Question

use only elementry operation and find the inverse of Matrix given below.
A=
3 3 4
2 -3 4
0 -1 1​

Answer 1

$\large\underline{\sf{Given- }}$

$\: \: \: \: \: \: \: \: \: \: \: \bull \: \: \: \begin{gathered}\sf A=\left[\begin{array}{ccc}3&3&4\\2& - 3&4\\0&-1&1\end{array}\right]\end{gathered}$

$\large\underline{\sf{To\:Find - }}$

$\: \: \: \: \: \: \: \: \: \: \: \: \bull \: \sf \: {A}^{ - 1} \: using \: elementary \: trasformation \: method$

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:Let \: A \: = \: IA$

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

$\rm :\longmapsto\: \bf \: OP \: R_1 \to R_1-R_2$

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

$\rm :\longmapsto\: \bf \: OP \: R_2 \to R_2-2R_1$

$\begin{gathered}\sf \left[\begin{array}{ccc}1& 6&0\\0& - 15&4\\0&-1&1\end{array}\right]\end{gathered} = \begin{gathered}\sf \left[\begin{array}{ccc}1& - 1&0\\ - 2& 3&0\\0&0&1\end{array}\right]\end{gathered}A$

$\rm :\longmapsto\: \bf \: OP \: R_2 \longleftrightarrow R_3$

$\begin{gathered}\sf \left[\begin{array}{ccc}1& 6&0\\0& - 1&1\\0& - 15&4\end{array}\right]\end{gathered} = \begin{gathered}\sf \left[\begin{array}{ccc}1& - 1&0\\ 0& 0&1\\ - 2&3&0\end{array}\right]\end{gathered}A$

$\rm :\longmapsto\: \bf \: OP \: R_2 \to -R_2$

$\begin{gathered}\sf \left[\begin{array}{ccc}1& 6&0\\0& 1& - 1\\0& - 15&4\end{array}\right]\end{gathered} = \begin{gathered}\sf \left[\begin{array}{ccc}1& - 1&0\\ 0& 0& - 1\\ - 2&3&0\end{array}\right]\end{gathered}A$

$\rm :\longmapsto\: \bf \: OP \: R_1 \to R_1 - 6R_2 \\ \rm :\longmapsto\: \bf \: OP \: R_3 \to R_3 + 15R_2$

$\begin{gathered}\sf \left[\begin{array}{ccc}1& 0& 6\\0& 1& - 1\\0&0& - 11\end{array}\right]\end{gathered} = \begin{gathered}\sf \left[\begin{array}{ccc}1& - 1& 6\\ 0& 0& - 1\\ - 2&3& - 15\end{array}\right]\end{gathered}A$

$\rm :\longmapsto\: \bf \: OP \: R_3 \to - \dfrac{1}{11} R_3$

$\begin{gathered}\sf \left[\begin{array}{ccc}1& 0& 6\\0& 1& - 1\\0&0&1\end{array}\right]\end{gathered} = \begin{gathered}\sf \left[\begin{array}{ccc}1& - 1& 6\\ 0& 0& - 1\\ \dfrac{2}{11} & - \dfrac{3}{11} &\dfrac{15}{11} \end{array}\right]\end{gathered}A$

$\rm :\longmapsto\: \bf \: OP \: R_2 \to R_2 + R_3 \\ \rm :\longmapsto\: \bf \: OP \: R_1 \to R_1 - 6R_3$

$\begin{gathered}\sf \left[\begin{array}{ccc}1& 0& 0\\0& 1& 0\\0&0&1\end{array}\right]\end{gathered} = \begin{gathered}\sf \left[\begin{array}{ccc} - \dfrac{1}{11} & \dfrac{7}{11} & - \dfrac{24}{11} \\ \dfrac{2}{11} & - \dfrac{3}{11} & \dfrac{4}{11} \\ \dfrac{2}{11} & - \dfrac{3}{11} &\dfrac{15}{11} \end{array}\right]\end{gathered}A$

$\rm :\longmapsto\:We\: know,\: A {A}^{ - 1} = I$

$\bf\implies \: {A}^{ - 1} = \begin{gathered}\sf \left[\begin{array}{ccc} - \dfrac{1}{11} & \dfrac{7}{11} & - \dfrac{24}{11} \\ \dfrac{2}{11} & - \dfrac{3}{11} & \dfrac{4}{11} \\ \dfrac{2}{11} & - \dfrac{3}{11} &\dfrac{15}{11} \end{array}\right]\end{gathered}$