00:01
In this problem we are given three matrices and we are going to compute a dear inverse using the inversion algorithm if the inverse ever exists.
00:13
So let's get started with the first matrix.
00:20
We have let's say m equal to minus 1, minus 4, 3, 4, 2, minus 4, 2, minus 4, 2, minus 4, 2, minus 9.
00:32
So the condition for the inverse to exist is that the determinant should be, should not be equal to zero.
00:43
So we first compute the determinant of this guy.
00:47
So okay.
00:49
I'm going to use the first row to get the determined.
00:54
So minus one times the determinant of the minor, corresponding minor.
01:00
So 4 .1 to minus 9.
01:03
Minus three times the minor, the determinant of the corresponding minor, so 2 -1, minus 4, minus 9, plus minus 4 times the determinant of the corresponding minor.
01:22
Like that.
01:24
So, minus, minus 36, minus 2, minus 2, minus 3, minus 18, like that.
01:33
4 minus 4 4 plus 16 and the result of this calculation will be 0 therefore this matrix is non -invertible okay that was easy now let's to the second part we have m equal to 222 6 7 7 6 7 let us compute to the tournament again using the first row.
02:19
So we have two times the determinant of the minor 7 -7 -67 minus 6 times the termament of the minor 2 to 6 -7 plus 6 times terminate of the minor 2 to 7 7.
02:38
So we have 2 times 49 minus 42 minus 6 times 14 minus 12 plus 6 times plus 6 times 14 minus 14 the calculation the result of its calculation is equal to 2 which is not equal to 0 therefore this guy is invertible.
03:06
Okay now the inverse of a matrix is given by the edge of the matrix divided by its determinant.
03:21
So its edge of the matrix is defined by the transpose of the co -factor matrix and the co -factor is given entry by in a symbolic manner or schematically let's say minus one to the power i plus j for this ij entries and the minor the determinant of the corresponding minor so okay let's write the following so this is schematically it will look like this we will have this plus minus pattern here like an alternating pattern and we will have all these determinants in these entries since we have a 3x3 matrix all these miners will be 2 by 2 which are easy to compute indeed and the entries go like this i want to copy re -express this guy over here so that we see what we are dealing with just as a reminder m equal to 2 2 2 677 667 6 667 okay so the minor of the first matrix 7 7 7 minor of the first entry i'm 1 1 entry is 7 7 67 minor of the entry 1 2 you have 2 2 6 6 7 and the minor of the entry one three two two seven seven now the minor of the entry two one six six seven for two two we have two six two seven and for two three two six two seven and four three one six seven four three one six seven four three two 2626 and for 3 3 23 2627 these determinants are really easy to compute and i will just write down the result 7 -0 minus 6 minus 2 2 0 minus 2 2 we have the by taking transpose we will get the edge sugar so 7 minus 2 0 0 2 minus 2 minus 6 0 2 and finally by dividing by dividing this guy by the determinant of the m matrix we compute the inverse to be 1 half times times this guy okay so this is the inverse of the matrix that is given to us in part b now let's do part c so it will be a bit length because we have a 4 by 4 matrix so the minors will be 3 by 3 so it will take slight just very slightly more time to compute the inverse so we have m equal to 2 1 1 -00 minus 4 -2 -0 minus 1 012 2 minus 4 0 -0 minus 5 let us compute determined and let us hope that it is 0 but it will not be clearly so i will consider the third row of the co -factor expansion the because we have only one non -zero entry over there.
08:05
So following this, this checker, i think, checkered board pattern of the signs.
08:14
So plus minus, plus minus, plus like that.
08:19
We can compute the determinant of this guy using this entry alone as two times the determinant of the minor of this end.
08:30
So 2 minus 4 0 1 to 0 and 0 minus 1 minus 5.
08:43
Now this is an easy determinant because we have lots of zeros.
08:51
So 2 times...