Question

If $A=\left[\begin{array}{ccc}3& 2& 0\\ 1& 4& 0\\ 0& 0& 5\end{array}\right]$.Show that $A^2-7A+10=0$

Answer 1

$A^2=A.A=\left[\begin{array}{ccc}3& 2& 0\\ 1& 4& 0\\ 0& 0& 5\end{array}\right]\left[\begin{array}{ccc}3& 2& 0\\ 1& 4& 0\\ 0& 0& 5\end{array}\right]$

$=\left[\begin{array}{ccc}3\times 3+2\times 1+0\times 0& 3\times 2+2\times 4+0\times 0& 3\times 0+2\times 0+0\times 5\\ 1\times 3+4\times 1+0\times 0& 1\times 2+4\times 4+0\times 0& 1\times 0+4\times 0+0\times 5\\ 0\times 3+0\times 1+5\times 0& 0\times 2+0\times 4+5\times 0& 0\times 0+0\times 0+5\times 5\end{array}\right]$

$=\left[\begin{array}{ccc}9+2+0& 6+8+0& 0+0+0\\ 3+4+0& 2+16+0& 0+0+0\\ 0+0+0& 0+0+0& 0+0+25\end{array}\right]$

$A^2=\left[\begin{array}{ccc}11& 14& 0\\ 7& 18& 0\\ 0& 0& 25\end{array}\right]$

$-7A=-7\left[\begin{array}{ccc}3& 2& 0\\ 1& 4& 0\\ 0& 0& 5\end{array}\right]$

$=\left[\begin{array}{ccc}-21& -14& 0\\ -7& -28& 0\\ 0& 0& -35\end{array}\right]$

$10=10I_3=10\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$=\left[\begin{array}{ccc}10& 0& 0\\ 0& 10& 0\\ 0& 0& 10\end{array}\right]$

Now $A^2+(-7A)+10=\left[\begin{array}{ccc}11& 14& 0\\ 7& 18& 0\\ 0& 0& 25\end{array}\right]+\left[\begin{array}{ccc}-21& -14& 0\\ -7& -28& 0\\ 0& 0& -35\end{array}\right]\left[\begin{array}{ccc}10& 0& 0\\ 0& 10& 0\\ 0& 0& 10\end{array}\right]$

$=\left[\begin{array}{ccc}11-21+10& 14-14+0& 0+0+0\\ 7-7+0& 18-28+10& 0+0+0\\ 0+0+0& 0+0+0& 25-35+10\end{array}\right]$

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

$=0$

Hence Proved.