00:01
In this problem we are given this matrix a.
00:05
1, 2, 1, 1, 3, 5, 3, 3, 1, 2, 8, 2, 1, 2, 9, 2.
00:19
And we are going to decide whether this matrix is invertible and if so, we will find its inverse using the adjoint method.
00:30
To decide whether it is invertible or not, we should compute the determinant.
00:37
If it is equal to 0, then it is non -invertible and if it is some other number than 0, then it is invertible.
00:47
Let us expand the determinant with respect to the first row.
00:51
So we have 1 times the determinant of the remaining matrix.
00:58
So we have 5, 3, 3, 2, 8, 2, 2, 9, 2 minus 3 times the determinant of the rest, 2, 1, 1, 2, 8, 2, 2, 9, 2 plus 1 times the determinant of this remaining matrix, 2, 1, 1, 5, 3, 3, 2, 9, 2.
01:41
Finally, we have minus 1 times the determinant of this remaining matrix.
01:48
We have 2, 1, 1, 5, 3, 3, 2, 8, 2.
01:57
Then we take one more step to simplify the rest of these determinants.
02:06
For the first term, we have 5 times the determinant of 8, 2, 9, 2 minus 2 times the determinant of 3, 3, 9, 2.
02:20
This is 9.
02:23
Plus 2 times 3, 3, 8, 2 minus 3 times.
02:31
Let's zoom out.
02:35
2 times 8, 2, 9, 2 minus 2 times 1, 1, 8, 2 plus 2 times.
02:46
Again, this is 9.
02:51
Finally, 2 times 1, 1, 8, 2 here.
02:56
Plus, we have 1 times.
02:59
As a rule of factor, 2, 3, 3, 9, 2 minus 5, 1, 1, 9, 2 plus 2, 1, 1, 3, 3...