If $A = \left[ {\begin{array}{cccccccccccccccccccc}1&0&2\\0&2&1\\2&0&3\end{array}} \right]$, prove that ${A^3} - 6{A^2} + 7A + 2I = O$.

.:

We are given that, $A = \left[ {\begin{array}{cccccccccccccccccccc}1&0&2\\0&2&1\\2&0&3\end{array}} \right]$

$\Rightarrow$ ${A^2} = A \cdot A = \left[ {\begin{array}{cccccccccccccccccccc}1&0&2\\0&2&1\\2&0&3\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}1&0&2\\0&2&1\\2&0&3\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{1 \times 1 + 0 \times 0 + 2 \times 2}&{0 \times 1 + 0 \times 2 + 0 \times 2}&{1 \times 2 + 0 \times 1 + 2 \times 3}\\{0 \times 1 + 2 \times 0 + 1 \times 2}&{0 \times 0 + 2 \times 2 + 1 \times 0}&{0 \times 2 + 2 \times 1 + 3 \times 1}\\{2 \times 1 + 0 \times 0 + 3 \times 2}&{2 \times 0 + 0 \times 2 + 3 \times 0}&{2 \times 2 + 0 \times 1 + 3 \times 3}\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{1 + 0 + 4}&{0 + 0 + 0}&{2 + 0 + 6}\\{0 + 0 + 2}&{0 + 4 + 0}&{0 + 2 + 3}\\{2 + 0 + 6}&{0 + 0 + 0}&{4 + 0 + 9}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}5&0&8\\2&4&5\\8&0&{13}\end{array}} \right]$

Now, ${A^3} = {A^2} \cdot A = \left[ {\begin{array}{cccccccccccccccccccc}5&0&8\\2&4&5\\8&0&{13}\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}1&0&2\\0&2&1\\2&0&3\end{array}} \right]$
$= \left[ {\begin{array}{cccccccccccccccccccc}{5 \times 1 + 0 \times 0 + 2 \times 8}&{5 \times 0 + 0 \times 2 + 8 \times 0}&{5 \times 2 + 0 \times 1 + 8 \times 3}\\{2 \times 1 + 4 \times 0 + 5 \times 2}&{2 \times 0 + 4 \times 2 + 5 \times 0}&{2 \times 2 + 4 \times 1 + 5 \times 3}\\{8 \times 1 + 0 \times 0 + 13 \times 2}&{8 \times 0 + 0 \times 2 + 13 \times 0}&{2 \times 8 + 1 \times 0 + 13 \times 3}\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{5 + 0 + 16}&{0 + 0 + 0}&{10 + 0 + 24}\\{2 + 0 + 10}&{0 + 8 + 0}&{4 + 4 + 15}\\{8 + 0 + 26}&{0 + 0 + 0}&{16 + 0 + 39}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{21}&0&{34}\\{12}&8&{23}\\{34}&0&{55}\end{array}} \right]$

Now L.H.S. $= {A^3} - 6{A^2} + 7A + 2I$

$= \left[ {\begin{array}{cccccccccccccccccccc}{21}&0&{34}\\{12}&8&{23}\\{34}&0&{55}\end{array}} \right] - 6\left[ {\begin{array}{cccccccccccccccccccc}5&0&8\\2&4&5\\8&0&{13}\end{array}} \right] + 7\left[ {\begin{array}{cccccccccccccccccccc}1&0&2\\0&2&1\\2&0&3\end{array}} \right] + 2\left[ {\begin{array}{cccccccccccccccccccc}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{21}&0&{34}\\{12}&8&{23}\\{34}&0&{55}\end{array}} \right] - \left[ {\begin{array}{cccccccccccccccccccc}{30}&0&{48}\\{12}&{24}&{30}\\{48}&0&{78}\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}7&0&{14}\\0&{14}&7\\{14}&0&{21}\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}2&0&0\\0&2&0\\0&0&2\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{21 - 30}&{0 - 0}&{34 - 48}\\{12 - 12}&{8 - 24}&{23 - 30}\\{34 - 48}&{0 - 0}&{55 - 78}\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}{7 + 2}&{0 + 0}&{14 + 0}\\{0 + 0}&{14 + 2}&{7 + 0}\\{14 + 0}&{0 + 0}&{21 + 2}\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{ - 9}&0&{ - 14}\\0&{ - 16}&{ - 7}\\{ - 14}&0&{ - 23}\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}9&0&{14}\\0&{16}&7\\{14}&0&{23}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}0&0&0\\0&0&0\\0&0&0\end{array}} \right] = 0 =$R.H.S

Hence, proved.