$R_{1} \rightarrow R_{1}+2 R_{2}$

$\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{3} & \frac{-5}{3} & -1 \\ -4 & 23 & 12 \\ -\frac{1}{3} & \frac{11}{3} & 2\end{array}\right]A$

$R_{2} \rightarrow-\frac{1}{3} R_{2}$

$\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{3} & -\frac{5}{3} & -1 \\ \frac{4}{3} & -\frac{23}{3} & -4 \\ -\frac{1}{3} & \frac{11}{3} & 2\end{array}\right]$

$A^{-1}=\left[\begin{array}{ccc}\frac{1}{3} & \frac{-5}{3} & -1 \\ \frac{4}{3} & -\frac{23}{3} & -4 \\ -\frac{1}{3} & \frac{11}{3} & 2\end{array}\right]$