Question

By using elementary transformation find the inverse matrix of A = $\left[ \begin{matrix}
   1 & 2 \\
   2 & -1 \\
\end{matrix} \right]$

Answer 1

Hint: In question it is mentioned by elementary transformations. Elementary transformation is nothing but row transformations. We use the condition matrix multiplied by inverse gives identity. By row transformation we convert our matrix into identity similarly applying those to the identity matrix we get our inverse found directly because the equation before now changes to identity multiplied by inverse which in turn gives us inverse.

Complete step-by-step answer:

By basic knowledge of matrix, we know that this is true:

$A{{A}^{-1}}=I$

Here, A is given matrix

${{A}^{-1}}$ is inverse matrix

I is identity matrix

We write an augmented matrix by combining the matrix and identity matrix.

augmented matrix = $\left[ \left. A \right|I \right]$

By applying elementary transformation:

There are no column operations. So, the elementary transformation in turn say only row operations are applied. We convert A into I. So, this in turn converts I to ${{A}^{-1}}$ .After applying we get a new augmented matrix.

New augmented matrix = $\left[ \left. I \right|{{A}^{-1}} \right]$

Given in the question: the matrix A is given by following:

$A=\left[ \begin{matrix}

   1 & 2 \\
   2 & -1 \\
\end{matrix} \right]$

By adding an identity to the above matrix, we get the augmented matrix = $\left[ \left. \begin{matrix}

   1 & 2 \\

   2 & -1 \\

\end{matrix} \right|\begin{matrix}

   1 & 0 \\

   0 & 1 \\

\end{matrix} \right]$

We apply row transformation: (2nd row) – 2(1st row) = 2nd row

Mathematical: ${{R}_{2}}\to {{R}_{2}}-2{{R}_{1}}$ , we get:

Assume augmented matrix to be X

$X=\left[ \left. \begin{matrix}

   1 & 2 \\

   0 & -5 \\

\end{matrix} \right|\begin{matrix}

   1 & 0 \\

   -2 & 1 \\

\end{matrix} \right]$

Now again we apply row transformation: $\left( {{1}^{st}}row \right)=\left( {{1}^{st}}row \right)+\dfrac{\left( {{2}^{nd}}row \right)}{5}\times 2$

Mathematical: ${{R}_{1}}\to {{R}_{1}}+\dfrac{2}{5}{{R}_{2}}$ , we get:

$X=\left[ \begin{matrix}

   1 & 0 \\

   0 & -5 \\

\end{matrix}\left| \begin{matrix}

   \dfrac{1}{5} & \dfrac{2}{5} \\

   -2 & 1 \\

\end{matrix} \right. \right]$

Now this time we apply the row transformation: $\left( {{2}^{nd}}row \right)=\left(

\dfrac{{{2}^{nd}}row}{-5} \right)$

Mathematical: ${{R}_{2}}\to \dfrac{{{R}_{2}}}{-5}$ , we get:

$X=\left[ \begin{matrix}

   1 & 0 \\

   0 & 1 \\

\end{matrix}\left| \begin{matrix}

   \dfrac{1}{5} & \dfrac{2}{5} \\

   \dfrac{2}{5} & -\dfrac{1}{5} \\

\end{matrix} \right. \right]$

From the conditions we know when A gets converted into an identity matrix, the identity matrix gets converted into the inverse of A directly.

Here we converted A to identity. So, the second part is the inverse.

Here the second part of augmented matrix is ${{A}^{-1}}$

By above, we get:

${{A}^{-1}}=\left[ \begin{matrix}

   \dfrac{1}{5} & \dfrac{2}{5} \\

   \dfrac{2}{5} & -\dfrac{1}{5} \\

\end{matrix} \right]$

Take $-\dfrac{1}{5}$ common from the matrix we get

${{A}^{-1}}=-\dfrac{1}{5}\left[ \begin{matrix}

   -1 & -2 \\

   -2 & 1 \\

\end{matrix} \right]$

Hence this is inverse of A.

Note: As we can see the determinant of A is -5 that’s why we took $-\dfrac{1}{5}$ common at the last step to make the inverse look right.