Find the inverse of the following matrix without using a calculator 1-1 2 - 3 2 1 0 4 - 25

18 - (17/3) (5/3)

25 (25/3) (7/3)

4 (4/3) (1/3)

Step-by-step explanation:

You can solve this problem by using the Gauss-Jordan method.

You have the original matrix and then the Identity matrix.

So:

Original Identity

1 - 1 2 1 0 0

-3 2 1 0 1 0

0 4 - 25 0 0 1

By the Gauss-Jordan method, in the original place you will have the identity and in the place that the identity currently is you will have the inverse matrix:

So, let's start by setting the first row element to 0 in the second and the third line.

The first row element of the third line is already at zero, so no changes there. In the second line, we need to do:

L2 = L2 + 3L1

So now we have the following matrixes.

1 - 1 2 | 1 0 0

0 - 1 7 | 3 1 0

0 4 - 25 | 0 0 1

Now we need the element in the second line, second row to be 1. So we do:

L2 = - L2

1 - 1 2 | 1 0 0

0 1 - 7 | - 3 - 1 0

0 4 - 25 | 0 0 1

Now, in the second row, we need to make the elements at the first and third line being zero. So, we have the following operations:

L1 = L1 + L2

L3 = L3 - 4L2

Now our matrixes are:

1 0 - 5 | - 2 - 1 0

0 1 - 7 | - 3 - 1 0

0 0 3 | 12 4 1

Now we need the element in the third line, third row being one. So we do:

L3 = - L3

1 0 - 5 | - 2 - 1 0

0 1 - 7 | - 3 - 1 0

0 0 1 | 4 (4/3) (1/3)

Now, in the third row, we need the elements in the first and second line being zero. So we do:

L1 = L1 + 5L3

L2 = L2 + 7L3

So we have:

1 0 0 | 18 - (17/3) (5/3)

0 1 0 | 25 (25/3) (7/3)

0 0 1 | 4 (4/3) (1/3)

So the inverse matrix is:

18 - (17/3) (5/3)

25 (25/3) (7/3)

4 (4/3) (1/3)