Question

Find the adjoint of the matrix $A=\left( \begin{matrix}
   -1 & -2 & -2 \\
   2 & 1 & -2 \\
   2 & -2 & 1 \\
\end{matrix} \right)$ and hence show that $A.\left( adjA \right)=\left| A \right|{{I}_{3}}$

Answer 1

Hint: In the above question we have to prove that $A.\left( adjA \right)=\left| A \right|{{I}_{3}}$ for that firstly we need to determine the minor of matrix A and then find the $\left( adjA \right)$ after finding the $\left( adjA \right)$ we need to multiply it with the matrix A then find the determinant of matrix A which comes to be 27 multiply it with the identity matrix after multiplying both equate both terms together they will be equal.

Complete step-by-step answer:
So, lets get into a solution,
To solve the question, we have to first know what is an adjoint matrix and before that the co-factor and minor of a matrix.
 A “minor” is the determinant of the square matrix from by deleting one row and one column from some larger matrix.
Example : $\begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix}$ assume this is the determinent.
So, the minor of the element \[{{a}_{11}}=\begin{matrix}
   {{a}_{22}} & {{a}_{23}} \\
   {{a}_{32}} & {{a}_{33}} \\
\end{matrix}\]. Where the row and column releted to ${{a}_{11}}$is deleted for finding the minor.
The Cofactor is the number you get when you remove the column and row of a designated element in a matrix.
Example: Co-factor of the element ${{a}_{11}}$is ${{A}_{11}}={{\left( -1 \right)}^{1+1}}\begin{matrix}
   {{a}_{22}} & {{a}_{23}} \\
   {{a}_{32}} & {{a}_{33}} \\
\end{matrix}=\begin{matrix}
   {{a}_{22}} & {{a}_{23}} \\
   {{a}_{32}} & {{a}_{33}} \\
\end{matrix}$
Now, The transpose of a co-factor matrix is called adjoint of a matrix. Here one term comes that’s call transpose that mean when the row and column of a matrix is interchanged.
Example: ${{\left( \begin{matrix}
   {{a}_{11}} & {{a}_{12}} \\
   {{a}_{21}} & {{a}_{22}} \\
\end{matrix} \right)}^{T}}=\left( \begin{matrix}
   {{a}_{11}} & {{a}_{21}} \\
   {{a}_{12}} & {{a}_{22}} \\
\end{matrix} \right)$

Now, for finding the adjoint of a matrix we have to find the each element of the matrix’s co-factor and minor of the matrix.
Given matrix is $A=\left( \begin{matrix}
   -1 & -2 & -2 \\
   2 & 1 & -2 \\
   2 & -2 & 1 \\
\end{matrix} \right)$
So, the co-factor of the element
$\begin{align}
  & -1={{a}_{11}}={{\left( -1 \right)}^{1+1}}\begin{matrix}
   1 & -2 \\
   -2 & 1 \\
\end{matrix} \\
 & =1-\left( 4 \right) \\
 & =-3
\end{align}$
Continuing the process for each element,
$\begin{align}
  & -2={{a}_{12}}={{\left( -1 \right)}^{1+2}}\begin{matrix}
   2 & -2 \\
   2 & 1 \\
\end{matrix} \\
 & =-\left( 2-\left( -4 \right) \right) \\
 & =-6
\end{align}$
$\begin{align}
  & -2={{a}_{13}}={{\left( -1 \right)}^{1+3}}\begin{matrix}
   2 & 1 \\
   2 & -2 \\
\end{matrix} \\
 & =\left( -4 \right)-\left( 2 \right) \\
 & =-6
\end{align}$
$\begin{align}
  & 2={{a}_{21}}={{\left( -1 \right)}^{2+1}}\begin{matrix}
   -2 & -2 \\
   -2 & 1 \\
\end{matrix} \\
 & =-\left( \left( -2 \right)-\left( 4 \right) \right) \\
 & =6
\end{align}$
$\begin{align}
  & 1={{a}_{22}}={{\left( -1 \right)}^{2+2}}\begin{matrix}
   -1 & -2 \\
   2 & 1 \\
\end{matrix} \\
 & =-1-\left( -4 \right) \\
 & =3
\end{align}$
$\begin{align}
  & -2={{a}_{23}}={{\left( -1 \right)}^{2+3}}\begin{matrix}
   -1 & -2 \\
   2 & -2 \\
\end{matrix} \\
 & =-\left( 2-\left( -4 \right) \right) \\
 & =-6
\end{align}$
$\begin{align}
  & 2={{a}_{31}}={{\left( -1 \right)}^{3+1}}\begin{matrix}
   -2 & -2 \\
   1 & -2 \\
\end{matrix} \\
 & =4-\left( -2 \right) \\
 & =6
\end{align}$
$\begin{align}
  & -2={{a}_{32}}={{\left( -1 \right)}^{3+2}}\begin{matrix}
   1 & -2 \\
   -2 & 1 \\
\end{matrix} \\
 & =-\left( 1-\left( 4 \right) \right) \\
 & =3
\end{align}$
$\begin{align}
  & 1={{a}_{33}}={{\left( -1 \right)}^{3+3}}\begin{matrix}
   -1 & -2 \\
   2 & 1 \\
\end{matrix} \\
 & =-1-\left( -4 \right) \\
 & =3
\end{align}$
So, these are the co-factors of the determinant.
Now put the value in the matrix for finding the adjoint matrix. But remember the factor that the values must put in the transpose manner.
So, the adjoint matrix is $adjA=\left( \begin{matrix}
   -3 & 6 & 6 \\
   -6 & 3 & 3 \\
   -6 & -6 & 3 \\
\end{matrix} \right)$
Now, we have to prove the $A.\left( adjA \right)=\left| A \right|{{I}_{3}}$
So, L.H. S$=A.\left( adjA \right)$
So, we have to multiply the matrix $A$ with the adjoint matrix.
So, the product is $=A.\left( adjA \right)$
$\begin{align}
  & =\left( \begin{matrix}
   -1 & -2 & -2 \\
   2 & 1 & -2 \\
   2 & -2 & 1 \\
\end{matrix} \right)\times \left( \begin{matrix}
   -3 & 6 & 6 \\
   -6 & 3 & 3 \\
   -6 & -6 & 3 \\
\end{matrix} \right) \\
 & =\left( \begin{matrix}
   3+12+12 & -6-6+12 & -6-6-6 \\
   -6-6+12 & 12+3+12 & 12+3-6 \\
   -6+12-6 & 12-6-6 & 12-6+3 \\
\end{matrix} \right) \\
 & =\left( \begin{matrix}
   27 & 0 & 0 \\
   0 & 27 & 0 \\
   0 & 0 & 27 \\
\end{matrix} \right) \\
\end{align}$
Now we have to find out the determinant value of $A$Matrix.
$A=\left( \begin{matrix}
   -1 & -2 & -2 \\
   2 & 1 & -2 \\
   2 & -2 & 1 \\
\end{matrix} \right)$
$\begin{align}
  & \left| A \right|=\left( -1 \right)\left| \begin{matrix}
   1 & -2 \\
   -2 & 1 \\
\end{matrix} \right|-2\left| \begin{matrix}
   -2 & -2 \\
   -2 & 1 \\
\end{matrix} \right|+2\left| \begin{matrix}
   -2 & -2 \\
   1 & -2 \\
\end{matrix} \right| \\
 & =\left( -1 \right)\left( 1-4 \right)-2\left( -2-4 \right)+2\left( 4+2 \right) \\
 & =3+12+12 \\
 & =27
\end{align}$
Now in right hand side we see the unit matrix which is of order $3$
That means ${{I}_{3}}=\left( \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right)$
So, the R.H.S is $\left| A \right|{{I}_{3}}$
$\begin{align}
  & =27\left( \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right) \\
 & =\left( \begin{matrix}
   27 & 0 & 0 \\
   0 & 27 & 0 \\
   0 & 0 & 27 \\
\end{matrix} \right) \\
\end{align}$
So, as the L.H.S is equal to R.H.S so hence proved that $A.\left( adjA \right)=\left| A \right|{{I}_{3}}$

Note: Here students have to take care of two or three factors. Once the adjoint matrix is evaluated there the value must be in the transpose of the value that is obtained. Another thing is whenever the adjoint matrix is evaluated we have to take care of cofactor and minor of that matrix with proper sign. And always whenever the step of multiplication students have to take row column and row method.