Video Transcript
Consider the shown matrices. Matrix 𝐴 equals one, two, three,
four. And matrix 𝐵 equals one, two,
three, 𝑘. Is it possible to choose 𝑘 such
that 𝐴𝐵 equals 𝐵𝐴? If so, what should 𝑘 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 𝑎, 𝑏, 𝑐, 𝑑 is one
matrix and 𝑒, 𝑓, 𝑔, ℎ is the other matrix. Now, if we look at the first
element, what we’ve got is 𝑎𝑒 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 𝐴𝐵 equal to 𝐵𝐴 if we have a value of 𝑘. Well, on first inspection, you
might think, well, 𝐴𝐵 is going to be the same as 𝐵𝐴 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 𝑘 should be. Because by just having a look at
our matrices, we can see that 𝑘 is gonna be equal to four. And why is that? Well, the reason that 𝑘 would be
equal to four is because if 𝑘 is four, then matrix 𝐴 and matrix 𝐵 are
identical. So therefore, if we multiply them
together, either 𝐴𝐵 or 𝐵𝐴, it won’t affect the outcome.
But what we’re gonna do now is just
prove this by completing 𝐴𝐵 and 𝐵𝐴. Well, if we have 𝐴𝐵, this is 𝐴
multiplied by 𝐵. So we have one, two, three, four
multiplied by one, two, three, 𝑘. 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 𝑘. Then we’d have three multiplied by
one plus four multiplied by three and then three multiplied by two plus four 𝑘. So therefore, we can say that 𝐴𝐵
is gonna be equal to the two-by-two matrix seven, two plus two 𝑘, 15, six plus four
𝑘.
So now what we can do is move on to
𝐵𝐴. Well, 𝐵𝐴 is gonna be the matrix
one, two, three, 𝑘 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 𝑘, and finally three
multiplied by two plus four 𝑘. So therefore, the result would have
been a two-by-two matrix seven, 10, three plus three 𝑘, and six plus four 𝑘.
Okay, great. So what would we have done now? Well, what we can do is find 𝑘 by
equating corresponding elements and then substitute this back in to see if it give
us the same result for 𝐴𝐵 and 𝐵𝐴. Well, if we start with two plus two
𝑘 equals 10 and then subtract two from each side of the equation, we’re gonna get
two 𝑘 equals eight. Then divide by two, we get 𝑘
equals four. Okay, right. So now if we substitute 𝑘 equals
four into the other elements, all we’re gonna get the same for 𝐴𝐵 and 𝐵𝐴. Well, if we take a look at the
bottom-left element of the 𝐵𝐴, then what we’re gonna have is three plus three 𝑘,
which is the same as three plus three multiplied by four if 𝑘 is equal to four,
which is gonna be equal to 15. Great, that’s what we got in
𝐴𝐵.
And for the bottom-right element of
𝐴𝐵 and 𝐵𝐴, we’ve got six plus four 𝑘, 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 𝑘 was equal to
four, the result for 𝐴𝐵 and 𝐵𝐴 would both be the two-by-two matrix seven, 10,
15, 22. So therefore, the answer to the
question, “is it possible to choose 𝑘 such that 𝐴𝐵 equals 𝐵𝐴,” is yes. And the value of 𝑘 which would
make this possible is four.
