For the matrix $A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&1&1\\1&2&{ - 3}\\2&{ - 1}&3\end{array}} \right]$ , show that ${A^3} - 6{A^2} + 5A + 11I = O.$ Hence, find ${A^{ - 1}}$ .

We have $A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&1&1\\1&2&{ - 3}\\2&{ - 1}&3\end{array}} \right]$ $\therefore$ ${A^2} = AA = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&1&1\\1&2&{ - 3}\\2&1&3\end{array}} \right]\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&1&1\\1&2&{ - 3}\\2&{ - 1}&3\end{array}} \right]$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{1 \cdot 1 + 1 \cdot 1 + 1 \cdot 2}&{1 \cdot 1 + 1 \cdot 2 + 1 \cdot ( - 1)}&{1 \cdot 1 + 1 \cdot ( - 3) + 1 \cdot 3}\\{1 \cdot 1 + 2 \cdot 1 + ( - 3) \cdot 2}&{1 \cdot 1 + 2 \cdot 2 + ( - 3) \cdot ( - 1)}&{1 \cdot 1 + 2 \cdot ( - 3) + ( - 3) \cdot ( - 3)}\\{2 \cdot 1 + ( - 1) \cdot 1 + 3 \cdot 2}&{2 \cdot 1 + ( - 1) \cdot 2 + 3 \cdot ( - 1)}&{2 \cdot 1 + ( - 1) \cdot ( - 3) + 3 \cdot 3}\end{array}} \right]$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{1 + 1 + 2}&{1 + 2 - 1}&{1 - 3 + 3}\\{1 + 2 - 6}&{1 + 4 + 3}&{1 - 6 - 9}\\{2 - 1 + 6}&{2 - 2 - 3}&{2 + 3 + 9}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&2&1\\{ - 3}&8&{ - 14}\\7&{ - 3}&{14}\end{array}} \right]$

and ${A^3} = {A^2}A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&2&1\\{ - 3}&8&{ - 14}\\7&{ - 3}&{14}\end{array}} \right]\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&1&1\\1&2&{ - 3}\\2&{ - 1}&3\end{array}} \right]$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{4 \cdot 1 + 2 \cdot 1 + 1 \cdot 2}&{4 \cdot 1 + 2 \cdot 2 + 1 \cdot ( - 1)}&{4 \cdot 1 + 2 \cdot ( - 3) + 1 \cdot 3}\\{ - 3 \cdot 1 + 8 \cdot 1 + ( - 14) \cdot 2}&{ - 3 \cdot 1 + 8 \cdot 2 + ( - 14) \cdot ( - 1)}&{ - 3 \cdot 1 + 8 \cdot ( - 3) + ( - 14) \cdot 3}\\{7 \cdot 1 + ( - 3) \cdot 1 + 14 \cdot 2}&{ \cdot 1 + ( - 1) \cdot 2 + 3 \cdot ( - 1)}&{7 \cdot 1 + ( - 3) \cdot ( - 3) + 14 \cdot 3}\end{array}} \right]$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}8&7&1\\{ - 23}&{27}&{ - 69}\\{32}&{ - 13}&{58}\end{array}} \right]$

Now, L.H.S. $= {A^3} - 6{A^2} + 5A + 11I$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}8&7&1\\{ - 23}&{27}&{ - 69}\\{32}&{ - 13}&{58}\end{array}} \right] - 6\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&2&1\\{ - 3}&8&{ - 14}\\7&{ - 3}&{14}\end{array}} \right] + 5\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&1&1\\1&2&{ - 3}\\2&{ - 1}&3\end{array}} \right] + 11\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}8&7&1\\{ - 23}&{27}&{ - 69}\\{32}&{ - 13}&{58}\end{array}} \right] + \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 24}&{ - 12}&{ - 6}\\{18}&{ - 48}&{84}\\{ - 42}&{18}&{ - 84}\end{array}} \right] + \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}5&5&5\\5&{10}&{ - 15}\\{10}&{ - 5}&{15}\end{array}} \right] + \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{11}&0&0\\0&{11}&0\\0&0&{11}\end{array}} \right]$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{8 - 24 + 5 + 11}&{7 - 12 + 5 + 0}&{1 - 6 + 5 + 0}\\{ - 23 + 18 + 5 + 0}&{27 - 48 + 10 + 11}&{ - 69 + 84 - 15 + 0}\\{32 - 42 + 10 + 0}&{ - 13 + 18 - 5 + 0}&{58 - 84 + 15 + 11}\end{array}} \right]$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&0&0\\0&0&0\\0&0&0\end{array}} \right] = O \Rightarrow {A^3} - 6{A^2} + 5A + 11I = O$

Hence, proved.
Now, ${A^3} - 6{A^2} + 5A + 11I = O$
$\Rightarrow$ $11I = - {A^3} + 6{A^2} - 5A$

…(i)
Multiplying (i) by ${A^{ - 1}}$,

we get
$11{A^{ - 1}}I = - {A^{ - 1}}{A^3} + 6{A^{ - 1}}{A^2} - 5{A^{ - 1}}A \Rightarrow 11{A^{ - 1}} = - {A^2} + 6A - 5I$

$\Rightarrow$ ${A^{ - 1}} = - \cfrac{1}{{11}}{A^2} + \cfrac{6}{{11}}A - \cfrac{5}{{11}}I$

$\Rightarrow$ $= - \cfrac{1}{{11}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&2&1\\{ - 3}&8&{ - 14}\\7&{ - 3}&{14}\end{array}} \right] + \cfrac{6}{{11}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&1&1\\1&2&{ - 3}\\2&{ - 1}&3\end{array}} \right] - \cfrac{5}{{11}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$

$= \cfrac{1}{{11}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 4 + 6 - 5}&{ - 2 + 6 + 0}&{ - 1 + 6 + 0}\\{3 + 6 + 0}&{ - 8 + 12 - 5}&{14 - 18 + 0}\\{ - 7 + 12 + 0}&{3 - 6 + 0}&{ - 14 + 18 - 5}\end{array}} \right]$

$= \cfrac{1}{{11}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 3}&4&5\\9&{ - 1}&{ - 4}\\5&{ - 3}&{ - 1}\end{array}} \right]$