Hi I was working on inverting 2by2 matrix in general form by using a,b,c,d.
I know the formula (which is below) but I have questions in the process of getting the formula.
1) To get rid of the second column, it seems like the order of subtraction matter. How can you tell which one should be subtracted by the other? Because to get zero, it's possible either way.
Clarified in the comments:
I meant that to get (0,1) vector for column 2, I used linear transformation (same as I did for column 1). And the first element needs to be zero. In order to get zero, I have two ways. $b(b2)-(ad-cd)(b1)$ or $(ad-cd)(b1)-b(b2)$ for $T(b1,b2)$. But looking at the result, I can't go with $b(b2)-(ad-cd)(b1)$. Why?
2) What is limitation/restriction for linear transformation here? For example, can we use zero and one to make the calculation a lot easier? ( I know it won't work but want to know why it doesn't work).
Thanks!
To find the inverse of $A$ in general you row reduce the augmented matrix $\begin{bmatrix} A & I \end{bmatrix}$ into the form $\begin{bmatrix} I & B \end{bmatrix}$, then $B$ is the resulting inverse. So in the general $2 \times 2$ case you start with
$$\begin{bmatrix} a & b & 1 & 0 \\ c & d & 0 & 1 \end{bmatrix}.$$
Now you want to use row operations to cancel the $c$. Multiply the first row by $c/a$ to get
$$\begin{bmatrix} c & bc/a & c/a & 0 \\ c & d & 0 & 1 \end{bmatrix}.$$
Now subtract the second row from the first row:
$$\begin{bmatrix} c & cb/a & c/a & 0 \\ 0 & d-bc/a & -c/a & 1 \end{bmatrix}$$
For convenience let's clear the fractions:
$$\begin{bmatrix} ac & bc & c & 0 \\ 0 & ad-bc & -c & a \end{bmatrix}.$$
Now we need to cancel the $bc$, so we multiply the second row by $\frac{bc}{ad-bc}$, getting
$$\begin{bmatrix} ac & bc & c & 0 \\ 0 & bc & -\frac{bc^2}{ad-bc} & \frac{abc}{ad-bc} \end{bmatrix}.$$
Now we subtract the second row from the first row:
$$\begin{bmatrix} ac & 0 & c+\frac{bc^2}{ad-bc} & -\frac{abc}{ad-bc} \\ 0 & bc & -\frac{bc^2}{ad-bc} & \frac{abc}{ad-bc} \end{bmatrix}.$$
Now when you divide out the $ac$ and $bc$ in the first and second rows you get the result (some simplification is required in one place).
Note that in this argument I am assuming that $a,b,c$ and of course $ad-bc$ are all nonzero (since I divide by each of them at some point). Some small modifications are required when any of them are zero (in particular, row exchanges may be required).