Question

if 3A= (1 2 2)
(2 1 -2)
(-2 2 -1)
then show that A- 1 =AT​

Answer 1

If $A=\left[\begin{array}{ccc}-1&-2&-2\\2&1&-2\\2&-2&1\end{array}\right]$

then show that the adjoint of A is 3A'. Find $A^{-1}.$

We have:

Solution:

$A=\left[\begin{array}{ccc}-1&-2&-2\\2&1&-2\\2&-2&1\end{array}\right]$

[A] = - 1(1 - 4) + 2(2 + 4) - 2( - 4 - 2)

   = 3 + 12 + 12 = 27 ≠ 0,  $A^{-1}$ exists.

∴ Adjoint of $A=\left[\begin{array}{ccc}-3&-6&-6\\6&3&-6\\-6&-6&3\end{array}\right]'$

Adjoint of $A=\left[\begin{array}{ccc}-3&6&6\\-6&3&-6\\-6&-6&3\end{array}\right]$

$A'=\left[\begin{array}{ccc}-1&2&2\\-2&1&-2\\-2&-2&1\end{array}\right]$

∴ $3A'=3\left[\begin{array}{ccc}-1&2&2\\-2&1&-2\\-2&-2&1\end{array}\right]$

$3A'=\left[\begin{array}{ccc}-3&6&6\\-6&3&-6\\-6&-6&3\end{array}\right]$

Thus, adjoint of A = $3A'=\left[\begin{array}{ccc}-3&6&6\\-6&3&-6\\-6&-6&3\end{array}\right]$ , shown.

We know that:

∴$A^{-1}=\dfrac{1}{9} \left[\begin{array}{ccc}-1&2&2\\-2&1&-2\\-2&-2&1\end{array}\right]$

$A^{-1}= \left[\begin{array}{ccc}\dfrac{-1}{9} &\dfrac{2}{9}&\dfrac{2}{9}\\\dfrac{-2}{9}&\dfrac{1}{9}&\dfrac{-2}{9}\\\dfrac{-2}{9}&\dfrac{-2}{9}&\dfrac{1}{9}\end{array}\right]$

∴ $A^{-1}= \left[\begin{array}{ccc}\dfrac{-1}{9} &\dfrac{2}{9}&\dfrac{2}{9}\\\dfrac{-2}{9}&\dfrac{1}{9}&\dfrac{-2}{9}\\\dfrac{-2}{9}&\dfrac{-2}{9}&\dfrac{1}{9}\end{array}\right]$

Answer 2

It is proved using Matrix Inverse and Transpose calculation that the inverse and transpose are the same i.e. $A^{-1} = A^{T}$.

The given matrix is :

$3A = \left[\begin{array}{ccc}1&2&2\\2&1&-2\\-2&2&-1\end{array}\right]$

For finding the inverse of the given matrix, we perform row and column transformations to convert it to an identity matrix.

Step 1 : Subtract Row 1 multiplied twice from Row 2 and update in Row 2

So, the matrix becomes :

             $\left[\begin{array}{ccc}1&2&2\\0&-3&-6\\-2&2&-1\end{array}\right]$

And the identity augment becomes :

            $\left[\begin{array}{ccc}1&0&0\\-2&1&0\\0&0&1\end{array}\right]$          

Step 2 : Add Row 1 multiplied twice to Row 3 and divide Row 2 by -3.

So, the matrix becomes :

             $\left[\begin{array}{ccc}1&2&2\\0&1&2\\0&6&3\end{array}\right]$

And the identity augment becomes :

            $\left[\begin{array}{ccc}1&0&0\\2/3&-1/3&0\\2&0&1\end{array}\right]$          

Step 3 : Add Row 2 multiplied twice from Row 1 and Row 2 multiplied by 6 from Row 3. Then, divide Row 3 by 9.

So, the matrix becomes :

             $\left[\begin{array}{ccc}1&0&-2\\0&1&2\\0&0&1\end{array}\right]$

And the identity augment becomes :

            $\left[\begin{array}{ccc}-1/3&2/9&-2/9\\2/3&-1/3&0\\2/9&-2/9&-1/9\end{array}\right]$          

Step 4 : Add Row 3 multiplied twice to Row 1 and Row 2.

So, the matrix becomes (Identity) :

             $\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$

And the identity augment becomes ( Inverse of 3A ) :

            $\left[\begin{array}{ccc}1/9&2/9&-2/9\\2/9&1/9&2/9\\2/9&-2/9&-1/9\end{array}\right]$          

So, finally, the inverse of A is ( above inverse divided by 3 ) :

           $A^{-1} = \left[\begin{array}{ccc}1/3&2/3&-2/3\\2/3&1/3&2/3\\2/3&-2/3&-1/3\end{array}\right]$

Now, we find the transpose of the given matrix by reversing the rows and the columns :

          $A = A^{-1} = \left[\begin{array}{ccc}1/3&2/3&2/3\\2/3&1/3&-2/3\\-2/3&2/3&-1/3\end{array}\right]$

        $A^{T} = \left[\begin{array}{ccc}1/3&2/3&-2/3\\2/3&1/3&2/3\\2/3&-2/3&-1/3\end{array}\right]$

        $A^{T} = A^{-1}$

Hence, proved that the inverse and transpose of the given matrix are the same.

         

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