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

By Mary

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

### 1 thought on “Find using elementary, the inverted of <br />A = 1 2 3<br /> 0 2 4<br /> 0 0 5​”

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,

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