In Exercises 1 - 12, find the products AB and BA to determine ...

Question

In Exercises 1 - 12, find the products AB and BA to determine whether B is the multiplicative inverse of A. - 4 0 - 2 4 A = B = 1 3 0 1

Matrices A and B for finding their products in college algebra, chapter 7.

Accepted Answer

everyone welcome back in this problem. We are given two matrices and we're asked to work out the products A. B and B. A. Okay. In order to show if B is the multiplication inverse of a matrix is We're given our A and B. Two by two matrices. A is negative 52 in the first row, 03 in the second row. And B is negative 76 in the first row and 50 in the second row. Alright. The first thing we want to do when we're multiplying matrices is check the dimension. Okay, so A is two by two, B is two by two. When you're doing matrix multiplication, you want the number of columns of the first matrix equal to the number of rows of the second matrix. Okay, so this second dimension of the first matrix and the first dimension of the second matrix, you want them to be equal? In this case? They are So we can do the multiplication. Okay, now if we're thinking about B being the multiplication inverse of A. In order for that to be true, we would need both B and A to have the exact same dimension. In this case. They're both two by two matrices. So it is possible based on dimension. Okay, Alright, so let's start with the product A. B. Hey, this is going to be matrix. A first column negative 50. 2nd column 23 times be first column negative 75 2nd column 60. Now, what is this going to equal? Okay, so when we do a matrix multiplication. What we do is we take the first row of the first matrix. We multiply it by the first column of the second matrix and we do that element wise. Okay we put that in the position of our resulting matrix of that particular row and column and then we move along to next. Okay, so what does this look like? Let's start with row one. Call him one. We're going to have negative five. Okay. The first element In the row we've chosen times -7. The first element in the column we've chosen then we're gonna add we're going to move along across the row down the column. And so we're going to get to the second thing in our chosen row, times five. The second thing in our chosen column. Okay, Alright, so that's it for that one there. Now we can stick with the same row and move across to the second column. So we have the first row of the first matrix times the second column on the second matrix. Okay, this is going to go in row one column two of the resulting matrix. So we're gonna start with the first thing in the rho times the first thing in the column, plus the second thing and the rho times the second thing in the column. So we get negative five times six plus two times zero. Alright, we can do the same thing now with row two. So we're moving on to row two Of the first matrix. And we're going to go back to column one. Okay this is gonna go in row two column one of our resulting matrix. We start with the first element of the row zero times the first element of the column negative seven. Let me fix that negative there, negative seven. There we go. Okay. Lost the second thing in the rho times the second element in the column. Alright, so that comes from here And the final element is going to be that last row times. The last column might get zero times six Plus three times 0. Alright so working out this math now we're down to one matrix. We've done the multiplication. All we need to do is add So negative five times negative seven is 35 plus 10. We are going to get 45. Zero times negative seven is zero plus three times five is 15. We're gonna get 15 negative five times six is negative 30 plus zero, negative 30 0 times six plus three times zero is zero. Okay and so this is our resulting matrix here. This is A B. Okay so we've done a B. Now we need to go ahead and do be a. Okay so let's give ourselves some more room and we're gonna move on and do B times A. So we're gonna have negative 75 in the first column 60 in the second column of the first matrix K times a negative 50 in the first column, 23 in the second column. Alright, so this is gonna be the same process. We're gonna start with the first row of the first matrix. First column of the second matrix. So we have negative seven times negative five Plus six times 0. Moving across the same row, column two we get negative seven times two plus six times three. Second row, first column we're gonna have five times negative five plus zero times zero and second row. Second column we're going to get five times two plus zero times three. Okay. Alright. And again this is down to one matrix. We just need to do the addition. So we end up with negative seven times negative five is plus six times zero is zero. So we get 35. Five times negative five is negative 25 plus zero, negative 25 negative seven times two is negative 14 plus six times three. Which is 18. So negative 14 plus 18. That's gonna be four. And this last 15 times two is 10 plus zero. We get 10 there. Okay. Alright. So we found A B. We found B. A. Now the question wants us to find out is be the multiplication inverse of A. Okay, well, let's look at our resulting matrices. Ok. We'll see that A. B is not equal to B. A. Okay, the product A B is not equal to the product B. A. And they are not equal to the identity matrix in this case in two dimensions. Okay, These are not equal. So be is not the multiplication inverse. Bye bye. Okay, so let's go back up. Answer this question. We found that B is not the multiplication inverse of A. Okay, Because that product A B is not equal to the product B. A. And is not equal to the identity matrix. Thanks everyone for watching. I hope this video helped see you in the next one.

In Exercises 1 - 12, find the products AB and BA to determine whether B is the multiplicative inverse of A. - 4 0 - 2 4 A = B = 1 3 0 1

Key Concepts

Matrix Multiplication

Multiplicative Inverse of a Matrix

Identity Matrix

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