Question

Find using elementary the inverse of A = 1 0 0 3 3 0 5 2 -1​

Answer 1

Answer:

I THINK IT IS VERY DIFFICULT TO SAY FOR ME.

Answer 2

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

Given matrix,

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

We know,

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

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

$\red{\rm :\longmapsto\:OP \: R_2 \:\to \: R_2 \: - 3R_1}$

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

$\red{\rm :\longmapsto\:OP \: R_3 \:\to \: R_3 \: - 5R_1}$

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

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

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

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

$\rm :\longmapsto\:\begin{gathered}\sf \left[\begin{array}{ccc}1&0&0\\0&1&1\\0&0& - 3\end{array}\right]\end{gathered} = \begin{gathered}\sf\left[\begin{array}{ccc}1&0&0\\ 2&1& - 1\\ - 9& - 2&3\end{array}\right]\end{gathered}A$

$\red{\rm :\longmapsto\:OP \: R_3 \: \to \: - \: \dfrac{1}{3} \: R_3}$

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

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

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

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

Hence,

$\bf\implies \: {A}^{ - 1} = \begin{gathered}\sf\left[\begin{array}{ccc}1&0&0\\- 1& \dfrac{1}{3} &0 \\3& \dfrac{2}{3} & - 1\end{array}\right]\end{gathered}$