Find the products AB and BA to determine whether B is the mult...

Question

Find the products AB and BA to determine whether B is the multiplicative inverse of A.

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

Accepted Answer

Hey, everyone in this problem, we're given two matrices and we're asked to work out the products A B and B A, OK. In order to show if B is the inverse, the multiplicative inverse K of A, all right. So the matrices we're given A and B are two by two matrices. OK. So A is two by two and B is two by two. OK. When we're doing mo matrix multiplication, we need the number of columns of the first matrix to equal the number of rows of the second matrix. OK. In this case, they are equal. So we're OK to do the multiplication. And then when we're talking about multiplicative inverses, if it's gonna be an inverse, it needs to have the exact same dimension which again, in this case it does. So we're OK there. All right. So let's start with A B OK. This is gonna be A has first column, one, third, one and second column, one half one A times B which has first column negative 66 and second column three negative two. All right. Now, when we're doing multiplication, it's a little bit different than the addition and subtraction of matrix CS K, we're not gonna just do it turn by turn, what we're gonna do is we're gonna take the first row of a right multiply it by this first column of B OK? And this is gonna go in row one column, one of our resulting matrix. OK. So we're gonna take the first thing in the first row of a one third, we're gonna multiply it by the first thing in the first column of B negative six. OK? And then we're gonna add the second. So first row, second column, one half of A K and then second thing in the first column times six. OK. So that's how we get this position here. Row one, column one. OK? Now we're gonna do the same thing for the next column for in row one, column two, we're gonna start with row one. OK? And now we're gonna multiply by column two. OK? So this is gonna give us one third times three. OK? So the first two things in those multiply it together and then we're gonna add second things. OK? One half times negative two. Hey, this comes from the blue. All right. So we have the first row of our matrix. Now, now we need to do the second. OK? So when we're doing the second row, we are gonna have the second row. Hey, time is the first column. So it's gonna go in the second row, first column of our resulting matrix OK. So second row of A comes the first column of B, we get one times negative six plus one time six next coming from the green and then we have the final position. OK? So the final row here and that's gonna be impossible to see. So let's just rewrite this so that we can change our color a little bit. So we're gonna have green, we're gonna kind of highlight this with both So we can see that we use it for both. OK? And we're gonna do the final row of a times the final column of B, this is gonna give us one times three plus one times negative two. OK? And so that's gonna be the last entry in our new matrix. Now we have a resulting matrix. All we have to do is do this addition, subtraction, multiplication within this matrix and we're gonna get the resulting matrix. So we have negative six times a third. This is gonna be negative two plus 6/2 is three. So we're gonna end up with one, negative six plus six is going to give us zero. OK? Three divided by three, that's one plus two, divided by two. That's gonna give us negative one. So we're gonna have zero there as well and then we have three minus two. That's gonna give us one. OK? So we're gonna get the results. A matrix A B is gonna be the identity matrix in two dimensions. OK? All right now we need to do the same but for B A, OK, I'm not gonna do all of the colors this time. Let's just keep it clean and let's see what we get. OK? So we're gonna have negative 663, negative two for B and for a first column, one third, one, second column, one half, one, writing this out. OK? And again, we're starting with B OK? So we're gonna take the first row of B and multiply it by the first column of A. So we're gonna get negative six times a third K working across the row in B and down the column in A. We're gonna add three times one. We're gonna do the same thing for row one, column two. OK? So we're gonna work across the first row of B and down the second column of A. We're gonna get negative six times a half plus three times one. All right, we've done our top row. Let's move to the second row in the second row. Again, we're gonna go second row. First column is gonna be the second row of B times the first column of A. So we're gonna get six times one third plus working across negative two and working down times one. And finally the second row of B, the second column of A, we get six times this a half plus negative two times one. OK? All right. Let's give ourselves some more room here. And again, now it's just a matter of adding subtracting and multiplying like we normally would. OK? So here we're gonna have negative two plus three. That's gonna give us one. Yeah. Now right here we have negative three plus three. That's gonna be zero two minus two. That's gonna be zero and three minus two. That's gonna be one. So we get the identity matrix again. OK? And what you see here is that we have A B, the product is equal to the product B A. OK? And they are both equal to the identity matrix in two dimension. OK? The two by two identity matrix. So what this means is that B is the multiplicative inverse of A OK? And so we have our answer. It's gonna be A B is the multiplicative inverse of A. OK? Thanks everyone for watching. I hope this video helped see you in the next one.

Find the products AB and BA to determine whether B is the multiplicative inverse of A.
$A = [\begin{matrix} - 2 & 1 \\ \frac{3}{2} & - \frac{1}{2} \end{matrix}],$

Key Concepts

Matrix Multiplication

Multiplicative Inverse of a Matrix

Identity Matrix

Watch next