# Homework Help: A matrix multiplied by it's inverse is the identity matrix, right?

1. Jan 28, 2010

### krtica

Matrix A= 2x2, R1= -1, -1, R2= -7, 3
Matrix b= 2x2, R1= 1,0, R2= 0, 1

A*?=b
____________

To solve, I put ? on the one side of the equation as ?=A^(-1)b. My answer is then just the inverse of A, because what is multiplied by the identity matrix is itself. It is shown to be incorrect.

?=2x2, R1= -3, -1, R2= -7, -1

Also, is there a way to format matrices on the forum?

2. Jan 28, 2010

### vela

Staff Emeritus
How did you calculate the inverse? The matrix you wrote for ? isn't the inverse of A.

3. Jan 28, 2010

### LCKurtz

You just forgot to divide by the determinant of A when you calculated the inverse. You can click on the equation below to see how to write a matrix in tex.

$$A^{-1}=\left [ \begin{array}{cc} -\frac 3 4&-\frac 1 4\\ -\frac 7 4&-\frac 1 4 \end{array} \right ]$$

4. Jan 29, 2010

### krtica

I did forget to divide the matrix by its determinate, 1/[ad-bc]. It would then be my previous answer divided by -1/10, right?

It turned up incorrect, still. I also tried your answer, but still the same response.

5. Jan 29, 2010

### LCKurtz

The determinant is -10, but you don't "divide by -1/10". You divide by -10. So you should get

$$A^{-1} = -\frac {1} {10}\left[ \begin{array}{cc} 3&1\\ 7&-1 \end{array} \right ]$$

6. Jan 29, 2010

### krtica

That is what I meant. The answer is still wrong.

7. Jan 29, 2010

### LCKurtz

$$A^{-1} = -\frac {1} {10}\left[ \begin{array}{cc} 3&1\\ 7&-1 \end{array} \right ]$$

is the correct inverse. If you are trying to solve a system like

Ax = b

where x and b are column vectors then the solution will be

x = A-1b

8. Jan 29, 2010

### vela

Staff Emeritus
Are you sure you're using the same matrix as LCKurtz? Your original inverse not only was missing the determinant factor but had the wrong sign on one of the elements as well.