Question

Find using elementary, the inverted of
A = 1 2 3
0 2 4
0 0 5​

Answer 1

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

Given matrix is

$\rm :\longmapsto\:\begin{gathered}\sf A=\left[\begin{array}{ccc}1&2&3\\0&2&4\\0&0&5\end{array}\right]\end{gathered}$

We know that,

$\red{\rm :\longmapsto\:A = IA}$

$\rm :\longmapsto\:\begin{gathered}\sf \left[\begin{array}{ccc}1&2&3\\0&2&4\\0&0&5\end{array}\right]\end{gathered} = \begin{gathered}\sf\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]\end{gathered}A$

$\red{\rm :\longmapsto\:OP \: R_1 \: \to \:R_1 - R_2}$

$\rm :\longmapsto\:\begin{gathered}\sf \left[\begin{array}{ccc}1&0& - 1\\0&2&4\\0&0&5\end{array}\right]\end{gathered} = \begin{gathered}\sf\left[\begin{array}{ccc}1& - 1&0\\0&1&0\\0&0&1\end{array}\right]\end{gathered}A$

$\red{\rm :\longmapsto\:OP \: R_2 \: \to \: \dfrac{1}{2}R_2}$

$\rm :\longmapsto\:\begin{gathered}\sf \left[\begin{array}{ccc}1&0& - 1\\0&1&2\\0&0&5\end{array}\right]\end{gathered} = \begin{gathered}\sf\left[\begin{array}{ccc}1& - 1&0\\0& \dfrac{1}{2} &0\\0&0&1\end{array}\right]\end{gathered}A$

$\red{\rm :\longmapsto\:OP \: R_3 \: \to \: \dfrac{1}{5}R_3}$

$\rm :\longmapsto\:\begin{gathered}\sf \left[\begin{array}{ccc}1&0& - 1\\0&1&2\\0&0&1\end{array}\right]\end{gathered} = \begin{gathered}\sf\left[\begin{array}{ccc}1& - 1&0\\0& \dfrac{1}{2} &0\\0&0 & \dfrac{1}{5} \end{array}\right]\end{gathered}A$

$\red{\rm :\longmapsto\:OP \: R_2 \: \to \:R_2 - 2R_3}$

$\rm :\longmapsto\:\begin{gathered}\sf \left[\begin{array}{ccc}1&0& - 1\\0&1&0\\0&0&1\end{array}\right]\end{gathered} = \begin{gathered}\sf\left[\begin{array}{ccc}1& - 1&0\\0& \dfrac{1}{2} & - \dfrac{2}{5} \\0&0 & \dfrac{1}{5} \end{array}\right]\end{gathered}A$

$\red{\rm :\longmapsto\:OP \: R_1 \: \to \:R_1 + R_3}$

$\rm :\longmapsto\:\begin{gathered}\sf \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]\end{gathered} = \begin{gathered}\sf\left[\begin{array}{ccc}1& - 1& \dfrac{1}{5} \\0& \dfrac{1}{2} & - \dfrac{2}{5} \\0&0 & \dfrac{1}{5} \end{array}\right]\end{gathered}A$

$\bf\implies \:A {A}^{ - 1} = I$

Hence,

$\bf\implies \: {A}^{ - 1} = \begin{gathered}\sf\left[\begin{array}{ccc}1& - 1& \dfrac{1}{5} \\0& \dfrac{1}{2} & - \dfrac{2}{5} \\0&0 & \dfrac{1}{5} \end{array}\right]\end{gathered}$