Video: Applying Operations on Matrices to Find the Values of Unknowns

Consider the matrices 𝐴 = (0, βˆ’4 and 2, βˆ’2), 𝐡 = (βˆ’5, 6 and π‘₯, 𝑦). If 𝐴𝐡 = 𝐡𝐴, what are the values of π‘₯ and 𝑦?

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Video Transcript

Consider the matrices 𝐴 equals zero, negative four, two, negative two, 𝐡 equals negative five, six, π‘₯, 𝑦. If 𝐴𝐡 is equal to 𝐡𝐴, what are the values of π‘₯ and 𝑦?

To answer this question, we’ll need to begin by evaluating 𝐴𝐡 and 𝐡𝐴. Remember, these aren’t always the same due to the fact that matrix multiplication is not commutative. It can’t be performed in any order. So we’ll need to work them out separately. To find 𝐴𝐡, we find the dot product of the rows in the first matrix and the columns in the second. Let’s see what that looks like.

To find the first element in the first row, we’re going to find the dot product of the row zero, negative four and the column with entries negative five, π‘₯. That’s zero multiplied by negative five plus negative four multiplied by π‘₯. Zero multiplied by negative five is zero. So this is simply negative four π‘₯. To find the second element of our first row, we repeat this process, finding the dot product of the first row in the first matrix and the second column in the second. This time that zero multiplied by six plus negative four multiplied by 𝑦, which is negative four 𝑦.

To find the first entry of the second row, we find the dot product of the elements in the second row and first column. This time, that’s two multiplied by negative five plus negative two multiplied by π‘₯, which is negative 10 minus two π‘₯. And finally, we find the dot product of the elements on the second row in the first matrix and the second column in the second. That’s two multiplied by negative six [six] plus negative two multiplied by 𝑦, which is 12 minus two 𝑦.

Let’s repeat this process for 𝐡𝐴. The first entry is negative five multiplied by zero plus six multiplied by two, which is 12. The second entry is negative five multiplied by negative four, which is 20, plus six multiplied by negative two, which is negative 12. And that simplifies to make eight. Now, in fact, we don’t need to do anything more. We can actually solve this problem. But let’s complete this matrix.

π‘₯ multiplied by zero plus 𝑦 multiplied by two is simply two 𝑦. And π‘₯ multiplied by negative four plus 𝑦 multiplied by negative two is negative four π‘₯ minus two 𝑦. And we’re told these matrices are identical. This means each of their individual elements must be the same. And we can say that negative four π‘₯ is equal to 12. And negative four 𝑦 is equal to eight. And we’ll solve these equations to find π‘₯ and 𝑦, respectively.

To solve this first equation, we divide by negative four. That gives us that π‘₯ is equal to negative three. And similarly, we solve the second equation by dividing by negative four. And this time, we get 𝑦 is equal to negative two. Once we have these, it’s a really nice way to check our answers. We can substitute the values of π‘₯ and 𝑦 that we’ve worked out into the individual elements in our equation.

Substituting π‘₯ is equal to negative three into the expression negative 10 minus two π‘₯ gives us negative four. And substituting 𝑦 is equal to negative two into the expression two 𝑦 also gives us negative four. Remember, we said the individual elements had to be equal. So this is a good way to check what we’ve done is correct.

π‘₯ is equal to negative three. And 𝑦 is equal to negative two.

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