For each pair of matrices A and B, find (a) AB and (b) BA. See Ex...

Question

For each pair of matrices A and B, find (a) AB and (b) BA. See Example 7.A= , B =

Accepted Answer

Hello, everyone. We are asked to find the value of A B and B A for the following matrices. Matrix A is a two by two matrix with the first row of 78. And the second row of negative 51 B is a two by two matrix with a first row of 10 0. And a second row of three negative four, we have four answer choices all of which have 22 by two matrices because both A and B are two by two matrixes matrices. I know that my solution matrix will be a two by two matrix. So I'm gonna start with A B and the first element of the first row of our solution matrix will come from multiplying the first row of matrix A by the first column of matrix B. So the first element of the first row of A is seven. We're gonna multiply that by the first element of the first column of B which is 10. And we're gonna add to that the product of the second element of the first row of A which is eight multiplied by the second element of the first column of B which is three. And now we move to the second element in the first row of our answer matrix. And that's gonna be from the first row of a multiplied by the second column of B. So the first element in the first row is seven, the first element of the first column of B is zero, we're gonna add to that the product of the second element of the first row of A which is eight. And the second element of the second column of B which is negative four, moving down to the second row in our answer matrix, we are now going to do the second row of A multiplied by the first column of B. So the first element of the second row of A is negative five multiplying that by the first element of the first column of B which is 10, we're gonna add to that the second element of the second row of A which is one multiplied by the second element of the first column of B which is three. And then the last element we need to find the values for is the second element of the second row in our solution. And that's the second row of matrix A multiplied by the second column of matrix B. So negative five from matrix A multiplied by the first element of the second column of B which is zero added to the second element of the second row of A which is one multiplied by the second element of the second column of B which is negative four. Now that we have all our values, we are going to multiply and add. So for the first element in our solution matrix in the first row, seven multiplied by 10 is 70. And we're gonna add that to the product of eight and three, which is 24. So that should be 94. The second element in that row seven multiplied by zero is zero eight, multiplied by negative four is negative 32. Going down to the second row. Our first element we have negative five multiplied by 10 which is negative 50. And we're gonna add to that three. So that'd be negative 47. And then our second element in the second row negative five multiplied by zero is 01 multiplied by negative four is negative four. So we have our solution matrix for A B, it's a two by two matrix where the first row is negative 32 and the second row is negative 47 negative four. Now, we need to do this in the opposite direction. B A I personally need to rewrite these matrices. So B is listed first or I will choose the wrong values every time I try to multiply. So rewriting the original matrixes or I'm sorry, matrices B and then the A that we were given and this just helps me keep it in order. Otherwise like I said I would lose track of which row, which column I need to see it in the order I'm multiplying it in. So the first element in our solution matrix in the first row will be the first row of matrix B multiplied by the first column of matrix A. So the element 10 multiplied by seven plus the second element of the first row, which is zero multiplied by the second element of the first column which is negative five. The second element in the first row of our solution matrix is again the first row of B now multiplied by the second column of A. So 10 multiplied by the first element of the second column which is eight plus zero multiplied by the second element of the second column which is one going down to the second row of our answer answer matrix be the second row of B multiplied by the first column of A. So the first element of each of those three multiplied by seven plus the second element of the second row is negative four multiplied by the second element of the first column which is negative five. Moving to the second element of our second row of our solution matrix, we now have the second row of B multiplied by the second column of A. So three multiplied by eight plus the second element of the second row. So negative four multiplied by the second element of the second column which is one. All right. Again, I'm gonna do my addition and multiplication in this step. So starting at the top left corner, my first element I have seven multiplied by 10, which is 70 plus zero is still second element I have 10 multiplied by eight, which is 81 multiplied by zero is zero. So this is 80 going down to the second row, I have three multiplied by seven which is plus negative four multiplied by negative five, which is 2020 plus 21 is 41. For the second element I have three multiplied by eight, which is 24 plus negative four multiplied by one which is negative four. So this would be a positive 20. So here is our solution matrix for B A to get a two by two matrix, the first row is 70 80. The second row is 41 20. So now looking at my answer choices trying to match those up, um It looks like either A or C is right for A B. So now I need to check for B A and it's going to be answer choice. C have a nice day.

For each pair of matrices A and B, find (a) AB and (b) BA. See Example 7.A= , B =

Key Concepts

Matrix Multiplication

Dimensions of Matrices

Associative and Distributive Properties

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