# If $A=\begin{vmatrix}2 & 3\\1 & 2\end{vmatrix}$ prove that $A^3-4A^2+A=0$

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 ≤ i ≤ m and 1 ≤ j ≤ n.
Step 1:
$A=\begin{bmatrix} 2 & 3\\1 & 2\end{bmatrix}$
$A^3=\begin{bmatrix} 2 & 3\\1 & 2\end{bmatrix}\times \begin{bmatrix} 2 & 3\\1 & 2\end{bmatrix}\times \begin{bmatrix} 2 & 3\\1 & 2\end{bmatrix}$
$\quad=\begin{bmatrix} 4+3 & 6+6\\2+2 & 3+4\end{bmatrix}\times \begin{bmatrix} 2 & 3\\1 & 2\end{bmatrix}$
$\quad=\begin{bmatrix} 7 & 12\\4 & 7\end{bmatrix}\times \begin{bmatrix} 2 & 3\\1 & 2\end{bmatrix}$
$\quad=\begin{bmatrix} 26 & 45\\15 & 26\end{bmatrix}$
Step 2:
$A^2=\begin{bmatrix} 2 & 3\\1 & 2\end{bmatrix}\times \begin{bmatrix} 2 & 3\\1 & 2\end{bmatrix}$
$A^2=\begin{bmatrix} 7 & 12\\4 & 7\end{bmatrix}$
$\therefore 4A^2=\begin{bmatrix} 28 & 48\\16 & 28\end{bmatrix}$
Step 3:
$A^3-4A^2+A=\begin{bmatrix} 26 & 45\\15 & 26\end{bmatrix}-\begin{bmatrix} 28 & 48\\16 & 28\end{bmatrix}+\begin{bmatrix} 2 & 3\\1 & 2\end{bmatrix}$
$\qquad\qquad\;\;\;\;\;=\begin{bmatrix} 0 & 0\\0 & 0\end{bmatrix}$
Hence $A^3-4A^2+A=0$