00:01
So we're given two matrices, a and b.
00:03
And we want to show that a is the inverse of v.
00:06
That is, we want to show that a is b raised to the negative 1.
00:11
So if a is the inverse matrix of b, then we know that a times b must equal the identity matrix for a3 by 3, which is just represented by a diagonal of 1, zeros on this side, and zeros on this 1.
00:30
So this is the identity matrix for a 3x3.
00:34
So if we multiply a times b and we obtain this matrix right here, we have proven that a is the inverse matrix of b.
00:44
So to perform these multiplications, we're going to take this first row of a and multiply it times the first column of b to obtain the first term of the first row, sorry, which is this one right here.
01:04
Then we're going to multiply the first row over here times the second column to obtain the second term of the first row on the new matrix.
01:14
Finally, we're going to repeat the process where our third column to find the third element in our first row.
01:22
And we're going to repeat this process with the remaining rows in matrix a.
01:29
So, this multiplication will look like 3 times negative 6 is just negative 18.
01:37
Plus, so we're going to add all of these terms, 8 times 7, which is just 56, and 2 times negative 1, which is just negative 2.
01:47
I'm going to draw a line over here, just so we can have enough space to determine which terms belong to which row and column.
01:57
Now we're going to multiply 3 times 84, which is just 252.
02:06
Now we're going to multiply 8 times negative 26, which is just negative 208.
02:14
And finally, 2 times negative 22, which is just negative 44.
02:21
Now if we do that with the third column, we're going to obtain 3 times negative 6 is just negative 18.
02:29
8 times 1 is just 8.
02:31
And then 2 times 5 is just 10, sorry.
02:40
Now for the second row, sorry, we have 1...