Video: Identifying the Conditions for Two Matrices to Be Equal

Consider the shown matrices 𝐴 = [1, 2 and 3, 4], 𝐡 = [1, 2 and 3, π‘˜]. Is it possible to choose π‘˜ such 𝐴𝐡 = 𝐡𝐴? If so, what should π‘˜ be equal to?

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Video Transcript

Consider the shown matrices. Matrix 𝐴 equals one, two, three, four. And matrix 𝐡 equals one, two, three, π‘˜. Is it possible to choose π‘˜ such that 𝐴𝐡 equals 𝐡𝐴? If so, what should π‘˜ be equal to?

So in this problem, the first thing we’re gonna look at is how we multiply matrices together. So what we have is two-by-two matrices. So how would we multiply two two-by-two matrices? Well, let’s have a look at this. We’ve got π‘Ž, 𝑏, 𝑐, 𝑑 is one matrix and 𝑒, 𝑓, 𝑔, β„Ž is the other matrix. Now, if we look at the first element, what we’ve got is π‘Žπ‘’ plus 𝑏𝑔. So what we do is we multiply the first element in the first row of the first matrix by the first element of the first column of the second matrix. And then we add on the second element of the first row in the first matrix multiplied by the second element in the first column of the second matrix. And then what we do is complete this pattern for the other elements of our matrix.

Okay, great. So now what we need to do is work out whether it is possible to have 𝐴𝐡 equal to 𝐡𝐴 if we have a value of π‘˜. Well, on first inspection, you might think, well, 𝐴𝐡 is going to be the same as 𝐡𝐴 because if we just multiply things, it doesn’t matter which way round they are. However, it does like already shown when we’re dealing with matrices. Now, there are two ways of solving this problem. First of all, just by taking a look at our two matrices, we could probably see what the value of π‘˜ should be. Because by just having a look at our matrices, we can see that π‘˜ is gonna be equal to four. And why is that? Well, the reason that π‘˜ would be equal to four is because if π‘˜ is four, then matrix 𝐴 and matrix 𝐡 are identical. So therefore, if we multiply them together, either 𝐴𝐡 or 𝐡𝐴, it won’t affect the outcome.

But what we’re gonna do now is just prove this by completing 𝐴𝐡 and 𝐡𝐴. Well, if we have 𝐴𝐡, this is 𝐴 multiplied by 𝐡. So we have one, two, three, four multiplied by one, two, three, π‘˜. So then, if we use the multiplication rule we showed earlier, what we’d have is, for the first element, one multiplied by one plus two multiplied by three, then the second element, one multiplied by by two plus two π‘˜. Then we’d have three multiplied by one plus four multiplied by three and then three multiplied by two plus four π‘˜. So therefore, we can say that 𝐴𝐡 is gonna be equal to the two-by-two matrix seven, two plus two π‘˜, 15, six plus four π‘˜.

So now what we can do is move on to 𝐡𝐴. Well, 𝐡𝐴 is gonna be the matrix one, two, three, π‘˜ multiplied by the matrix one, two, three, four. So then this would be equal to one multiplied by one plus two multiplied by three, then one multiplied by two plus two multiplied by four, then three multiplied by one plus three π‘˜, and finally three multiplied by two plus four π‘˜. So therefore, the result would have been a two-by-two matrix seven, 10, three plus three π‘˜, and six plus four π‘˜.

Okay, great. So what would we have done now? Well, what we can do is find π‘˜ by equating corresponding elements and then substitute this back in to see if it give us the same result for 𝐴𝐡 and 𝐡𝐴. Well, if we start with two plus two π‘˜ equals 10 and then subtract two from each side of the equation, we’re gonna get two π‘˜ equals eight. Then divide by two, we get π‘˜ equals four. Okay, right. So now if we substitute π‘˜ equals four into the other elements, all we’re gonna get the same for 𝐴𝐡 and 𝐡𝐴. Well, if we take a look at the bottom-left element of the 𝐡𝐴, then what we’re gonna have is three plus three π‘˜, which is the same as three plus three multiplied by four if π‘˜ is equal to four, which is gonna be equal to 15. Great, that’s what we got in 𝐴𝐡.

And for the bottom-right element of 𝐴𝐡 and 𝐡𝐴, we’ve got six plus four π‘˜, so the same in each anyway. But if we did want to work that out, we could get six plus four multiplied by four, which is six plus 16, which is 22. So therefore, if π‘˜ was equal to four, the result for 𝐴𝐡 and 𝐡𝐴 would both be the two-by-two matrix seven, 10, 15, 22. So therefore, the answer to the question, β€œis it possible to choose π‘˜ such that 𝐴𝐡 equals 𝐡𝐴,” is yes. And the value of π‘˜ which would make this possible is four.

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