Answer:
18 -(17/3) (5/3)
25 (25/3) (7/3)
4 (4/3) (1/3)
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)