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. BC

Accepted Answer

Hello, everyone. We are asked to perform multiplication on the matrices BC as they are defined B is a three by two matrix with the top row as 85, the second row as zero negative four, the third row as 19 and matrix C as a two by three matrix with the top row negative 278. And the second row as 04 12, we are also given matrix A but the question does not ask us to multiply that. So we will not be worrying about it. We have four answer choices all of which are three by three matrices with slightly different values and signs looking at matrix B again, we know this is a three by two matrix and matrix C is a two by three matrix. Recall that when we multiply the columns of the first matrix must match the rows of the second matrix. And in this case, they do, we also can tell from here that our solution matrix will be a three by three as that's how many rows are in the first matrix and columns are in the second matrix. So I'm gonna start by making our matrix Um I'm gonna leave all of the actual multiplication in addition to a later step. Right now, I'm just setting everything up. The first element of the first row in our solution matrix will be from the first row multiplied by the first column. So the first row of B multiplied by the first column of C which is negative 20. So the first element of the row is eight multiplied by the first element of the column, which is negative two plus the second element of the row, which is five multiplied by the second element of the common column which is zero continuing horizontally. I again, am gonna use the first row of B but now I'm gonna multiply it by the second column of C. So I now have the value eight as the first element of the row multiplied by seven as the first element of the second column plus five as the second element of the first row multiplied by four, which is the second element of the second column. And then moving over to my third element in that top row. Again, I'm gonna use the first row of B which is eight multiplied by the first element of the third column of C which is eight plus the second element from that first row, which is five multiplied by the second element of that third column, which is 12. Next my second row, we'll be using the second row of B so zero, negative four and multiplied by each column. So the first element is zero, multiplied by the first element of the first column which is negative two plus negative four, multiplied by the second element of the first column which is zero. My next element will be again the second row of, of matrix B and the second column of matrix C so zero multiplied by the first element which is seven plus negative four, multiplied by the second element of the second column which is four, moving over to the third element in that row. Again, it will be zero this time multiplied by the first element of that third column. So eight plus negative four multiplied by the second element of that third column which is third row of our solution matrix. We will use the third row of B 19. And again, each element will come from multiplying by a column. So we all do. The first element is one multiplied by the first element of the first column which is negative two plus nine multiplied by the second element of the first column which is zero, the next element of our row. And our solution matrix will again be one now will be multiplied by the first element of the second column which is seven plus nine multiplied by the second element of the second column, which is four, moving over to our final element. To find out we again have one multiplied by the first element of the first column which is eight plus nine, multiplied by the second element of the third column, which is 12. So we have all of our multiplication put together. Now we need to multiply and then we can add and find our solution. So doing out just the multiplication. My first element in my first row comes to negative 16. Um The second element I have negative uh sorry, that's a positive 56 plus 20. And at this step, I'm just doing the multiplication and then the third element 64 plus 60 in my second row. Um I noticed they're both zero. So I'm gonna put it as zero. Second element of the second row would be zero plus negative 16. Third element of the second row is zero plus negative 48. Looking at my third round, I will have negative two plus zero and then I have seven plus 36 and eight plus 108. So now we are down to addition. So this should be where we get our final matrix. So I already did. The negative 16 plus zero is negative 16. 56 plus 20 is 76. 64 plus 60 is 124. First element of the second row is zero. Second element would be negative 16, 3rd, negative 48 third row, we get negative two and then we get 43 and 116. And I'm gonna close my matrix. So this three by three matrix where the top row is negative 16 124. The second row is zero, negative 16, negative 48. The third row is negative 2 43. 116 is our solution matrix and that matches answer choice A 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. BC

Key Concepts

Matrix Dimensions and Compatibility for Multiplication

Matrix Multiplication Procedure

Interpreting Matrix Products and Dimensions

Watch next