00:01
Okay, for this problem, we're given two matrices, a and b.
00:04
We want to find what values, if any, of k will make a times b the same as b times a.
00:12
So the way we want to do, the way that we want to do this is first calculate a times b.
00:18
So that's going to be doing the row column method.
00:23
We have 2 times 4 plus 3 times 5, 3 times 15, so we have 8 plus 15 in there.
00:28
Then for the next row we have negative 3 times 4, which is going to be negative 12, plus 3 times 1.
00:38
Then for the first row, second column element, we have 2 times negative 5, which is negative 10, plus 5k.
00:48
Then we have, for the last element, negative 3 times negative 5, which is going to be positive 15 plus k.
00:57
So that's our ab.
00:59
Actually, i guess i should calculate out exactly those values.
01:03
A plus 15 is 23, negative 12 plus 3 is 9, but we can't actually say the negative 10 plus 5k.
01:11
Oh, actually, i guess the factor, a common factor of 5.
01:15
That turns into 5 times k minus 2.
01:22
And then we have 15 plus k.
01:25
Nothing fancy we can do there.
01:28
Next, looking at doing b times a.
01:32
For the logic of it i'd like to have the matrices next to each other visibly makes it a little bit easier to keep track of things so we have four times two going to be eight plus negative five times negative three so we have eight plus 15 looks familiar then we have for the next row we're going to have yeah sorry we're going to have three times two so so 6 minus 3 times k.
02:13
Then for the first row, second column element, we'll have 4 times 5.
02:19
That's 20 minus 5...