How do I manipulate matrix equations to find the inverse of a 2x2 matrix?

[MurdocJensen]

Homework Statement

Find the inverse of matrix A

02
30

The Attempt at a Solution

I was thinking of doing a row swap to get a diagonal matrix with nonzero diagonal entries, PA (a.k.a. B). I want this matrix's inverse, B-inverse (easily found by dividing the ones of the identity matrix by the diagonal entries) to serve as a means to get to A-inverse

I want to use this relationship specifically: (B-inverse)(PA)=(A-inverse)(A)=I. I want to multiply all sides by A-inverse to show that (B-inverse)(P)=(A-inverse), but I am really shaky as to how I properly utilize the multiplication rules for matrices in this case.

For example: Would multiplying both sides by A-inverse cancel out A? Wouldn't I be applying A-inverse to the outermost matrix and not even hit A?

To sum: Where I really get lost is how to properly manipulate matrix equations in order to cancel.

Let me know if you have any questions. Sorry I couldn't make the post more visual. Don't know how to draw out matrices.

 

[w3390]

You could just use the identity that any matrix times its inverse equals the identity matrix.

 

[JonF]

You could just use the identity that any matrix times its inverse equals the identity matrix.

To use this identity you augment your matrix with the identity matrix. So you start with (A|I) where I is the identity matrix. You use matrix operations to transform A into I and your augmented matrix will be A^-1.

So you start with:

02|10
30|01

And you want to use elementary row operations to make your matrix to look like:

10|ab
01|cd

Where a,b,c,d will make up you’re a^-1

 

[MurdocJensen]

This is Gauss-Jordan, no? So even though I am permutating the original matrix to get another matrix, the end result is A^-1, not the inverse of the matrix after permutation?

 

[HallsofIvy]

Or just use $$\begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}0 & 2 \\ 3 & 0\end{bmatrix}= \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$$
to get the equations 0a+ 3b= 3b= 1, 2a+ 0b= 2a= 0, 0c+ 3d= 3d= 0, and 2c+ 0d= 2c= 1. Those equations are pretty close to begin trivial, aren't they?

 

[Mark44]

There's a formula that can be used to find the inverse of a 2x2 (only) matrix.
$$\text{If} A = \begin{bmatrix}a & b \\ c & d\end{bmatrix},$$
$$A^{-1} = \frac{1}{|A|}\begin{bmatrix}d & -b \\ -c & a\end{bmatrix}$$

Of course, it must be true that |A| is not 0.

 

[MurdocJensen]

At the moment we are ignoring the determinant formula for the 2x2 case. That's what that is, right, Mark 44?

Thanks for all the help. I don't know why I was going the direction I was with this. For some reason I keep thinking this class requires a ridiculously abstract approach when, in most cases, I can use simpler mechanisms.