Need to find inverse of this matrix:
$
\begin {bmatrix}
1 & 3/5\\
0 & 1\\
\end {bmatrix}
$
This is how it has been solved:
$
\begin {bmatrix}
1 & 3/5\\
0 & 1\\
\end {bmatrix}
$ $
\begin {bmatrix}
x_1 & x_2\\
x_3 & x_4\\
\end {bmatrix}
$ = $
\begin {bmatrix}
1 & 0\\
0 & 1\\
\end {bmatrix}
$ – step 1
$
\begin {bmatrix}
1 & 0\\
0 & 1\\
\end {bmatrix}
$ $
\begin {bmatrix}
x_1 & x_2\\
x_3 & x_4\\
\end {bmatrix}
$ = $
\begin {bmatrix}
1 & -3/5\\
0 & 1\\
\end {bmatrix}
$ – step 2
From step 1 to step 2, -3/5 added to first row, second column of $
\begin {bmatrix}
1 & -3/5\\
0 & 1\\
\end {bmatrix}
$
Also, -3/5 to first row,second column of $
\begin {bmatrix}
1 & 0\\
0 & 1\\
\end {bmatrix}
$
My query is if it is correct to add or deduct a particular number from say first column, first row on the right side and on the left side and both leave the equation intact?
Best Answer
One may find the inverse matrix by row operations on the augmented matrix:
$$ \left[\begin{array}{rr|rr} 1 & 3/5 & 1 &0 \\ 0 & 1 & 0 &1 \\ \end{array}\right]\overset{R_1'=-3/5R2+R1}{\longrightarrow} \left[\begin{array}{rr|rr} 1 & 0 & 1 &-3/5 \\ 0 & 1 & 0 &1 \\ \end{array}\right], $$
so $$ \begin {bmatrix} 1 & 3/5\\ 0 & 1\\ \end {bmatrix}^{-1} = \begin {bmatrix} 1 & -3/5\\ 0 & 1\\ \end {bmatrix}. $$
It is also useful to remember the general expression for the inverse of a 2x2 matrix:
$$ \begin {bmatrix} a & b\\ c & d\\ \end {bmatrix}^{-1}=\frac{1}{ad-bc} \begin {bmatrix} d & -b\\ -c & a\\ \end {bmatrix} $$
if the determinant $ad-bc\neq 0.$