Video Transcript
Consider the matrix equation eight,
one, nine, negative three is equal to π times three, zero, two, negative one plus
negative one, one, three, zero. Find the value of π which solves
this equation.
In this question, weβre given an
equation which involves matrices. And we need to determine the value
of π which solves this equation. Remember, when we say that a value
of π solves this equation, it means our equation must be true. Both sides of the equation need to
be equal. And to make both sides of this
equation equal, letβs start by evaluating the right-hand side of this equation.
The right-hand side of the equation
is as shown. And it includes two things. First, weβre multiplying a matrix
by the value π. This is scalar multiplication. Next, weβre adding these two values
together. This will be the addition of two
two-by-two matrices. So this is matrix addition. So to answer this question, we need
to recall what both of these mean.
Letβs start with what it means to
multiply matrix by a scalar. We need to recall when we multiply
a matrix by a scalar, we need to multiply every single entry inside of our matrix by
this scalar. In this case, weβre going to need
to multiply every single entry inside of our matrix by π.
So we can start with the entry in
row one column one. Thatβs three. We need to multiply this by π. This gives us a new entry of three
π. We can do exactly the same in row
two column one. The entry in this position is
two. And we need to multiply this by π,
giving us a new entry of two π. In row one column two, the entry in
our matrix is zero. So when we multiply this by π,
weβll get zero π, which is of course just equal to zero. And finally, in row two column one,
weβre going to get negative one multiplied by π, which weβll just write as negative
π.
So by evaluating the scalar
multiplication of our first matrix, we get the two-by-two matrix three π, zero, two
π, negative π. And remember, we still need to add
our other matrix.
Now to simplify this expression
even further, weβre going to need to recall how we add together two matrices. To add two matrices together, we
first need to recall we can only do this if they are of the same order. And remember, saying the two
matrices of the same order is just saying they have the same number of rows and
columns. This is because to add two matrices
together, all we do is add together the corresponding entries.
In our case, both of our matrices
are two-by-two square matrices. So we can add these together. When we add two matrices together,
we want to add the corresponding entries together. So in row one column one, weβll get
the two entries in row one column one added together. Weβll get three π plus negative
one. To find the entry in row one column
two, we need to add together the two entries in row one and column two. Itβs going to be zero plus one. We can do the same in row two
column one. We add together two π and three,
giving us two π plus three. And we can do exactly the same in
row two column two. We add together negative π and
zero to get negative π plus zero. This gives us the following
two-by-two matrix, which we can simplify.
Simplifying each entry inside of
our matrix, we get the two-by-two matrix three π minus one, one, two π plus three,
negative π.
Now remember, the question wants us
to find the value of π which solves our equation. So both sides of the equation given
to us in the question should be equal. And we found this two-by-two matrix
by simplifying the right-hand side of this equation. So it should be equal to the matrix
given in the left-hand side of this equation. Therefore, for π to solve this
equation, these two matrices have to be equal. And to find the value of π which
makes these matrices equal, we need to recall what it means for two matrices to be
equal.
We recall that we say that two
matrices are equal if they have exactly the same order and all of their
corresponding entries are equal. So letβs start by checking that
these two matrices have the same order. Both of them are square two-by-two
matrices. So they do indeed have the same
order.
Next, we need that all of their
corresponding entries are equal. So for these two matrices to be
equal, the entries in row one column one need to be equal. In other words, we must have that
three π minus one is equal to eight. Similarly, the entries in row two
column one must be equal. We must have that two π plus three
is equal to nine. Next, we must also have that the
entries in row one and column two are equal. However, both of these are just
equal to one. So this just gives us that one is
equal to one. And this is already true for any
value of π, so it doesnβt help us find the value of π. Finally, the entries in row two
column two must be equal. And this is only true if negative
π is equal to negative three.
So for our matrices to be equal,
all three of these equations must be equal. This is what we call a system of
equations. In fact, because all of these are
linear, we can also call this a system of linear equations. Thereβs a lot of different ways we
could solve this. The easiest way is just to solve
the equation negative π is equal to negative three. By dividing both sides of this
equation through by negative one, we see this is only true when π is equal to
three. And in fact, this is the correct
answer. However, itβs also worth
substituting this value into our other linear equations to check that these are also
correct.
Substituting π is equal to three
into our first linear equation, we get three multiplied by three minus one, which we
can calculate is nine minus one, which is eight. Substituting π is equal to three
into our second linear equation, we get two times three plus three, which is six
plus three, which is of course just equal to nine. Therefore, because all of our
linear equations are satisfied when π is equal to three, the two matrices must be
equal. And therefore, the value of π is
equal to three solves this matrix equation.
Thereβs one last thing worth
pointing out. We could have also checked our
answer by substituting the value of π is equal to three directly into our matrix
equation. We could then evaluate the
right-hand side of this equation and check itβs equal to the left-hand side of the
equation.