Question

If $A=\left[\begin{array}{ccc}1& 0& -3\\ 2& 1& 3\\ 0& 1& 1\end{array}\right]$, then verify that A² + A = A(A + I), where I is the identity matrix.

Answer 1

To verify: A² + A = A(A + I),

Given: $A=\left[\begin{array}{ccc}1& 0& -3\\ 2& 1& 3\\ 0& 1& 1\end{array}\right]$
$$\begin{gathered} A^2=\left[\begin{array}{ccc}1& 0& -3\\ 2& 1& 3\\ 0& 1& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& -3\\ 2& 1& 3\\ 0& 1& 1\end{array}\right] \\ =\left[\begin{array}{ccc}1+0+0& 0+0-3& -3+0-3\\ 2+2+0& 0+1+3& -6+3+3\\ 0+2+0& 0+1+1& 0+3+1\end{array}\right] \\ =\left[\begin{array}{ccc}1& -3& -6\\ 4& 4& 0\\ 2& 2& 4\end{array}\right] \end{gathered}$$

LHS:
$$\begin{gathered} A^2+A=\left[\begin{array}{ccc}1& -3& -6\\ 4& 4& 0\\ 2& 2& 4\end{array}\right]+\left[\begin{array}{ccc}1& 0& -3\\ 2& 1& 3\\ 0& 1& 1\end{array}\right] \\ =\left[\begin{array}{ccc}1+1& -3+0& -6-3\\ 4+2& 4+1& 0+3\\ 2+0& 2+1& 4+1\end{array}\right] \\ =\left[\begin{array}{ccc}2& -3& -9\\ 6& 5& 3\\ 2& 3& 5\end{array}\right] \end{gathered}$$

RHS:
$$\begin{gathered} A+I=\left[\begin{array}{ccc}1& 0& -3\\ 2& 1& 3\\ 0& 1& 1\end{array}\right]+\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right] \\ =\left[\begin{array}{ccc}1+1& 0+0& -3+0\\ 2+0& 1+1& 3+0\\ 0+0& 1+0& 1+1\end{array}\right] \\ =\left[\begin{array}{ccc}2& 0& -3\\ 2& 2& 3\\ 0& 1& 2\end{array}\right] \\ A\left(A+I\right)=\left[\begin{array}{ccc}1& 0& -3\\ 2& 1& 3\\ 0& 1& 1\end{array}\right]\left[\begin{array}{ccc}2& 0& -3\\ 2& 2& 3\\ 0& 1& 2\end{array}\right] \\ =\left[\begin{array}{ccc}2+0+0& 0+0-3& -3+0-6\\ 4+2+0& 0+2+3& -6+3+6\\ 0+2+0& 0+2+1& 0+3+2\end{array}\right] \\ =\left[\begin{array}{ccc}2& -3& -9\\ 6& 5& 3\\ 2& 3& 5\end{array}\right] \end{gathered}$$

Therefore, LHS = RHS.

Hence, A² + A = A(A + I) is verified.