Question

If $A=\left[\begin{array}{ccc}2& 0& 1\\ 2& 1& 3\\ 1& -1& 0\end{array}\right]$ find $A^2-5A+4I$ and hence find a matrix X such that $A^2-5A+4I+X=0$.

Answer 1

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

So,

$A^2-5A+4I$

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

$=\left[\begin{array}{ccc}2\left(2\right)+0+1\left(1\right)& 2\left(0\right)+0+1(-1)& 2\left(1\right)+0+0\\ 2\left(2\right)+1\left(2\right)-3\left(1\right)& 2\left(0\right)+1\left(1\right)-3(-1)& 2\left(1\right)+1\left(3\right)+3\left(0\right)\\ 1\left(2\right)+(-1)\left(2\right)+0& 0+1(-1)+0& 1\left(1\right)+(-1)\left(3\right)+0\end{array}\right]\left[\begin{array}{ccc}-10& 0& -5\\ -10& -5& -15\\ -5& 5& 0\end{array}\right]+\left[\begin{array}{ccc}4& 0& 0\\ 0& 4& 0\\ 0& 0& 4\end{array}\right]$

$=\left[\begin{array}{ccc}5-10+4& -1+0+0& 2-5+0\\ 9-10+0& -2-5+4& 5-15+0\\ 0-5+0& -1+5+0& -2+0+4\end{array}\right]$

So,

$A^2-5A+4I=\left[\begin{array}{ccc}-1& -1& -3\\ -1& -3& -10\\ -5& 4& 2\end{array}\right]$

So, Let, $A^2-5A+4I+x=0$

So, $X=-(A^2-5A+4I)$

$X=-1\left[\begin{array}{ccc}-1& -1& -3\\ -1& -3& -10\\ -5& 4& 2\end{array}\right]=\left[\begin{array}{ccc}1& 1& 3\\ 1& 3& 10\\ 5& -4& -2\end{array}\right]$