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For the matrix \(A = \left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 1&2&{ - 3} \\ 2&{ - 1}&3 \end{array}} \right]\), show that A3 - 6A2 + 5A + 11I = 0. Hence find A-1.
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For the matrix \(A = \left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 1&2&{ - 3} \\ 2&{ - 1}&3 \end{array}} \right]\), show that A3 - 6A2 + 5A + 11I = 0. Hence find A-1.

Answer

Given: \(A = \left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 1&2&{ - 3} \\ 2&{ - 1}&3 \end{array}} \right]\)
\(\therefore {A^2} = \left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 1&2&{ - 3} \\ 2&{ - 1}&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 1&2&{ - 3} \\ 2&{ - 1}&3 \end{array}} \right]\)
\( \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}} {1 + 1 + 2}&{1 + 2 - 1}&{1 - 3 + 3} \\ {1 + 2 - 6}&{1 + 4 + 3}&{1 - 6 - 9} \\ {2 - 1 + 6}&{2 - 2 - 3}&{2 + 3 + 9} \end{array}} \right]\)
\(= \left[ {\begin{array}{*{20}{c}} 4&2&1 \\ { - 3}&8&{ - 14} \\ 7&{ - 3}&{14} \end{array}} \right]\)
Now \({A^3} = {A^2}A = \left[ {\begin{array}{*{20}{c}} 4&2&1 \\ { - 3}&8&{ - 14} \\ 7&{ - 3}&{14} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 1&2&{ - 3} \\ 2&{ - 1}&3 \end{array}} \right]\) 
\(= \left[ {\begin{array}{*{20}{c}} {4 + 2 + 2}&{4 + 4 + 1}&{4 - 6 + 3} \\ { - 3 + 8 - 28}&{ - 3 + 16 + 14}&{ - 3 - 24 - 42} \\ {7 - 3 + 28}&{7 - 6 - 14}&{7 + 9 + 42} \end{array}} \right]\)
\(= \left[ {\begin{array}{*{20}{c}} 8&7&1 \\ { - 23}&{27}&{ - 69} \\ {32}&{ - 13}&{58} \end{array}} \right]\)
L.H.S. \( = {A^3} - 6{A^2} + 5A + 11I\) 
\(= \left[ {\begin{array}{*{20}{c}} 8&7&1 \\ { - 23}&{27}&{ - 69} \\ {32}&{ - 13}&{58} \end{array}} \right] - 6\left[ {\begin{array}{*{20}{c}} 4&2&1 \\ { - 3}&8&{ - 14} \\ 7&{ - 3}&{14} \end{array}} \right] + \)\(5\left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 1&2&{ - 3} \\ 2&{ - 1}&3 \end{array}} \right] + 11\left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right]\)
\(= \left[ {\begin{array}{*{20}{c}} 8&7&1 \\ { - 23}&{27}&{ - 69} \\ {32}&{ - 13}&{58} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} {24}&{12}&6 \\ { - 18}&{48}&{ - 84} \\ {42}&{ - 18}&{84} \end{array}} \right] \) \(+ \left[ {\begin{array}{*{20}{c}} 5&5&5 \\ 5&{10}&{ - 15} \\ {10}&{ - 5}&{15} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} {11}&0&0 \\ 0&{11}&0 \\ 0&0&{11} \end{array}} \right]\)
\( = \left[ {\begin{array}{*{20}{c}} {8 - 24 + 5}&{7 - 12 + 5}&{1 - 6 + 5} \\ { - 23 + 18 + 5}&{27 - 48 + 10}&{ - 69 + 84 - 15} \\ {32 - 42 + 10}&{ - 13 + 18 - 5}&{58 + 84 + 15} \end{array}} \right] \) \(+ \left[ {\begin{array}{*{20}{c}} {11}&0&0 \\ 0&{11}&0 \\ 0&0&{11} \end{array}} \right]\)
\(= \left[ {\begin{array}{*{20}{c}} { - 11}&0&0 \\ 0&{ - 11}&0 \\ 0&0&{ - 11} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} {11}&0&0 \\ 0&{11}&0 \\ 0&0&{11} \end{array}} \right] \) \(= \left[ {\begin{array}{*{20}{c}} 0&0&0 \\ 0&0&0 \\ 0&0&0 \end{array}} \right] = 0\)= R.H.S.
Now,   to find \({A^{ - 1}}\), multiplying \({A^3} - 6{A^2} + 5A + 11I = 0\) by \({A^{ - 1}}\) 
\(\Rightarrow {A^3}{A^{ - 1}} - 6{A^2}{A^{ - 1}} + 5A{A^{ - 1}} + 11I.{A^{ - 1}} = 0.{A^{ - 1}}\)
\( \Rightarrow {A^2} - 6A + 5I + 11{A^{ - 1}} = 0\)
\( \Rightarrow 11{A^{ - 1}} = 6A - 5I - {A^2}\)
\( \Rightarrow 11{A^{ - 1}} = 6\left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 1&2&{ - 3} \\ 2&{ - 1}&3 \end{array}} \right] \) \(- 5\left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 4&2&1 \\ { - 3}&8&{ - 14} \\ 7&{ - 3}&{14} \end{array}} \right]\)
\( \Rightarrow 11{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} 6&6&6 \\ 6&{12}&{ - 18} \\ {12}&{ - 6}&{18} \end{array}} \right] \) \(- \left[ {\begin{array}{*{20}{c}} 5&0&0 \\ 0&5&0 \\ 0&0&5 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 4&2&1 \\ { - 3}&8&{ - 14} \\ 7&{ - 3}&{14} \end{array}} \right]\)
\(\Rightarrow 11{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} {6 - 5 - 4}&{6 - 2}&{6 - 1} \\ {6 + 3}&{12 - 5 - 8}&{ - 18 + 14} \\ {12 - 7}&{ - 6 + 3}&{18 - 5 - 14} \end{array}} \right]\)
\( \Rightarrow 11{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} { - 3}&4&5 \\ 9&{ - 1}&{ - 4} \\ 5&{ - 3}&{ - 1} \end{array}} \right]\)
\( \Rightarrow {A^{ - 1}} = \frac{1}{{11}}\left[ {\begin{array}{*{20}{c}} { - 3}&4&5 \\ 9&{ - 1}&{ - 4} \\ 5&{ - 3}&{ - 1} \end{array}} \right]\)
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