00:01
In this problem, we have been given the two matrices and we have to determine the product if it exists.
00:07
So first we see that the first matrix that's a 3 by 3 matrix and the second one that's a 3 by 2 matrix.
00:15
And to understand whether the product exists or not, it's just sufficient that these two numbers will be and it should be same.
00:24
And we see that it's same here and hence the product exists and the resultant matrix that we will get.
00:30
Here will be 3 by 2 matrix.
00:32
So this will be equal to a 3 by 2 matrix in which we have three rows and two columns.
00:40
So we have created a vacant place here and the first element will be a11, the second one will be a12.
00:47
Similarly, in the second row it will be a21 -2, third row it will be a31 and a 3 -2.
00:54
So to get a1 for example, we will just take the first row of the first metrics and multiply that with the first column and we should multiply the corresponding elements like minus 1 is multiplied with 9 minus 9 is multiplied with minus 5 and 7 is multiplied with 5 and then we'll add them together so when we do that we get a 1 1 as minus 1 times 9 that's minus 9 minus 9 times minus 5 that's plus 45 7 times 5 that's 35 so here when we add 35 45 and minus 9, we get the result here as 71.
01:29
So let's keep filling that here.
01:32
So the first row, first element we got, let's get the first row second element.
01:37
So for that, we'll take the first row and the second column of the first and the second matrix respectively.
01:42
And when we multiply, we get 6 plus 0 plus 56.
01:47
And when we add 56 with 6, we get 62...