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Determinants — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDeterminants5 Marks
Question
If $A=\left[\begin{array}{ccc}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right]$, verify that $A^3-6 A^2+9 A-4 \mid=0$ and hence find $A^{-1}$
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Answer

Given: $A = \left[ {\begin{array}{*{20}{c}} 2&{ - 1}&1 \\ { - 1}&2&{ - 1} \\ 1&-1&2 \end{array}} \right]$
$\therefore {A^2} = \left[ {\begin{array}{*{20}{c}} 2&{ - 1}&1 \\ { - 1}&2&{ - 1} \\ 1&-1&2 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 2&{ - 1}&1 \\ { - 1}&2&{ - 1} \\ 1&{ - 1}&2 \end{array}} \right]$
$\Rightarrow {A^2}=\left[ {\begin{array}{*{20}{c}} {4 + 1 + 1}&{ - 2 - 2 - 1}&{2 + 1 + 2} \\ { - 2 - 2 - 1}&{1 + 4 + 1}&{ - 1 - 2 - 2} \\ {2 + 1 + 2}&{ - 1 - 2 - 2}&{1 + 1 + 4} \end{array}} \right]$ $=\left[ {\begin{array}{*{20}{c}} 6&{ - 5}&5 \\ { - 5}&6&{ - 5} \\ 5&{ - 5}&6 \end{array}} \right]$
Now $A^3 = A^2.A = \left[ {\begin{array}{*{20}{c}} 6&{ - 5}&5 \\ { - 5}&6&{ - 5} \\ 5&{ - 5}&6 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 2&{ - 1}&1 \\ { - 1}&2&{ - 1} \\ 1&{ - 1}&2 \end{array}} \right]$ 
$= \left[ {\begin{array}{*{20}{c}} {12 + 5 + 5}&{ - 6 - 10 - 5}&{6 + 5 + 10} \\ { - 10 - 6 - 5}&{5 + 12 + 5}&{ - 5 - 6 - 10} \\ {10 + 5 + 6}&{ - 5 - 10 - 6}&{5 + 5 + 12} \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} {22}&{ - 21}&{21} \\ { - 21}&{22}&{ - 21} \\ {21}&{ - 21}&{22} \end{array}} \right]$
$\ce{L.H.S.}   = {A^3} - 6{A^2} + 9A - 4I$
$= \left[ {\begin{array}{*{20}{c}} {22}&{ - 21}&{21} \\ { - 21}&{22}&{ - 21} \\ {21}&{ - 21}&{22} \end{array}} \right] - 6\left[ {\begin{array}{*{20}{c}} 6&{ - 5}&5 \\ { - 5}&6&{ - 5} \\ 5&{ - 5}&6 \end{array}} \right] $ $+ 9\left[ {\begin{array}{*{20}{c}} 2&{ - 1}&1 \\ { - 1}&2&{ - 1} \\ 1&{ - 1}&2 \end{array}} \right] - 4\left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} {22}&{ - 21}&{21} \\ { - 21}&{22}&{ - 21} \\ {21}&{ - 21}&{22} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} {36}&{ - 30}&{30} \\ { - 30}&{36}&{ - 30} \\ {30}&{ - 30}&{36} \end{array}} \right] $ $+ \left[ {\begin{array}{*{20}{c}} {18}&{ - 9}&9 \\ { - 9}&{18}&{ - 9} \\ 9&{ - 9}&{18} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 4&0&0 \\ 0&4&0 \\ 0&0&4 \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} {22 - 36}&{ - 21 + 30}&{21 - 30} \\ { - 21 + 30}&{22 - 36}&{ - 21 + 30} \\ {21 - 30}&{ - 21 + 30}&{22 - 36} \end{array}} \right] $ $+ \left[ {\begin{array}{*{20}{c}} {18 - 4}&{ - 9 - 0}&{9 - 0} \\ { - 9 - 0}&{18 - 4}&{ - 9 - 0} \\ {9 - 0}&{ - 9 - 0}&{18 - 4} \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} { - 14}&9&{ - 9} \\ 9&{ - 14}&9 \\ { - 9}&9&{14} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} { - 14}&9&{ - 9} \\ 9&{ - 14}&9 \\ { - 9}&9&{14} \end{array}} \right] $ $= \left[ {\begin{array}{*{20}{c}} 0&0&0 \\ 0&0&0 \\ 0&0&0 \end{array}} \right] = 0$= R.H.S.
Now, to find ${A^{ - 1}}$, multiplying ${A^3} - 6{A^2} + 9A - 4{I^{ - 1}} = 0.{A^{ - 1}}$by ${A^{ - 1}}$ 
$\Rightarrow {A^3}{A^{ - 1}} - 6{A^2}{A^{ - 1}} + 9A{A^{ - 1}} - 4I.{A^{ - 1}} = 0{A^{ - 1}}$
$ \Rightarrow {A^2} - 6A + 9I - 4{A^{ - 1}} = 0$
$\Rightarrow 4{A}^{-1} = {A^2} - 6A + 9I$
$\Rightarrow 4{A}^{-1} = \left[ {\begin{array}{*{20}{c}} 6&{ - 5}&5 \\ { - 5}&6&{ - 5} \\ 5&{ - 5}&6 \end{array}} \right]$ $ - 6\left[ {\begin{array}{*{20}{c}} 2&{ - 1}&1 \\ { - 1}&2&{ - 1} \\ 1&{ - 1}&2 \end{array}} \right] + 9\left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right]$
$\Rightarrow 4{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} 6&{ - 5}&5 \\ { - 5}&6&{ - 5} \\ 5&{ - 5}&6 \end{array}} \right] $ $- \left[ {\begin{array}{*{20}{c}} {12}&{ - 6}&6 \\ { - 6}&{12}&{ - 6} \\ 6&{ - 6}&{12} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 9&0&0 \\ 0&9&0 \\ 0&0&9 \end{array}} \right]$
$\Rightarrow 4{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} {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] $ $= \left[ {\begin{array}{*{20}{c}} 3&1&{ - 1} \\ 1&3&1 \\ { - 1}&1&3 \end{array}} \right]$
$ \Rightarrow {A^{ - 1}} = \frac{1}{4}\left[ {\begin{array}{*{20}{c}} 3&1&{ - 1} \\ 1&3&1 \\ { - 1}&1&3 \end{array}} \right]$

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