Question

If $A=\left[\begin{array}{ccc}2& 0& 1\\ 2& 1& 3\\ 1& -1& 0\end{array}\right]$, find A² − 5A + 4I and hence find a matrix X such that A² − 5A + 4I + X = 0.

Answer 1

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

$$\begin{gathered} A^2=\left[\begin{array}{ccc}2& 0& 1\\ 2& 1& 3\\ 1& -1& 0\end{array}\right]\left[\begin{array}{ccc}2& 0& 1\\ 2& 1& 3\\ 1& -1& 0\end{array}\right] \\ =\left[\begin{array}{ccc}4+0+1& 0+0-1& 2+0+0\\ 4+2+3& 0+1-3& 2+3+0\\ 2-2+0& 0-1-0& 1-3+0\end{array}\right] \\ =\left[\begin{array}{ccc}5& -1& 2\\ 9& -2& 5\\ 0& -1& -2\end{array}\right] \end{gathered}$$

Now,
$$\begin{gathered} A^2-5A+4I=\left[\begin{array}{ccc}5& -1& 2\\ 9& -2& 5\\ 0& -1& -2\end{array}\right]-5\left[\begin{array}{ccc}2& 0& 1\\ 2& 1& 3\\ 1& -1& 0\end{array}\right]+4\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right] \\ =\left[\begin{array}{ccc}5-10+4& -1-0+0& 2-5+0\\ 9-10+0& -2-5+4& 5-15+0\\ 0-5+0& -1+5+0& -2-0+4\end{array}\right] \\ =\left[\begin{array}{ccc}-1& -1& -3\\ -1& -3& -10\\ -5& 4& 2\end{array}\right] \end{gathered}$$

Now, A² − 5A + 4I + X = 0
⇒ X = −(A² − 5A + 4I)
$$\begin{gathered} \therefore X=-\left[\begin{array}{ccc}-1& -1& -3\\ -1& -3& -10\\ -5& 4& 2\end{array}\right] \\ =\left[\begin{array}{ccc}1& 1& 3\\ 1& 3& 10\\ 5& -4& -2\end{array}\right] \end{gathered}$$