If A = 1&0&2 0&2&1 2&0&3 , prove that A^3 - 6A^2 + 7A + 2I = 0.

Question

If A = $\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}$, prove that $A^3 - 6A^2 + 7A + 2I = 0$.

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Answer

$L.H.S = A^3 - 6A^2 + 7A + 2I$
$A^3$ $=\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}$
$A^3$ $=\begin{bmatrix}1+0+4&0+0+0&2+0+6\\0+0+2&0+4+0&0+2+3\\2+0+6&0+0+0&4+0+9\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}$
$\text{A}^3=\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}$
$A^3$ $=\begin{bmatrix}5+0+16&0+0+0&10+0+24\\2+0+10&0+8+0&4+4+15\\8+0+26&0+0+0&16+0+39\end{bmatrix}$$=\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}...\text{(i)}$
$6A^2$$=-6\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}$$= 6\begin{bmatrix}1+0+4&0+0+0&2+0+6\\0+0+2&0+4+0&0+2+3\\2+0+6&0+0+0&4+0+9\end{bmatrix}$
$6A^2$$=\begin{bmatrix}30&0&48\\12&24&30\\48&0&78\end{bmatrix}...\text{(ii)}$
7A + 2I $=7\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}+2\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
7A + 2I $=\begin{bmatrix}7+2&0+0&14+0\\0+0&14+2&7+0\\14+0&0+0&21+2\end{bmatrix}$$ =\begin{bmatrix}9&0&14\\0&16&7\\14&0&23\end{bmatrix}...\text{(iii)}$
Now from (i), (ii), and (iii) equation, we get
$A^3 - 6A^2 + 7A + 2I$
$=\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}-\begin{bmatrix}30&0&48\\12&24&30\\48&0&78\end{bmatrix}+\begin{bmatrix}9&0&14\\0&16&7\\14&0&23\end{bmatrix}$
$=\begin{bmatrix}21-30&0-0&34-48\\12-12&8-24&23-30\\34-48&0-0&55-78\end{bmatrix}\begin{bmatrix}9&0&14\\0&16&7\\14&0&23\end{bmatrix}$
$=\begin{bmatrix}-9&0&-14\\0&-16&-7\\-14&0&-23\end{bmatrix}+\begin{bmatrix}9&0&14\\0&16&7\\14&0&23\end{bmatrix}$
$=\begin{bmatrix}-9+9&0+0&-14+14\\0+0&-16+16&-7+7\\-14+14&0+0&-23+23\end{bmatrix}$
$=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$= 0 (Zero matrix) = R.H.S.

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