If A = [ ll 4 & 5 2 & 1 ], show that A ^ -1 =1 6 ( A -5 I ).

Question

If $A =\left[\begin{array}{ll}4 & 5 \\ 2 & 1\end{array}\right]$, show that $A ^{-1}=\frac{1}{6}( A -5 I )$.

✓

Answer

$ |A|=\left|\begin{array}{ll} 4 & 5 \\ 2 & 1 \end{array}\right|=4-10=-6 \neq 0$
$\therefore A ^{-1}$ exists.
Consider $AA ^{-1}=1$
$\therefore\left[\begin{array}{ll} 4 & 5 \\2 & 1 \end{array}\right] A ^{-1}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1
\end{array}\right]$
By $\left(\frac{1}{4}\right) R_1$, we get,
$ \left[\begin{array}{ll} 1 & \frac{5}{4} \\ 2 & 1 \end{array}\right] A ^{-1}=\left[\begin{array}{ll} \frac{1}{4} & 0 \\
0 & 1 \end{array}\right] $
By $R_2-2 R_1$ we get,
$ \left[\begin{array}{cc} 1 & \frac{5}{4} \\ 0 & -\frac{3}{2} \end{array}\right] A ^{-1}=\left[\begin{array}{cc} \frac{1}{4} & 0 \\ -\frac{1}{2} & 1 \end{array}\right]$
By $\left(-\frac{2}{3}\right) R _2$, we get,
$ {\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] A ^{-1}=\left[\begin{array}{cc} -\frac{1}{6} & \frac{5}{6} \\ \frac{1}{3} & -\frac{2}{3}\ \end{array}\right]} $
$ \therefore A ^{-1}=\frac{1}{6}\left[\begin{array}{cc} -1 & 5 \\ 2 & -4 \end{array}\right] \ldots .(1) $
$ \frac{1}{6}( A -5 I )=\frac{1}{6}\left\{\left[\begin{array}{ll} 4 & 5 \\ 2 & 1 \end{array}\right]-5\left[\begin{array}{ll}
1 & 0 \\ 0 & 1 \end{array}\right]\right\} $
$ =\frac{1}{6}\left\{\left[\begin{array}{ll} 4 & 5 \\ 2 & 1 \end{array}\right]-\left[\begin{array}{ll} 5 & 0 \\ 0 & 5
\end{array}\right]\right\} $
$ =\frac{1}{6}\left[\begin{array}{cc} -1 & 5 \\ 2 & -4 \end{array}\right]$
From $(1)$ and $(2)$, we get $A ^{-1}=\frac{1}{6}( A -5 I )$