Inverse of a 2x2 matrix with an example

Question

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?

Answer 1

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 $ad-bc\neq 0.$

Answer 2

What you actually do is multiplication by elementary matrices. Doing this you can perform elementary transformations. These are

1. swapping two rows / columns
2. multiplying a row / column by a scalar
3. 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.

Answer 3

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 $$\det(M)\ M^{-1} \:=\:\operatorname{trace}(M)\ \mathbb 1 \,-M\;=\; \begin{bmatrix} d & -b\\ -c & a\end {bmatrix}\,.$$