# Homework Help: Finding the inverse and finding a matrix * A = 0 matrix

1. Sep 29, 2005

### mr_coffee

Hello everyone!! Matrices fun! Anyways,
I'm trying to find a matrix that is equal to the idenity matrix if u multiply 2 matrices together. Well that matrix is the inverse So i'm trying to find the inverse but the answer is wrong:
A =
-5 3
2 -9

So i found the determinant and i switched the a and d, and negated the c and b . isn't htat the inverse?

1/35-6 = 1/29

-9 -3
-2 -5

-9/29 -3/29
-2/29 -5/29

isn't thtat the inverse of A?

I also have B =
-1 6
5 -30

I need to multiply that by some matrix C so the resultant matrix is
0 0
0 0
but i can't just say C is equal to
0 0
0 0
any ideas? thanjks!

I did notice, Row 1 is just 1/5 row 2

2. Sep 29, 2005

### TD

You have to switch a and d, also switch b and c and then change the signs of these last 2. After that, you have to divide by the determinant. So:

$$A = \left( {\begin{array}{*{20}c} a & b \\ c & d \\ \end{array} } \right) \Rightarrow A^{ - 1} = \frac{1} {{\det A}}\operatorname{adj} A = \frac{1} {{ad - bc}}\left( {\begin{array}{*{20}c} d & { - c} \\ { - b} & a \\ \end{array} } \right)$$

If you don't "see" it right away, work out the following matrix product to get a system of lineair equations.

$$\left( {\begin{array}{*{20}c} { - 1} & 6 \\ 5 & { - 30} \\ \end{array} } \right) \cdot \left( {\begin{array}{*{20}c} a & b \\ c & d \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} 0 & 0 \\ 0 & 0 \\ \end{array} } \right)$$

3. Sep 29, 2005

### Hurkyl

Staff Emeritus
Multiplying a matrix by something to get zero strongly reminds me of the notion of a null space...

4. Sep 29, 2005

### mr_coffee

Thanks TD, but for some reason its still wrong, i got:
-9/29 -2/29
-3/29 -5/29

5. Sep 29, 2005

### TD

Check your determinant again, that is ad-bc

6. Sep 29, 2005

### mr_coffee

lol what the f, (-5)(-9) - (3)(2)
1/45-6
1/39 right?

7. Sep 29, 2005

### TD

Well, the determinant is 39 (so not 29). Then, you have to divide by it indeed.
So 1/39 * adj(A)