Show that the matrix

Solution:

We have, $A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$

$\therefore A^{2}=A A$

$\Rightarrow A^{2}=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$

$\Rightarrow A^{2}=\left[\begin{array}{ll}4+3 & 6+6 \\ 2+2 & 3+4\end{array}\right]$

$\Rightarrow A^{2}=\left[\begin{array}{ll}7 & 12 \\ 4 & 7\end{array}\right]$

$\Rightarrow A^{3}=A^{2} A$

$\Rightarrow A^{3}=\left[\begin{array}{ll}7 & 12 \\ 4 & 7\end{array}\right]\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$

$\Rightarrow A^{3}=\left[\begin{array}{cc}14+12 & 21+24 \\ 8+7 & 12+14\end{array}\right]$

$\Rightarrow A^{3}=\left[\begin{array}{ll}26 & 45 \\ 15 & 26\end{array}\right]$

Now, $A^{3}-4 A^{2}+A$

$\Rightarrow A^{3}-4 A^{2}+A=\left[\begin{array}{ll}26 & 45 \\ 15 & 26\end{array}\right]-4\left[\begin{array}{cc}7 & 12 \\ 4 & 7\end{array}\right]+\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$

$\Rightarrow A^{3}-4 A^{2}+A=\left[\begin{array}{ll}26 & 45 \\ 15 & 26\end{array}\right]-\left[\begin{array}{cc}28 & 48 \\ 16 & 28\end{array}\right]+\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$

$\Rightarrow A^{3}-4 A^{2}+A=\left[\begin{array}{ll}26-28+2 & 45-48+3 \\ 15-16+1 & 26-28+2\end{array}\right]$

$\Rightarrow A^{3}-4 A^{2}+A=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=O$

Hence proved.