Inverse of a 2x2 matrix with an example 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?
 A: 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 $ad-bc\neq 0.$
A: Possibly more a commenting on your post, but worthwhile to be promoted to an answer.
Special$\;$ You want to invert a matrix which is upper triangular, thus the inverse will be of the same form, and your ansatz simplifies to
$$\begin {bmatrix}
1 & 0\\ 0 & 1
\end {bmatrix}
\:\stackrel{!}{=}\:
\begin {bmatrix} 1 & 3/5\\ 0 & 1\end {bmatrix}
\begin {bmatrix} x_1 & x_2\\ 0 & x_4\end {bmatrix}
\:=\:
\begin{bmatrix} x_1 & x_2+\frac35 x_4\\ 0 & x_4\end {bmatrix}$$
Hence $x_1=x_4=1\,$ and $\,x_2=-3/5\,$.
General$\;$ The general expression for the inverse of a $2\times 2$-matrix $M$ is invoked in Golden_Ratio's answer. A possibly to high-toned but instructive proof starts with the characteristic polynomial of $M$
$$\chi(\lambda)\:=\:\lambda^2 \,-\,\operatorname{trace}(M)\,\lambda \,+\, \det(M)$$
and exploits Cayley-Hamilton
$$M^2 \,-\,\operatorname{trace}(M)\,M \,+\, \det(M)\,\mathbb 1\;=\;
\begin{bmatrix} 0 & 0\\ 0 & 0\end {bmatrix}$$
Assume $\det M \neq 0$, then multiplying with $M^{-1}$ yields
$$\det(M)\ M^{-1} \:=\:\operatorname{trace}(M)\ \mathbb 1 \,-M\;=\;
\begin{bmatrix} d & -b\\ -c & a\end {bmatrix}\,.$$
A: What you actually do is multiplication by elementary matrices.
Doing this you can perform elementary transformations. These are

*

*swapping two rows / columns

*multiplying a row / column by a scalar

*adding a multiple of one row/column to another row /column
Depending if you multiply from the left or from the right performs the action of rows or columns.
In your case from step 1 to step 2, you add $(-3/5) \cdot (\text{row } 2)$
to $(\text{row }1).$
And this you do on both sides.

