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)

3 is greater then both of those numbers