Question

Find $A^{2}-5A+6I$ if A=$\begin{bmatrix}2&0&1\\\4&5&3\\\6&5&1\end{bmatrix}$

Answer 1

We have$A^{2}=A\times A$

$A^{2}$=$A\times A$=$\begin{bmatrix}2&0&1\\\4&5&3\\\6&5&1\end{bmatrix}$
=[2(2)+0(2)+1(1) 2(0)+0(1)+1(-1) 2(1)+0(3)+1(0) 2(2)+1(2)+3(1) 2(0)+1(1)+3(-1) 2(2)+1(3)+3(0) 1(2)+(-1)(2)+0(1) 1(0)+(-1)(1)+0(-1) 1(1)+(-1)(3)+0(0)]
=[4+0+1 0+0-1 2+0+0 4+2+3 0+1-3 2+3+0 2-2+0 0-1+0 1-3+0]
=[5-12 9-25 0-1-2]

∴ $A^{2}$-5A+6I
=$\begin{bmatrix}2&0&1\\\4&5&3\\\6&5&1\end{bmatrix}$
=[5-12 9-25 0-1-2] -[1005 10515 5-50] +[600 060 006]
=[-5+6 -1+0 -3+0 9-10 -2-5 5-15 0-5 -1+5 -2-0] +[600 060 006]
=[-5-1-3 -1-7-10 -54-2]+[600 060 006]
=[-5+6 -1+0 -3+0 -1+0 -7+6 -10+0 -5+0 4+0 -2+6]
=[1-1-3 -1-1-10 -544]