Determinants — Class 12 Maths Solution

We have

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

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{4 + 1 + 1}&{ - 2 - 2 - 1}&{2 + 1 + 2}\\{ - 2 - 2 - 1}&{1 + 4 + 1}&{ - 1 - 2 - 2}\\{2 + 1 + 2}&{ - 1 - 2 - 2}&{1 + 1 + 4}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}6&{ - 5}&5\\{ - 5}&6&{ - 5}\\5&{ - 5}&6\end{array}} \right]$

and ${A^3} = {A^2}A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}6&{ - 5}&5\\{ - 5}&6&{ - 5}\\5&{ - 5}&6\end{array}} \right]\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 1}&1\\{ - 1}&2&{ - 1}\\1&{ - 1}&2\end{array}} \right]$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{12 + 5 + 5}&{ - 6 - 10 - 5}&{6 + 5 + 10}\\{ - 10 - 6 - 5}&{5 + 12 + 5}&{ - 5 - 6 - 10}\\{10 + 5 + 6}&{ - 5 - 10 - 6}&{5 + 5 + 12}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{22}&{ - 21}&{21}\\{ - 21}&{22}&{ - 21}\\{21}&{ - 21}&{22}\end{array}} \right]$

Now, ${A^3} - 6{A^2} + 9A - 4I$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{22}&{ - 21}&{21}\\{ - 21}&{22}&{ - 21}\\{21}&{ - 21}&{22}\end{array}} \right] - 6\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}6&{ - 5}&5\\{ - 5}&6&{ - 5}\\5&{ - 5}&6\end{array}} \right] + 9\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 1}&1\\{ - 1}&2&{ - 1}\\1&{ - 1}&2\end{array}} \right] - 4\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{22 - 36 + 18 - 4}&{ - 21 + 30 - 9 + 0}&{21 - 30 + 9 + 0}\\{ - 21 + 30 - 9 - 0}&{22 - 36 + 18 - 4}&{ - 21 + 30 - 9 + 0}\\{21 - 30 + 9 + 0}&{ - 21 + 30 - 9 + 0}&{22 - 36 + 18 - 4}\end{array}} \right]$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&0&0\\0&0&0\\0&0&0\end{array}} \right] = O$

Hence, ${A^3} - 6{A^2} + 9A - 4I = O$

Now, ${A^3} - 6{A^2} + 9A - 4I = O \Rightarrow 4I = {A^3} - 6{A^2} + 9A$

Multiplying both sides by ${A^{ - 1}}$,

we get
$\Rightarrow$ ${A^{ - 1}} = \cfrac{1}{4}{A^2} - \cfrac{6}{4}A + \cfrac{9}{4}I$

$= \cfrac{1}{4}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}6&{ - 5}&5\\{ - 5}&6&{ - 5}\\5&{ - 5}&6\end{array}} \right] - \cfrac{6}{4}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 1}&1\\{ - 1}&2&{ - 1}\\1&{ - 1}&2\end{array}} \right] + \cfrac{9}{4}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$

$= \cfrac{1}{4}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{6 - 12 + 9}&{ - 5 + 6 + 0}&{5 - 6 + 0}\\{ - 5 + 6 + 0}&{6 - 12 + 9}&{ - 5 + 6 + 0}\\{5 - 6 + 0}&{ - 5 + 6 + 0}&{6 - 12 + 9}\end{array}} \right] = \cfrac{1}{4}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&1&{ - 1}\\1&3&1\\{ - 1}&1&3\end{array}} \right]$