Question

If \[A = \left[ {\begin{array}{*{20}{c}}
  1&2&2 \\
  2&1&2 \\
  2&2&1
\end{array}} \right]\] , then prove that ${A^2} - 4A - 5I = 0$ , also find ${A^{ - 1}}$ .

Answer 1

Hint: First we find ${A^2} = A \times A$ , we multiply two matrices in some step. In the first step we make sure that the number of columns in the first one equals the number of rows in the second one. In the second step we multiply the elements of each row of the first matrix by the elements of each in the second matrix. At the last step we add the product. After completing the multiplication prove the given condition and find the inverse of the matrix A.

Complete step-by-step answer:
Given matrix \[A = \left[ {\begin{array}{*{20}{c}}
  1&2&2 \\
  2&1&2 \\
  2&2&1
\end{array}} \right]\]
Now we find the matrix ${A^2}$
Therefore ${A^2} = A \times A$
$ = \left[ {\begin{array}{*{20}{c}}
  1&2&2 \\
  2&1&2 \\
  2&2&1
\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
  1&2&2 \\
  2&1&2 \\
  2&2&1
\end{array}} \right]$
Multiplying matrices by using the matrix multiplication rule and we get
$ = \left[ {\begin{array}{*{20}{c}}
  {1 \times 1 + 2 \times 2 + 2 \times 2}&{1 \times 2 + 2 \times 2 + 2 \times 2}&{1 \times 2 + 2 \times 2 + 2 \times 1} \\
  {2 \times 1 + 1 \times 2 + 2 \times 2}&{2 \times 2 + 1 \times 1 + 2 \times 2}&{2 \times 2 + 1 \times 2 + 2 \times 1} \\
  {2 \times 1 + 2 \times 2 + 1 \times 2}&{2 \times 2 + 2 \times 1 + 1 \times 2}&{2 \times 2 + 2 \times 2 + 1 \times 1}
\end{array}} \right]$
Calculating and simplifying and we get
$ = \left[ {\begin{array}{*{20}{c}}
  {1 + 4 + 4}&{2 + 2 + 4}&{2 + 4 + 2} \\
  {2 + 2 + 4}&{4 + 1 + 4}&{4 + 2 + 2} \\
  {2 + 4 + 2}&{4 + 2 + 2}&{4 + 4 + 1}
\end{array}} \right]$
\[ = \left[ {\begin{array}{*{20}{c}}
  9&8&8 \\
  8&9&8 \\
  8&8&9
\end{array}} \right]\]
We know the identity matrix of $3 \times 3$ matrix $I = \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]$
Now we calculate the L.H.S.
${A^2} - 4A - 5I$
Substitute the matrix and calculate
$ = \left[ {\begin{array}{*{20}{c}}
  9&8&8 \\
  8&9&8 \\
  8&8&9
\end{array}} \right] - 4\left[ {\begin{array}{*{20}{c}}
  1&2&2 \\
  2&1&2 \\
  2&2&1
\end{array}} \right] - 5\left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]$
\[ = \left[ {\begin{array}{*{20}{c}}
  9&8&8 \\
  8&9&8 \\
  8&8&9
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  4&8&8 \\
  8&4&8 \\
  8&8&4
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  5&0&0 \\
  0&5&0 \\
  0&0&5
\end{array}} \right]\]
Take common of $' - '$ sign and we get
\[ = \left[ {\begin{array}{*{20}{c}}
  9&8&8 \\
  8&9&8 \\
  8&8&9
\end{array}} \right] - \left( {\left[ {\begin{array}{*{20}{c}}
  4&8&8 \\
  8&4&8 \\
  8&8&4
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
  5&0&0 \\
  0&5&0 \\
  0&0&5
\end{array}} \right]} \right)\]
Adding the second and third matrices and we get
\[ = \left[ {\begin{array}{*{20}{c}}
  9&8&8 \\
  8&9&8 \\
  8&8&9
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  {4 + 5}&{8 + 0}&{8 + 0} \\
  {8 + 0}&{4 + 5}&{8 + 0} \\
  {8 + 0}&{8 + 0}&{4 + 5}
\end{array}} \right]\]
Simplifying and we get
\[ = \left[ {\begin{array}{*{20}{c}}
  9&8&8 \\
  8&9&8 \\
  8&8&9
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  9&8&8 \\
  8&9&8 \\
  8&8&9
\end{array}} \right]\]
\[ = \left[ {\begin{array}{*{20}{c}}
  {9 - 9}&{8 - 8}&{8 - 8} \\
  {8 - 8}&{9 - 9}&{8 - 8} \\
  {8 - 8}&{8 - 8}&{9 - 9}
\end{array}} \right]\]
$ = \left[ {\begin{array}{*{20}{c}}
  0&0&0 \\
  0&0&0 \\
  0&0&0
\end{array}} \right]$
We know that it is a null matrix.
$ = 0$
$ = $ R.H.S.
The given condition ${A^2} - 4A - 5I = 0$ is proved.
From the above we get ${A^2} - 4A = 5I$
Post multiply the above equation by ${A^{ - 1}}$ and we get
${A^2}{A^{ - 1}} - 4A{A^{ - 1}} = 5I{A^{ - 1}}$
$ \Rightarrow A - 4I = 5{A^{ - 1}}$
Put the matrix in the above equation and we get
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  1&2&2 \\
  2&1&2 \\
  2&2&1
\end{array}} \right] - 4\left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right] = 5{A^{ - 1}}$
Multiply $4$ with the identity matrix
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  1&2&2 \\
  2&1&2 \\
  2&2&1
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  4&0&0 \\
  0&4&0 \\
  0&0&4
\end{array}} \right] = 5{A^{ - 1}}$
Calculate using matrix subtraction and we get
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  {1 - 4}&{2 - 0}&{2 - 0} \\
  {2 - 0}&{1 - 4}&{2 - 0} \\
  {2 - 0}&{2 - 0}&{1 - 4}
\end{array}} \right] = 5{A^{ - 1}}$
Simplifying and get
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  { - 3}&2&2 \\
  2&{ - 3}&2 \\
  2&2&{ - 3}
\end{array}} \right] = 5{A^{ - 1}}$
Divide both sides by $5$ and we get
$ \Rightarrow {A^{ - 1}} = \dfrac{1}{5}\left[ {\begin{array}{*{20}{c}}
  { - 3}&2&2 \\
  2&{ - 3}&2 \\
  2&2&{ - 3}
\end{array}} \right]$
Simplifying and we get
$ \Rightarrow {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
  { - \dfrac{3}{5}}&{\dfrac{2}{5}}&{\dfrac{2}{5}} \\
  {\dfrac{2}{5}}&{ - \dfrac{3}{5}}&{\dfrac{2}{5}} \\
  {\dfrac{2}{5}}&{\dfrac{2}{5}}&{ - \dfrac{3}{5}}
\end{array}} \right]$
Therefore the inverse of $A$ is ${A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
  { - \dfrac{3}{5}}&{\dfrac{2}{5}}&{\dfrac{2}{5}} \\
  {\dfrac{2}{5}}&{ - \dfrac{3}{5}}&{\dfrac{2}{5}} \\
  {\dfrac{2}{5}}&{\dfrac{2}{5}}&{ - \dfrac{3}{5}}
\end{array}} \right]$

Note: Make sure that the determinant of the matrix for which you are finding the inverse is not $0$ . Because if it is zero, then the matrix will not be invertible, hence we cannot find the inverse of that particular matrix. Make sure there are no calculation mistakes as there are calculation processes. We know that from the property of inverse $A{A^{ - 1}} = I$ , where $A,I$ are matrices of the same order.