### Video Transcript

Scalar Multiplication of
Matrices

In this video, weβre going to learn
how to multiply any matrix by any scalar, thatβs a real number, and weβll see the
properties and results that arise from this definition. Before we begin with scalar
multiplications of matrices, we can recall weβve seen the type of scalar
multiplication before. We know how to use scalar
multiplication and vectors. Consider the vector three, negative
four. We know how we would multiply this
by any real number. For example, we could multiply
through by five. We then know to multiply a vector
by five, we need to multiply it component-wise. In other words, we need to multiply
each component through by five.

So multiplying each component by
five, we get a new horizontal component of five times three and a new vertical
component of five multiplied by negative four. This gives us the new vector with
horizontal component 15 and vertical component negative 20. In other words, weβve scaled our
vector up by a factor of five. It turns out we can define exactly
the same operation in terms of matrices. To start, letβs consider the matrix
of order π by π, which is given as follows. That is, the entry in row π and
column π of this matrix is given by π sub ππ. Just as we did with multiplying a
vector by a scalar, we want to multiply every entry in our matrix by our scalar. So if we let π be our scalar, we
can define π times π΄ as follows.

We want to multiply every single
entry inside of our matrix by π. This is often referred to as
scaling our matrix by a factor of π because every single entry inside of our matrix
is multiplied by π. And remember we can also represent
this by what happens to every entry in our matrix. We can also represent this as the
matrix whose entry in row π and column π is π times π ππ. Letβs now see some examples of how
we would calculate scalar multiplication of matrices.

Consider the matrix π΄. Find nine times π΄, where π΄ is
equal to the one-by-two matrix two, negative one.

In this question, weβre given
matrix π΄ and we need to find nine times π΄. Remember, nine is just a real
number, so this is scalar multiplication of our matrix. So to answer this question, all we
need to do is recall how we multiply a matrix by a scalar. We do this by multiplying every
single entry inside of our matrix by our scalar. So we need to multiply the entry in
the first row and first column by nine and the entry in the first row and second
column by nine. This gives us the following matrix,
and we can just evaluate this.

In the first row and first column,
we have nine times two, which is of course just equal to 18. And in the first row and second
column, we have nine multiplied by negative one. This is of course just equal to
negative nine, and this gives us our final answer. Therefore, we were able to show if
π΄ is equal to the one-by-two matrix two, negative one, then nine π΄ will be equal
to the one-by-two matrix 18, negative nine.

Letβs now see an example which will
show us how we can solve equations by using our new definition to multiply matrices
by scalars.

Given that π₯ times the two-by-two
matrix negative two, zero, negative three, negative five is equal to the two-by-two
matrix 14, zero, 21, 35, find the value of π₯.

Weβre given an equation involving
two matrices and π₯, and we need to determine the value of π₯. First, we need to notice by looking
at our equation, π₯ is a number. Itβs not a matrix. This is because π₯ is given in
lowercase. Another reason we can see this is
weβre asked to find the value of π₯. Weβre only asked to do this when π₯
represents a number. So in our equation, when we
multiply our matrix by π΄, this is scalar multiplication of a matrix. So the first thing weβll do is
start by π₯ multiplied by our matrix and apply scalar multiplication to rewrite this
in a different form.

To do this, we need to recall that
when we multiply a matrix by a scalar, we multiply every single entry inside of our
matrix by our scalar. In this case, we need to multiply
all of our entries by π₯. By doing this, we get the
two-by-two matrix of entries π₯ times negative two, π₯ times zero, π₯ times negative
three, and π₯ times negative five. And of course, we can evaluate or
simplify all of these entries. Doing this, we get the two-by-two
matrix negative two π₯, zero, negative three π₯, negative five π₯.

But remember, in the question,
weβre told that this matrix is exactly equal to the two-by-two matrix 14, zero, 21,
35. So weβve now shown that these two
matrices are equal. To find our value of π₯, weβre
going to need to recall what it means for two matrices to be equal. Recall that we say that two
matrices are equal if every single entry in both of our matrices are equal and they
have the same order. Of course, both of these are
two-by-two matrices. So all we need to do is check that
all of their entries are equal.

Doing this, we get a series of
equations. By equating the entries in row one
and column one, we get 14 should be equal to negative two π₯. By equating the entries in row one
and column two of our matrices, we get that zero should be equal to zero. By equating the entries in row two
and column one of both of our matrices, we get that 21 should be equal to negative
three π₯. And finally, by equating the
entries in row two and column two of our matrices, we get that 35 should be equal to
negative five π₯. This gives us a system of equations
we need to solve. In fact, all of these are linear
equations, so we could call this a system of linear equations.

Firstly, we can see that the second
equation is true for all values of π₯. Next, we can just solve all three
of the remaining linear equations. Weβll divide the first one through
by negative two, the second one through by negative three, and the third one through
by negative five. And by doing this, we can see that
all of these are solved by π₯ being equal to negative seven. This means we were able to show
that the value of π₯ should be negative seven. However, itβs also worth pointing
out we can confirm this answer either by substituting the value of π₯ is equal to
negative seven into our matrix or substituting π₯ is equal to negative seven into
our original equation. We could then verify that this
gives us the correct solution to the equation.

Therefore, given that π₯ times the
two-by-two matrix negative two, zero, negative three, negative five was equal to the
two-by-two matrix 14, zero, 21, 35, we were able to show that the value of π₯ must
be equal to negative seven.

Letβs now see an example of a
useful result which comes from our definition of scalar matrix multiplication.

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.

Letβs now see how we can use these
results to help us answer questions.

Let π be a two-by-three matrix
whose entries are all zero. If π΄ is any two-by-three matrix,
which of the following is equivalent to five π΄ minus three π? Option (A) two times π times π΄,
option (B) negative two times π΄ times π, option (C) negative three π, option (D)
five π΄, or option (E) two π΄.

In this question, weβre given an
expression in terms of two matrices, and we need to determine which of five options
is this expression equal to. And in fact, thereβs a lot of
different ways of answering this question. However, thereβs one important
thing we do need to notice. Both matrix π and matrix π΄ are
two-by-three matrices. This means the number of rows on
one matrix and the number of columns on the other matrix are never equal. And when this happens, this means
we canβt multiply π΄ by π, and we canβt multiply π by π΄. So options (A) and (B) canβt be
correct.

Letβs now see what we can do with
the expression weβre given. Letβs start by recalling that π is
a two-by-three matrix, where every entry is equal to zero. Now we could write this out in
terms of matrices. However, we can also write this as
the two-by-three zero matrix, zero, two, three. Letβs now write out matrix π΄ and
the zero matrix in full. To do this, we need to recall a
matrix of order two by three will have two rows and three columns. Weβll also call the entry in matrix
π΄ in row π, column π π ππ. This gives us the following
expression.

We can now simplify this expression
either by using our definition for scalar multiplication of a matrix or by using the
fact that for any number π, π multiplied by the π-by-π zero matrix is just equal
to the π-by-π zero matrix. This gives us that three π is just
equal to the two-by-three zero matrix. Now we see weβre subtracting the
zero matrix from our matrix five π΄. Thereβs a few different things we
could do. For example, we could use our
definition of scalar multiplication to bring five inside of our matrix. Then we could use our definition of
matrix subtraction to answer this question.

However, this is not necessary
because weβre subtracting the zero matrix of the same order. And when we do this, we subtract
zero from every entry inside of our matrix. Of course, subtracting zero is not
going to change any of the values, so this is just equal to five π΄. And this gives us that option (D)
is the correct answer. Itβs also worth pointing out we
could check that option (C) and option (E) are not correct in all scenarios. For example, if we set π΄ equal to
the two-by-three matrix where all entries are one, then weβve already shown that
five π΄ minus three π should be equal to five π΄. And, of course, five π΄ is every
entry in π΄ multiplied by five. Itβs the two-by-three matrix where
every entry is five.

This is, of course, not equal to
two π΄, since this would be the two-by-three matrix where every entry is two, and
itβs also not equal to the matrix negative three, π since weβve already shown that
this will be equal to the two-by-three zero matrix. Therefore, we were able to show the
only correct option is option (D); five π΄ minus three π will be equal to five
π΄.

Letβs now look at one last property
that we can get from scalar multiplication of matrices. Weβll first show this property by
looking at it in action.

Let π΄ and π΅ be the following two
matrices. We want to calculate the following
two expressions. First, we want to calculate three
multiplied by the sum of π΄ and π΅. We then want to calculate three π΄
plus three π΅.

In this case, weβre checking
whether we can distribute three over our parentheses in this case. And if we can do this in general,
this would be called a distributive property.

Letβs start with the first
expression. Letβs find three times π΄ plus
π΅. Remember, we first need to
calculate the expression inside of our parentheses. So weβll start by writing our
matrices π΄ and π΅ out in full. And remember, when weβre adding two
matrices of the same order together, we just add their entries together. So in row one, column one, weβll
get negative four plus three which we can calculate as negative one. In row one, column two, weβll get
six plus seven which is equal to 13. And we can do the same to find the
rest of the entries. We get negative five, 14, 13, and
one.

And remember, we need to multiply
this matrix by three. So weβre multiplying this matrix by
the scalar three. Remember to do this, we multiply
every entry inside our matrix by three. For the entry in row one, column
one of our matrix, we have three multiplied by negative one, which we can calculate
as negative three. For the entry in row one, column
two, we have three multiplied by 13, which we can calculate as 39. And we can do the same to find the
rest of the entries. We get negative 15, 42, 39, and
three.

Letβs now see what happens if we
calculate three π΄ plus three π΅. Once again, weβll start by writing
our matrices π΄ and π΅ out in full. This time, we first need to use our
scalar multiplication to multiply both of our matrices by three. Remember, we do this by multiplying
every single entry inside of our matrix by three. Weβll start by calculating three
π΄. Multiplying every entry by three
and simplifying, we get the two-by-three matrix negative 12, 18, zero, 24, 21,
six. We can then do the same to find
three π΅; itβs equal to the following two-by-three matrix.

Finally, all we need to do is add
these two together. Remember, we do this by adding all
the entries together. By adding the entries in row one
and column one together, we get negative 12 plus nine, which is equal to negative
three. By adding the entries in row one,
column two together, we get 39. And we can do the same to find the
rest of the entries. We get the following two-by-three
matrix, which we can see is exactly equal to the one we found before. In other words, three times π΄ plus
π΅ is equal to three π΄ plus three π΅. And in fact, this is true in
general. For any number π and π-by-π
matrices π΄ and π΅, we have π times π΄ plus π΅ is equal to ππ΄ plus ππ΅. And this is called the distributive
property because we can distribute π over our parentheses.

Letβs now go over the key points of
this video. First, we showed that multiplying a
matrix π΄ by a scalar π means multiplying every single entry of π΄ by π. And in fact, itβs worth pointing
out π can be any number. It could be complex. However, weβve only used real
numbers in this video. And we can write this as ππ΄. Finally, we showed five useful
properties for π-by-π matrices π΄ and π΅. The first four of these are
identities we directly derive from the definition of scalar multiplication and
matrix addition. Finally, we showed the scalar
multiplication is distributive when combined with matrix addition.