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# If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$ show that $A^2 - 5A + 7I = 0$

Toolbox:
• If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix whose entries are given by dot product of the corresponding row of A and the corresponding column of B: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
• The sum / difference $A(+/-)B$ of two $m$-by-$n$ matrices $A$ and $B$ is calculated entrywise: $(A (+/-) B)_{i,j} = A_{i,j} +/- B_{i,j}$ where 1 $\leq$ i $\leq$ m and 1 $\leq$ j $\leq$ n.
Step 1: Given
$A=\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}$
$A^2=\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}$
$\;\;\;\;=\begin{bmatrix}9-1 & 3+2\\-3-2 & -1+4\end{bmatrix}\Rightarrow \begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}$
Step 2: LHS:
$A^2-5A+7I=0$
$\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}-5\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}+7\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$
$\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}-\begin{bmatrix}15 & 5\\-5 & 10\end{bmatrix}+\begin{bmatrix}7 & 0\\0 & 7\end{bmatrix}$
$\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}+\begin{bmatrix}-15 &- 5\\5 & -10\end{bmatrix}+\begin{bmatrix}7 & 0\\0 & 7\end{bmatrix}$
$\begin{bmatrix}8-15+7 & 5-5+0\\-5+5+0 & 3-10+7\end{bmatrix}=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}=0\Rightarrow RHS$
edited Mar 19, 2013