Given

Question

Given

A = [\begin{matrix} 4 & - 2 \\ 3 & 1 \end{matrix}],

and

C = \left[ \begin{matrix} -5 & 4 & 1 \ 0 & 3 & 6 \end{matrix} ight]

, find each product, if possible. See Examples 5–7. BA

Accepted Answer

Hello, everyone. We are asked to perform the multiplication of the matrices B A as they are defined B is a three by two matrix with a first row of 85, a second row of zero negative four, a third row of 19 A is a two by two matrix with the first row as negative six, negative five, second row is 72. We're also given matrix C but the problem does not ask for it. So we will not be using it all of our solutions. There are four possible answer choices all of which are three by two matrices. They have varying values and signs. So in this order we are asked B multiplied by A. So B is a three by two matrix where A is a two by two matrix. So just verifying that the columns from the first matrix match the rows of the second matrix. And as they do, we can multiply these and we'll have a solution that is a three by two matrix. Now I cannot make my brain work backwards. So I need to rewrite this so that I have matrix B 850 negative 419 written before matrix A which is negative six, negative 572. Now, I know I can see how to do this problem a lot better. Again, knowing that our answer is gonna be a three by two matrix. I'm gonna start setting it up over here and I recall I have three rows and I'm gonna have two columns. So my first row, the first element will come from the first row of matrix B multiplied by the first column of matrix A. So the first element of B which is eight multiplied by the first element of the first column of A which is negative six. And that gets added to the second element of the first row of B which is five multiplied by the second element of the first column of A which is seven continuing across my second element. In my first row of my solution will be the first row from B multiplied by the second column from A. So again, that would be eight multiplied by the first element of the second column which is negative five plus five multiplied by the second element of the second column which is two, second row will be the second row of B. So zero negative four. And we're gonna multiply it by the first column of A. So the first element of the first of the second row is zero multiplied by the first element of the first column which is negative six plus the second element of the second row of B is negative four. And that'll be multiplied by the second element of the first column of A, which is seven. Next I do again. The second row from B now multiplied by the second column of A. So zero multiplied by negative five plus negative four, multiplied by the second element of the second column which is two. And now I can begin on my third row of my solution, which will be using the third row of B which is 19. And for the first element, we are multiplying it by the first column of A. So one multiplied by negative six plus nine multiplied by seven. And then our last element we need to come up with would be the third row of B multiplied by the second column of A. So one multiplied by negative five plus nine multiplied by two. Now that I have it all written out, I will do the multiplication. So I get negative 48 plus 35. For my first element, next one, I have negative 40 plus 10, second row, I get zero plus negative 28. And next element I get zero plus negative eight, third row, negative six plus 56 and then negative five plus 18. And now I am ready to combine things. Oops actually, one little mistake there. Let's go just back and fix that nine times seven is 63. So that third element in the first column of our solution should be negative six plus 63. Now, when we add those together, the first row looks like it will be negative 13, negative 30. The second row will be negative 28 negative eight. The third row will be 57 13 and that should be our three by two matrix. That is our solution. And I'm gonna, again, the first row is negative 13, negative 32nd row is negative 28 negative eight, third row is 57 13. And that looks like it matches with answer choice. D have a nice day.

Given $A = [\begin{matrix} 4 & - 2 \\ 3 & 1 \end{matrix}],$ and $C = \(\left\)[ \(\begin{matrix}\) -5 & 4 & 1 \\ 0 & 3 & 6 \(\end{matrix}\) \(\right\)]$ , find each product, if possible. See Examples 5–7. BA

Key Concepts

Matrix Dimensions and Compatibility

Matrix Multiplication Process

Non-Commutativity of Matrix Multiplication

Watch next