Find each product, if possible. See Examples 5–7.

Question

Find each product, if possible. See Examples 5–7. <3x2 Matrix><2x3 Matrix>

Accepted Answer

Hello, everyone. We are asked to perform multiplication of the two matrices given below. We are given a two by three matrix with the top row of radical five radical five negative radical and a second row of radical seven radical 343 and zero. And then we have a three by two matrix with the first row as 12 negative 14 second row as 16 third row has 04. We have four answer choices all of which are two by two matrices with slightly different signs and values. First recall that in order to multiply, we need to first check the dimensions. So we have a two by three matrix and a three by two matrix. And as long as the columns of the first matrix match the number of rows in the second matrix, we can multiply it. In this case they do. And then the rows of the first matrix and the columns of the second matrix tell us the size of our solution. In this case, it's a two by two. So in my first step, I'm gonna just write all of the multiplication. In addition, I will do it out in the next step. So for the first element of our solution matrix, we are going to multiply the first row of the first matrix by the first column of the second matrix. So we get radical five, which is the first element multiplied by 12, which is the first element of the column plus radical five as the second element of the first row multiplied by seven as the second element of the first column plus negative radical 20 multiplied by zero as they are the third elements in both rows and columns. Next, I'm going to do the first row of the first matrix multiplied by the second column of the second matrix. And that'll go in my second element of the top row of my solution matrix. So I get radical five multiplied by the first element of that second column which is negative 14 plus radical five multiplied by the second element which is 16 plus negative radical 20 multiplied by the fourth element of that column which is four next going into the second row in my solution matrix. And I'm gonna multiply the second row of the first matrix by the first column of the second matrix. So we get radical seven multiplied by the first element in the column which is 12 plus radical 343 multiplied by the second element of the column which is seven plus the third element of the row which is zero multiplied by the third element of the column, which is also zero. Next, I'm going to do the last element we need to put in our solution matrix. And that is the second row of the first matrix multiplied by the second column of the second matrix. So we get radical seven multiplied by negative plus radical multiplied by 16 plus zero, multiplied by four. So now that I have written down all of the multiplication I have to do, I'm gonna do out what I can and see what we get. Um I'm not gonna do the addition yet, just the multiplication. So when I do this, I get 12 radical five plus seven radical five and I'm not gonna rate plus zero. Uh next element in the top row, I have negative 14 radical five plus 16 radical five minus four radical in the second row, I have 12 radical seven plus seven radical 343. And the next element I have a negative 14 radical seven plus 16 radical 343. So before I try to do the addition and subtraction here, I wanna see if I can simplify the radical 343 or the radical I know that I can simplify radical 20 as I can break this into factors of radical four multiplied by radical five, which would be two radical five. So I'm gonna plug that back in for radical 20 when I simplify this in terms of the square root of that one's gonna take me a second to think about. But knowing there are radical sevens in our problem, I'm gonna see what multiplied by seven gives me 343. And it's actually 49. So radical 49 multiplied by radical seven is the same thing as saying seven radical seven. So we're gonna use that to simplify a little bit more as we go. So I will plug that in when it is time. So my first element of my first row, 12 radical five plus seven radical five gives me 19 radical five. Second element I have negative 14 radical five plus 16 radical five which is two radical five. And now I want to find out what this is when I put in two radical five for radical 20. So hold on, I have two radical five and then I'm gonna treat this as minus four times two radical five. So that's the same as minus eight radical five. So I can see that this element actually is a negative six radical five in the bottom row. Uh If I'm gonna do it off to the side for half a second seven multiplied by the square root of 343 is like saying seven multiplied by seven radical seven. So that'd be 49 radical seven. So I can rewrite this as 49 radical seven. And when I do that this will give me 61 radical seven as the first element in my second row. And same thing when I do that over here, this is like 16 now times seven radical seven which gives me 112 radical seven. That sounds right. So now when I do negative 14 radical seven plus 112 radical seven, that will give me 98 radical seven. So we have our answer. Our solution matrix is a two by two matrix where the top row is 19 radical five negative six radical five. The second row is 61 radical seven radical seven. And this matches answer choice B have a nice day.

Find each product, if possible. See Examples 5–7. <3x2 Matrix><2x3 Matrix>

Key Concepts

Matrix Multiplication

Dimensions of Matrices

Element-wise Calculation

Watch next