00:01
For this question, we have a, which is an n -by -n matrix, and we're asked to prove that a times in is equal to in times a, which is equal to a.
00:08
Let's remember what in is.
00:11
I -n is an m -by -n matrix such that it's ij's entry, denoted by i -j, is equal to 0 if i is not equal to j, and it's equal to 1 if i is equal to j.
00:25
To visualize, i .n is a matrix.
00:30
Such that we have ones on the diagonals and zero everywhere else.
00:37
Something like this.
00:41
So how do we show a times i n is equal to in equal to a is equal to a? let's show it by steps.
00:48
So let's first show a times i n is equal to a.
00:52
To make it more convenient, let's call this matrix b.
00:57
So two matrices are equivalent if they're equal to each other, if every entry is equal to each other.
01:05
So if we can show that the ij's entry of b is the ij's entry of a, we are done.
01:10
So let's see what is the ij entry of b, b ij, what is it? using the definition, using the formula for matrix multiplication, we see that since b is equal to a times in, b ij is equal to a, i1, times i1j, plus a, i2, plus i2j plus a i3 oh sorry this is a typo this is supposed to be multiplication right here a i2 times i2 j um plus a i3 times i 3 j plus all the way to a i n times i n j so looking at the entries for i n if i is not equal to j, the i ij term will be zero.
02:11
So for every single one of these terms, besides one, the i ij term will be zero.
02:19
So we're left with the only term that's not zero is when i is equal to j.
02:24
And that is the case if we have i j, j, j.
02:33
Because we're always working with the j column and what is the corresponding a term? well, if we're in the j -th column, we must have a -i -j.
02:49
Now, j is equal to j, so this term becomes 1.
02:55
So essentially we're dealing with a -i -j times 1, which is just a -i -j...