# Homework Help: Why can't i figure out inverses! ahh! simple 3X3 matrix

1. Oct 15, 2005

### mr_coffee

Hello everyone, i have no idea why i can't grasp this simple concept...
i have:
A =
1 4 9
0 1 9
0 0 1

I have to find A^-1, A inverse.
So I found the determinant along row 3,
1*det(B) = 1;
B =
1 4
0 1
det(B) = (1)(1) - (4)(0) = 1;
so i take 1/det * A now wouldn't that just be A itself?
A inverse =
1 4 9
0 1 9
0 0 1

it said,
1 1 correct
4 4 incorrect
9 9 incorrect
0 0 correct
1 1 correct
9 9 incorrect
0 0 correct
0 0 correct
1 1 correct

2. Oct 15, 2005

### tmc

simply write the matrix [A|I] and row-reduce until you get [I|A^-1]

3. Oct 15, 2005

### mr_coffee

thats a harder method, he taught us that one, he is showing us a new way, by multiplying the adjoint by 1/determinant, ur way will work but its very time consuming.

4. Oct 15, 2005

### GDogg

No, it's the adjoint of A you have to multiply, not A...

5. Oct 15, 2005

### mr_coffee

how do u find the adjoint of A if its more t hen a 2x2?
I know how to find it if its a 2x2, by negating the b and c, and switching the a and d. But how would i find the adjoint of a 3 x3?

6. Oct 15, 2005

### GDogg

7. Oct 15, 2005

### HallsofIvy

You start by knowing the definition of "adjoint"!
The "i,j" entry in the adjoint is the "cofactor" of the "j,i" entry in the original matrix.
The "cofactor" of an "j,i" entry is the determinant of the matrix formed by dropping the row and column that entry is in, times 1 or -1 depending on the parity of i+j.
When you have more experience, especially with larger order matrices than "2 by 2", you will realize that "row reduction" is far easier!

Last edited by a moderator: Oct 15, 2005
8. Oct 16, 2005

### mr_coffee

Thanks everyone! :)

9. Oct 16, 2005

### mr_coffee

Ivy.... Say you have a huge matrix, your allowed to use ur calculator, wouldn't it be easier to find the determinant, because a calcualtor can do that quite easily, rather then row reducing?

10. Oct 16, 2005

### Disar

Matrix Inversion

So, letting matrix A =A

A= 1 4 9
0 1 9
0 0 1

The Determinant of A = 1

Taking the cofactors of A:
c(1,1) = '+' * determinant of | 1 9 |
| 0 1 |
= 1*1 - 0*9 = +1 (indicating positive 1)

c(1,2) = '-' * determinant of | 0 9 |
| 0 1 |
= '-' 0 = 0

Following the same logic ie + - + - + - + , etc till c(3,3)
The cofactors in matrix form are:
| 1 0 0 |
| -4 1 0 |
| 27 -9 1 |

Switching the rows with columns:

| 1 -4 27|
| 0 1 -9 |
| 0 0 1 |

Then dividing by the determinant of A:

| 1 -4 27|
| 0 1 -9 |
| 0 0 1 |

gives you the inverse of A

Checking with my handy dandy TI89

It checks out!