Question
Find the matrix $X$ such that $A X=I$ where $A=\left[\begin{array}{cc}6 & 17 \\ 1 & 3\end{array}\right]$
✓
Answer
Given, $A X=1$
$\therefore\left[\begin{array}{cc}6 & 17 \\1 & 3\end{array}\right] X =\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right]$
Applying $R_1 \leftrightarrow R_2$, we get
$\left[\begin{array}{cc}1 & 3 \\6 & 17\end{array}\right] X=\left[\begin{array}{ll}0 & 1 \\1 & 0\end{array}\right]$
Applying $R_2 \rightarrow R_2-6 R_1$, we get
$\left[\begin{array}{cc}1 & 3 \\0 & -1\end{array}\right] X=\left[\begin{array}{cc}0 & 1 \\1 & -6\end{array}\right]$
Applying $R_1 \rightarrow R_1+3 R_2$, we get
$\left[\begin{array}{cc}1 & 0 \\0 & -1\end{array}\right] X=\left[\begin{array}{cc}3 & -17 \\1 & -6\end{array}\right]$
Applying $R_2 \rightarrow(-1) R_2$, we get
$\begin{array}{l}{\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right] X=\left[\begin{array}{cc}3 & -17 \\-1 & 6\end{array}\right]}\end{array} $
$\therefore X=\left[\begin{array}{cc}3 & -17 \\-1 & 6\end{array}\right]$
$\therefore\left[\begin{array}{cc}6 & 17 \\1 & 3\end{array}\right] X =\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right]$
Applying $R_1 \leftrightarrow R_2$, we get
$\left[\begin{array}{cc}1 & 3 \\6 & 17\end{array}\right] X=\left[\begin{array}{ll}0 & 1 \\1 & 0\end{array}\right]$
Applying $R_2 \rightarrow R_2-6 R_1$, we get
$\left[\begin{array}{cc}1 & 3 \\0 & -1\end{array}\right] X=\left[\begin{array}{cc}0 & 1 \\1 & -6\end{array}\right]$
Applying $R_1 \rightarrow R_1+3 R_2$, we get
$\left[\begin{array}{cc}1 & 0 \\0 & -1\end{array}\right] X=\left[\begin{array}{cc}3 & -17 \\1 & -6\end{array}\right]$
Applying $R_2 \rightarrow(-1) R_2$, we get
$\begin{array}{l}{\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right] X=\left[\begin{array}{cc}3 & -17 \\-1 & 6\end{array}\right]}\end{array} $
$\therefore X=\left[\begin{array}{cc}3 & -17 \\-1 & 6\end{array}\right]$
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