Question

Show that the matrix $A=\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]$ satisfies the equation $A^3-4A^2+A=0$.

Answer 1

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

$A^2=AA$

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

$=\left[\begin{array}{cc}4+3& 6+6\\ 2+2& 3+4\end{array}\right]$

$=\left[\begin{array}{cc}7& 12\\ 4& 7\end{array}\right]$

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

$=\left[\begin{array}{cc}7& 12\\ 4& 7\end{array}\right]\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]$

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

$=\left[\begin{array}{cc}26& 45\\ 15& 26\end{array}\right]$

Now,

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

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

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

$=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

$\therefore$ $A^3-4A^2+A=0$