Question

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

Answer 1

Hint: Start by finding out the values of \[5A,{A^2}\& {A^3}\] by using the properties of matrix multiplication . Substitute these values in the given equation and find out ${A^{ - 1}}$ by rearranging the terms and using the identity $A \cdot {A^{ - 1}} = I$.

Complete step-by-step answer:
Given, $A = \left( {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&2&{ - 3} \\
  2&{ - 1}&3
\end{array}} \right)$
Now, we need to show that the matrix A satisfies the equation
i.e. ${A^3} - 6{A^2} + 5A + 11I = 0$
Firstly, let us compute 5A
$5A = 5 \cdot \left( {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&2&{ - 3} \\
  2&{ - 1}&3
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  5&5&5 \\
  5&{10}&{ - 15} \\
  {10}&{ - 5}&{15}
\end{array}} \right)$
Now, We know $A \cdot A = {A^2}$ and ${A^2} \cdot A = {A^3}$
Therefore , using the rule of multiplication of matrices i.e. Row by Column, we get

${A^2} = \left( {\begin{array}{*{20}{c}}
  {{a_{11}} \cdot {a_{11}} + {a_{12}} \cdot {a_{21}} + {a_{13}} \cdot {a_{31}}}&{{a_{11}} \cdot {a_{12}} + {a_{12}} \cdot {a_{22}} + {a_{13}} \cdot {a_{32}}}&{{a_{11}} \cdot {a_{13}} + {a_{12}} \cdot {a_{23}} + {a_{13}} \cdot {a_{33}}} \\
  {{a_{21}} \cdot {a_{11}} + {a_{22}} \cdot {a_{21}} + {a_{23}} \cdot {a_{31}}}&{{a_{21}} \cdot {a_{12}} + {a_{22}} \cdot {a_{22}} + {a_{23}} \cdot {a_{32}}}&{{a_{21}} \cdot {a_{13}} + {a_{22}} \cdot {a_{23}} + {a_{23}} \cdot {a_{33}}} \\
  {{a_{31}} \cdot {a_{11}} + {a_{32}} \cdot {a_{21}} + {a_{33}} \cdot {a_{31}}}&{{a_{31}} \cdot {a_{12}} + {a_{32}} \cdot {a_{22}} + {a_{33}} \cdot {a_{32}}}&{{a_{31}} \cdot {a_{13}} + {a_{32}} \cdot {a_{23}} + {a_{33}} \cdot {a_{33}}}
\end{array}} \right)$
Substituting the values , we get
${A^2} = \left( {\begin{array}{*{20}{c}}
  4&2&1 \\
  { - 3}&8&{ - 14} \\
  7&{ - 3}&{14}
\end{array}} \right)$

So , $6{A^2} = \left( {\begin{array}{*{20}{c}}
  {24}&{12}&6 \\
  { - 18}&{48}&{ - 84} \\
  {42}&{ - 18}&{84}
\end{array}} \right)$
Now , similarly we will find ${A^3} = {A^2} \cdot A$
Substituting the values of ${A^2}\& A$,
$
  {A^3} = {A^2} \cdot A = \left( {\begin{array}{*{20}{c}}
  4&2&1 \\
  { - 3}&8&{ - 14} \\
  7&{ - 3}&{14}
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&2&{ - 3} \\
  2&{ - 1}&3
\end{array}} \right) \\
   = \left( {\begin{array}{*{20}{c}}
  {4 \times 1 + 2 \times 1 + 1 \times 2}&{4 \times 1 + 2 \times 2 - 1 \times 1}&{4 \times 1 - 2 \times 3 + 1 \times 3} \\
  { - 3 \times 1 + 8 \times 1 - 14 \times 2}&{ - 3 \times 1 + 8 \times 2 + 14 \times 1}&{ - 3 \times 1 - 8 \times 3 - 14 \times 3} \\
  {7 \times 1 - 3 \times 1 + 14 \times 2}&{7 \times 1 - 3 \times 2 - 14 \times 1}&{7 \times 1 + 3 \times 3 + 14 \times 3}
\end{array}} \right) \\
  {A^3} = \left( {\begin{array}{*{20}{c}}
  8&7&1 \\
  { - 23}&{27}&{ - 69} \\
  {32}&{ - 13}&{58}
\end{array}} \right) \\
$
Now substituting all the values in the equation ${A^3} - 6{A^2} + 5A + 11I = 0$, we get
$\left( {\begin{array}{*{20}{c}}
  8&7&1 \\
  { - 23}&{27}&{ - 69} \\
  {32}&{ - 13}&{58}
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
  {24}&{12}&6 \\
  { - 18}&{48}&{ - 84} \\
  {42}&{ - 18}&{84}
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
  5&5&5 \\
  5&{10}&{ - 15} \\
  {10}&{ - 5}&{15}
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
  {11}&0&0 \\
  0&{11}&0 \\
  0&0&{11}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  0&0&0 \\
  0&0&0 \\
  0&0&0
\end{array}} \right)$
Now, from this we will find ${A^{ - 1}}$
$ - 11I = {A^3} - 6{A^2} + 5A$
Multiplying both side by ${A^{ - 1}}$ , we get
$ - 11{A^{ - 1}} = {A^2} - 6A + 5I\left( {\because A \cdot {A^{ - 1}} = I} \right)$
${A^{ - 1}} = \dfrac{{{A^2} - 6A + 5I}}{{ - 11}}$
Now , substitute the values of A and I , we get
$\therefore {A^{ - 1}} = \dfrac{{ - 1}}{{11}}\left( {\begin{array}{*{20}{c}}
  3&{ - 4}&{ - 5} \\
  { - 9}&1&4 \\
  { - 5}&3&1
\end{array}} \right)$

Note: Two matrices can be multiplied if and only if the number of columns of the pre matrix is equal to the number of rows of the post matrix (AB, A is pre matrix, B is post matrix). Students must remember all the properties and identities related to matrices and determinants , As they are very important from an exam point of view as well as for faster calculations.