### Video Transcript

If π΄ is equal to the one-by-three
matrix eight, negative three, one, what is the value of zero π΄.

Weβre given a matrix π΄ and weβre
asked to evaluate zero multiplied by π΄. Of course, zero is a number, so
this is scalar multiplication of our matrix. The first thing weβre going to need
to recall is how we multiply a matrix by a scalar. We recall scalar multiplication of
a matrix means we multiply every single entry by our scalar. In this case, weβre going to need
to multiply every entry by zero. Doing this, we get the one-by-three
matrix with entry in row one, column one zero times eight; row one, column two zero
times negative three; and row one, column three zero times one. And of course we can evaluate the
expressions of all of our entries. We know that zero times eight is
equal to zero, zero times negative three is equal to zero, and zero times one is
equal to zero. So this gives us the one-by-three
matrix with every entry equal to zero, which is our final answer.

Therefore, we were able to show if
π΄ is equal to the one-by-three matrix eight, negative three, one, then zero π΄ will
be equal to the one-by-three zero matrix.

We can notice a very useful result
from this question. We know that any number multiplied
by zero is equal to zero. So, in fact, it didnβt matter what
our matrix was. It always wouldβve given a matrix
with every entry equal to zero. So letβs confirm this result. If we have a matrix of order π by
π, and weβll call this matrix π΄, and weβll call the entry in row π and column π
of matrix π΄ π ππ, then if we multiply our matrix by the scalar zero, every entry
inside of our matrix should be zero. In other words, this should be
equal to the π-by-π zero matrix, represented by zero sub ππ.

And in fact, we can prove this
result. We could do this by writing the
matrix π΄ out in matrix notation. However, we already know the entry
in row π and column π. So when we multiply our matrix π΄
by the scalar zero, we multiply every single entry by zero. In other words, the entry in row π
and column π of this matrix is zero times π ππ. However, we know for any number,
zero times that number will always be equal to zero. In other words, for all π and all
π, zero times π ππ is equal to zero. So every entry inside of our matrix
is equal to zero. In other words, this is equal to
the zero matrix of order π by π.

And itβs important to remember we
need to keep the order of our matrix because scalar multiplication does not change
the order of our matrix. But this is not the only useful
result we can get from this definition of scalar multiplication. Another question we can ask is,
what is one multiplied by a matrix π΄? Remember, when we multiply a matrix
by a scalar, we multiply every single entry in that matrix by the scalar. So, in this case, weβre multiplying
every entry inside of our matrix by one. Of course, this isnβt going to
change the value of any of our entries. So this should be equal to π΄.

And in fact, we can prove this
using a very similar method to what we did above. When we multiply our matrix π΄ by
the scalar one, weβre multiplying every single entry in matrix π΄ by one. In other words, the entry in row π
and column π is going to be equal to one times π ππ because we know the entry in
row π and column π of matrix π΄ is π ππ. And of course, one multiplied by
any number is equal to that same number. So one times π ππ is going to be
equal to π ππ for any values of π and π. Therefore, the entries inside of
our matrix donβt change. The entry in row π and column π
is just π ππ. Therefore, weβve shown for any
matrix π΄, one π΄ is equal to π΄.

And thatβs another useful result
weβre going to need going forward. We want to ask the question, what
is negative one π΄ equal to? Of course we know how to do
this. We multiply every single entry
inside the matrix π΄ by negative one. So this seems to suggest that
negative one times π΄ is going to be equal to negative π΄. And to fully express this, weβre
going to need to recall exactly what we mean by negative π΄. The easiest way to do this is to
think what happens if you subtract a matrix from itself.

Remember, when you subtract
matrices, you do it component-wise. So when we calculate the matrix π΄
minus itself, every entry is going to be subtracted from itself. Every entry is going to be equal to
zero. And of course, it keeps the
dimension ππ. So perhaps a better way to write
this equation would be to add our matrix π΄ to both sides of the equation. This would give us the equivalent
statement π΄ plus negative one times π΄ is equal to π΄ minus π΄. And we know exactly what π΄ minus
π΄ is equal to. Itβs the zero matrix of order π by
π. And we could prove this in exactly
the same way we did above.

First, to evaluate negative one
multiplied by π΄, we need to use scalar multiplication. We multiply every entry inside of
our matrix by negative one. So in row π, column π, weβll have
π ππ plus negative one times π ππ. And of course we can then simplify
this. Negative one multiplied by π ππ
is going to be equal to negative π ππ for any values of π and π. But remember, when we add two
matrices together, we do this component-wise. So in row π, column π, weβre
going to get π ππ minus π ππ, and the number minus itself is equal to
zero. So every entry inside of our matrix
is going to be zero. This is going to be equal to the
π-by-π zero matrix.

And thereβs one final result which
we can get from this definition of scalar multiplication. Before we do this, letβs clear a
little bit of space. This result is going to be very
similar to our first result. However, this time, instead of
multiplying a matrix by the scalar zero, weβre instead going to multiply a zero
matrix by any scalar. If we let π be any number, then we
can consider what happens when we multiply the π-by-π zero matrix by π. Of course, multiplying by the
scalar π means we multiply every single entry inside of the zero matrix by π.

But all of the entries inside of
the zero matrix are zero. So weβre just going to get π
multiplied by zero for all of our entries. This is just going to be equal to
the π-by-π zero matrix. And we could prove this by using a
very similar method to what we did before. When we multiply a matrix by a
scalar, we need to multiply every single entry inside of our matrix by the
scalar. So in this case, weβre going to get
in row π, column π π multiplied by the entry in row π and column π of the zero
matrix. However, every single entry in the
zero matrix is zero. So in row π, column π, we get π
times zero which is equal to zero. So this is just equal to the
π-by-π zero matrix.