### Video Transcript

Consider the shown matrices. Matrix π΄ equals one, two, three,
four. And matrix π΅ equals one, two,
three, π. Is it possible to choose π such
that π΄π΅ equals π΅π΄? If so, what should π be equal
to?

So in this problem, the first thing
weβre 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, letβs have a look at
this. Weβve got π, π, π, π is one
matrix and π, π, π, β is the other matrix. Now, if we look at the first
element, what weβve got is ππ plus ππ. 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 π΄π΅ equal to π΅π΄ if we have a value of π. Well, on first inspection, you
might think, well, π΄π΅ is going to be the same as π΅π΄ because if we just multiply
things, it doesnβt matter which way round they are. However, it does like already shown
when weβre 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 π΄ and matrix π΅ are
identical. So therefore, if we multiply them
together, either π΄π΅ or π΅π΄, it wonβt affect the outcome.

But what weβre gonna do now is just
prove this by completing π΄π΅ and π΅π΄. Well, if we have π΄π΅, this is π΄
multiplied by π΅. 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 weβd 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 weβd 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 π΄π΅
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
π΅π΄. Well, π΅π΄ 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 π΄π΅ and π΅π΄. Well, if we start with two plus two
π equals 10 and then subtract two from each side of the equation, weβre 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 weβre gonna get the same for π΄π΅ and π΅π΄. Well, if we take a look at the
bottom-left element of the π΅π΄, then what weβre 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, thatβs what we got in
π΄π΅.

And for the bottom-right element of
π΄π΅ and π΅π΄, weβve 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 π΄π΅ and π΅π΄ would both be the two-by-two matrix seven, 10,
15, 22. So therefore, the answer to the
question, βis it possible to choose π such that π΄π΅ equals π΅π΄,β is yes. And the value of π which would
make this possible is four.