i need help solving these matrices by finding the inverse?

Question

[3 -1
-2 2 ]


[1 -1 0
1 0 -1
6 -2 -3]

Answer 1

There are a lot of ways to do it.

One way is called Gauss-Jordan Elimination.

To do it, you put an identity matrix on the right, and transform the original matrix into an identity matrix. What's left on the right is the inverse.

The three things you are allowed to do are:
1) You can swap rows.
2) You can multiply a row by a number.
3) You can multiply a row by a number and add or subtract it to another row.

+3 −1 | +1 +0
−2 +2 | +0 +1

R1 = R1 + R2:

+1 +1 | +1 +1
−2 +2 | +0 +1

R2 = 2R1 + R2:

+1 +1 | +1 +1
+0 +4 | +2 +3

R2 = R2 / 4:

+1.00 +1.00 | +1.00 +1.00
+0.00 +1.00 | +0.50 +0.75

R1 = R1 - R2:

+1.00 +0.00 | +0.50 +0.25
+0.00 +1.00 | +0.50 +0.75

So the matrix on the right:

+0.50 +0.25 = 1/2 1/4
+0.50 +0.75 = 1/2 3/4

Is the inverse.

- - - - - - - - - - - - - - -

+1 −1 +0 | +1 +0 +0
+1 +0 −1 | +0 +1 +0
+6 −2 −3 | +0 +0 +1

Swap R1 and R2:

+1 +0 −1 | +0 +1 +0
+1 −1 +0 | +1 +0 +0
+6 −2 −3 | +0 +0 +1

R2 = R2 - R1:

+1 +0 −1 | +0 +1 +0
+0 −1 +1 | +1 −1 +0
+6 −2 −3 | +0 +0 +1

R2 = R2(-1)

+1 +0 −1 | +0 +1 +0
+0 +1 −1 | −1 +1 +0
+6 −2 −3 | +0 +0 +1

R3 = R3 - 6R1:

+1 +0 −1 | +0 +1 +0
+0 +1 −1 | −1 +1 +0
+0 −2 +3 | +0 −6 +1

R3 = R3 + 2R2:

+1 +0 −1 | +0 +1 +0
+0 +1 −1 | −1 +1 +0
+0 +0 +1 | −2 −4 +1

R2 = R2 + R3:

+1 +0 −1 | +0 +1 +0
+0 +1 +0 | −3 −3 +1
+0 +0 +1 | −2 −4 +1

R1 = R1 + R3:

+1 +0 +0 | −2 −3 +1
+0 +1 +0 | −3 −3 +1
+0 +0 +1 | −2 −4 +1

So the inverse is:

−2 −3 +1
−3 −3 +1
−2 −4 +1

Answer 2

For small matrices, the adjoint method works well. This works for any size matrix but becomes tedious beyond 4. To use the adjoint method:

1. Pick any element in the original matrix, cross out the row and column that the element is in and figure the determinant of what is left. Write the result down in matrix form.

2. Take the transpose (fip the matrix around the diagonal)

3. For any element where the sum of the row and column number are odd, change the sign (that is, multiply by -1).

4. Divide the whole resulting matrix by the determinant of the original matrix.

That's it. The result is the inverse. I wil demonstrate:

+1 -1 +0
+1 +0 -1
+6 -2 -3

I've added plus sings to make things line up in Yahoo Answers.

Do the first element, cross out the top row and left column leaving:

+0 -1
-2 -3

The determinant is -2 so that goes in the upper left of the answer. Crossing out the first row second column leaves:

+1 -1
+6 -3

With a determinant of +3

Continute term by term and you get:

-2 +3 -2
+3 -3 +4
+1 -1 +1

Take the trranspose:

-2 +3 +1
+3 -3 -1
-2 +4 +1

Switching signs on the elements with an odd row+column sum:

-2 -3 +1
-3 -3 +1
-2 -4 +1

Figure the determinant of the origianl matrix:

(1 x 0 x -3) + (-1 x -1 x 6) + (1 x -2 x 0) - (6 x 0 x 0) - (1 x -1 x -2) - (1 x -1 x -3) = 1

Divde the above matrix elements by 1` giving the inverse:

-2 -3 +1
-3 -3 +1
-2 -4 +1

On a 2x2 the adjoint method is really easy since crossing out a row and column leaves a single element. You can combine the determinant step, the sign switching, ahd the transpose by switching elements along the diagonal and negating the two off diagonal elements, so:

+3 -1
-2 +2

Becomes:

+2 +1
+2 +3

The determinant is (2x3) - (2x1) = 4. Dividing by 4 gives the inverse:

+1/2 +1/4
+1/2 +3/4

Answer 3

To solve for the inverse of a matrix, all you have to do is put it side by side with the identity matrix, put the matrix in reduced row echelon form, and the matrix you attached it to will become the inverse. I'll show you what I mean.

[3 -1]
[-2 2]

Your first step would be to put it side by side with the identity matrix. I'll use a vertical bar to separate the matrices.

[3 -1 | 1 0]
[-2 2 | 0 1]

Since we want a leading 1 in the first row, we have to create one.
R1 -> (1/3)R1.

[1 -1/3 | 1/3 0]
[-2 2 | 0 1]

Now, we want the leading 1 to have 0s under it.

R2 -> R2 + 2R1

[1 -1/3 | 1/3 0]
[0 4/3 | 2/3 1]

Now we want a leading one in the second row.

R2 -> (3/4)R2

[1 -1/3 | 1/3 0]
[0 1 | 1/2 3/4]

R1 -> R1 + (1/3)R2

[1 0 | 1/2 1/4]
[0 1 | 1/2 3/4]

Therefore, your inverse is

[1/2 1/4]
[1/2 3/4]

Let's verify that this is indeed the inverse. If we multiply this by our matrix, we should get the identity matrix.

[1/2 1/4] [3 -1]
[1/2 3/4] [-2 2]

[3/2 - 1/2 -1/2 + 1/2]
[3/2 - 3/2 -1/2 + 3/2]

This is equal to

[1 0]
[0 1]

So this is correct

For the second question you do the exact same thing, except you put it side to side with a 3 x 3 identity matrix,

[1 0 0]
[0 1 0]
[0 0 1]