Question

Using the
the
inverse of the matrices
elementary transformation
transformation find the
E
2 0 -1
5 1 0
0 1 3​

Answer 1

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

Given matrix is

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

Let assume that

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

Using Elementary Row Transformation Method, we have

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

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

$\rm :\longmapsto\:OPR_1 \: \to \: 3R_1$

$\rm :\longmapsto\:\begin{gathered}\sf \left[\begin{array}{ccc}6&0& - 3\\5& 1&0\\ 0&1&3\end{array}\right]\end{gathered} = \begin{gathered}\sf \left[\begin{array}{ccc}3&0&0\\0& 1&0\\ 0&0&1\end{array}\right]\end{gathered}A$

$\rm :\longmapsto\:OPR_1 \: \to \: R_1 -R_2$

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

$\rm :\longmapsto\:OPR_2 \: \to \: R_2 -5R_1$

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

$\rm :\longmapsto\:OPR_2 \: \to \dfrac{1}{6} \: R_2$

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

$\rm :\longmapsto\:OP \: R_1 \: \to \: R_1 + R_2$

$\rm :\longmapsto\:OP \: R_3 \: \to \: R_3 - R_2$

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

$\rm :\longmapsto\:OPR_3 \: \to \: 2R_3$

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

$\rm :\longmapsto\:OP \: R_1 \: \to \: R_1 + \dfrac{1}{2} R_3$

$\rm :\longmapsto\:OP \: R_2 \: \to \: R_2 - \dfrac{5}{2} R_3$

$\rm :\longmapsto\:\begin{gathered}\sf \left[\begin{array}{ccc}1& 0& 0 \\0& 1& 0 \\ 0&0& 1 \end{array}\right]\end{gathered} = \begin{gathered}\sf \left[\begin{array}{ccc}3& - 1& 1\\ - 15& 6& - 5\\ 5& - 2&2\end{array}\right]\end{gathered}A$

We know,

$\underbrace{ \boxed{ \bf \: {AA}^{ - 1} = {A}^{ - 1}A = I}}$

So, On comparing with this, we get

$\red{\rm :\longmapsto\: {A}^{ - 1} = \begin{gathered}\sf \left[\begin{array}{ccc}3& - 1& 1\\ - 15& 6& - 5\\ 5& - 2&2\end{array}\right]\end{gathered}}$

Additional Information :-

Let A and B are square matrices of order n, then

$\rm :\longmapsto\: |AB| = |A| \: |B|$

$\rm :\longmapsto\: |kA| = {k}^{n} |A|$

$\rm :\longmapsto\: | {A}^{T} | = |A|$

$\rm :\longmapsto\: |I| = 1$

$\rm :\longmapsto\: |adj \: A| = { |A| }^{n - 1}$

$\rm :\longmapsto\: |A \: adj \: A| = { |A| }^{n}$

$\rm :\longmapsto\: | {A}^{ - 1} | = \dfrac{1}{ |A| }$