Video Transcript
Daniel guesses that any two-by-two
matrix 𝐴, where 𝐴 times the two-by-two matrix one, two, three, four is equal to
the two-by-two matrix one, two, three, four multiplied by 𝐴, must be a combination
of the two-by-two matrix one, zero, zero, one and the two-by-two matrix one, two,
three, four. In other words, it must be 𝐴 is
equal to 𝑠 times the two-by-two matrix one, zero, zero, one plus 𝑡 times the
two-by-two matrix one, two, three, four, for some numbers 𝑠 and 𝑡. James wants to challenge this,
since he sees that 𝐴 is equal to the matrix one, two, three, four squared produces
the same product, the matrix one, two three, four cubed, when multiplied on either
side. Help Daniel by finding 𝑠 and 𝑡 so
that the matrix one, two, three, four squared is equal to 𝑠 times the matrix one,
zero, zero, one plus 𝑡 times the matrix one, two, three, four.
In this question, we’re given a
property of some two-by-two matrices labeled 𝐴. Daniel believes if we have a
two-by-two matrix 𝐴, where multiplying 𝐴 on the right by the two-by-two matrix
one, two, three, four or on the left by the two matrix one, two, three, four is
equal, then 𝐴 must be a combination of two different matrices: the two-by-two
matrix one, zero, zero, one and the two-by-two matrix one, two, three, four. And when he says combination, he
means a linear combination. In other words, there exist some
numbers 𝑠 and 𝑡 such that 𝑠 times the matrix one, zero, zero, one plus 𝑡 times
the matrix one, two, three, four must be equal to 𝐴. And James has noticed one possible
matrix for 𝐴, the two-by-two matrix one, two, three, four squared. We can verify that 𝐴 satisfies
this property by noticing 𝐴 is a two-by-two matrix. And we can substitute it into both
sides of the equation. We recall squaring a matrix means
we multiply it by itself and cubing a matrix means we multiply it by itself
twice.
Therefore, when we substitute this
matrix into this equation, we get the product of the matrix one, two, three, four of
itself two times on both sides of the equation. Both sides are the matrix one, two,
three, four cubed. Therefore, since this matrix
satisfies the condition, Daniel says it must be a linear combination of the two
matrices given.
The question wants us to determine
the values of the two constants which make this true. That’s the values of 𝑠 and 𝑡
which satisfy the given equation. To do this, we’re going to need to
balance both sides of the equation. Let’s start with the left-hand side
of the equation. That’s the two-by-two matrix one,
two, three, four squared. To square a matrix, we multiply it
by itself. This gives us the following. And to multiply two matrices
together, we need to find the sum of the products of the corresponding entries in
the rows of the first matrix and the columns of the second matrix. For example, the entry in row one,
column one of the products of these two matrices will be one times one plus two
times three. And we can evaluate this. It’s one plus six, which is equal
to seven. So the entry in row one, column one
of the product of these matrices is seven.
We can do the same to find the
entry in row one, column two. It’s equal to one times two plus
two times four, which we can calculate is equal to 10. We can follow the same process to
find the entry in row two, column one. It’s three times one plus four
times three, which if we calculate is equal to 15. Finally, the entry in row two,
column two will be three times two plus four times four, which is equal to 22. Therefore, the left-hand side of
this equation is the two-by-two matrix seven, 10, 15, 22.
Now, let’s move on to simplifying
the right-hand side of this equation. To do this, we recall to multiply a
matrix by a scalar, we multiply every single entry of the matrix by the scalar. We’ll do this term by term. Let’s start with the first
term. We need to multiply every single
entry in this matrix by 𝑠. This gives us the two-by-two matrix
𝑠, zero, zero, 𝑠.
Next, in our second term, we need
to multiply all of the entries of the matrix by 𝑡. This gives us the two-by-two matrix
𝑡, two 𝑡, three 𝑡, four 𝑡. Then, the sum of these two matrices
is the right-hand side of our equation. And the two-by-two matrix seven,
10, 15, 22 is the left-hand side of this equation. We want to find the values of 𝑠
and 𝑡 which balance both sides of the equation. They need to be equal. So let’s simplify this
equation.
We’ll start by adding the two
matrices on the right-hand side of the equation together. And recall to add two matrices
together, they need to be of the same order and we just add the corresponding terms
together. Therefore, the entry in row one,
column one of the sum of these two matrices will be 𝑠 plus 𝑡. And we can keep going. The element in row one, column two
of the sum of these two matrices will be zero plus two 𝑡, which is of course just
two 𝑡. And we can follow the same pattern
to find the final two entries. We get the two-by-two matrix 𝑠
plus 𝑡, two 𝑡, three 𝑡, 𝑠 plus four 𝑡. And we need to find the values of
𝑠 and 𝑡 which make this matrix equal to the two-by-two matrix seven, 10, 15,
22.
And we recall for two matrices to
be equal, they need to have the same order and all of the corresponding entries of
the two matrices need to be equal. And in this equation, the matrix on
the left-hand side has two rows and two columns. And the matrix on the right-hand
side also has two rows and two columns. Therefore, they both have the same
order. So, for these two matrices to be
equal, we just need the corresponding entries to be equal. If we then set each of the
corresponding entries to be equal, we get four equations. We need to find the values of 𝑠
and 𝑡 such that seven is equal to 𝑠 plus 𝑡, 10 is equal to two 𝑡, 15 is equal to
three 𝑡, and 22 is equal to 𝑠 plus four 𝑡. All four of these equations need to
be true for our two matrices to be equal.
Therefore, we’re solving four
linear equations in two variables. We can note that the second
equation and the third equation only involve one variable 𝑡. So we can solve these for 𝑡. Dividing the second equation
through by two gives us that 𝑡 is equal to five. And we can use this to find the
value of 𝑠. Let’s substitute 𝑡 is equal to
five into our first equation. This then gives us that seven is
equal to 𝑠 plus five. And we can solve this for 𝑠 by
subtracting five from both sides of the equation. We get that 𝑠 must be equal to
two.
But we’re not done yet. Remember, these values of 𝑠 and 𝑡
need to balance both sides of our matrix equation. This means they need to satisfy all
four of the equations we found. We can do this by substituting them
into our four linear equations. Or we can substitute them into our
matrix equation. Let’s check this entry by
entry. If 𝑠 is two and 𝑡 is five, 𝑠
plus 𝑡 is equal to seven, which agrees with our first entry. Next, if 𝑡 is equal to five, two
times 𝑡 is equal to 10, which agrees with our next entry. Third, if 𝑡 is equal to five,
three 𝑡 is equal to 15, which agrees with our third entry. Finally, if 𝑠 is equal to two and
𝑡 is equal to five, 𝑠 plus four 𝑡 is 22, which agrees with our final entry. And therefore, we were able to show
that 𝑠 is equal to two and 𝑡 is equal to five.
