### Video Transcript

In this video, we will learn how to
use matrix multiplication to determine the square and cube of a square matrix. Whilst the methods we use can be
extended to higher powers, in this video, we will only deal with squaring and
cubing. It is important to note that we can
raise any π-by-π square matrix to any power. In this video, we will only deal
with two-by-two and three-by-three matrices.

Letβs begin with some key
definitions. For a square matrix π΄ and positive
integr π, the πth power of π΄ is defined by multiplying this matrix by itself
repeatedly. That is, π΄ to the πth power is
equal to π΄ multiplied by π΄ multiplied by π΄, and so on, multiplied by π΄, where
there are π instances of the matrix π΄. In this video, we will calculate
the matrix π΄ squared by multiplying matrix π΄ by matrix π΄. We will also calculate π΄ cubed by
multiplying matrix π΄ by itself and itself again. This is equivalent to the matrix π΄
multiplied by the matrix π΄ squared.

Alternatively, we can multiply the
matrix π΄ squared by the matrix π΄. It is important to note that taking
the power of a matrix is only well defined if the matrix is square. If matrix π΄ has order π by π,
then this order will be common to π΄ squared, π΄ cubed, and so on. Squaring and cubing a matrix does
not alter its order. As already mentioned, in this
video, we will focus on two-by-two and three-by-three square matrices. We will now look at an example
where we need to square a two-by-two matrix.

Given that matrix π΄ equals
negative six, one, negative five, five, find π΄ squared.

We recall that when dealing with
powers of matrices, the matrix π΄ squared will only be defined if the matrix π΄ is
square. In this case, π΄ is a two-by-two
matrix. To calculate π΄ squared, we need to
multiply the matrix π΄ by itself. This is equal to negative six, one,
negative five, five multiplied by negative six, one, negative five, five. When multiplying matrices, we begin
by multiplying the elements in the first row of the first matrix by the first column
in the second matrix. The top-left element or component
of our matrix will be equal to negative six multiplied by negative six plus one
multiplied by negative five.

Next, we multiply the first row of
the first matrix by the second column of the second matrix. This gives us negative six
multiplied by one plus one multiplied by five. We can then repeat this process for
the second row of the first matrix. This gives us negative five
multiplied by negative six plus five multiplied by negative five and negative five
multiplied by one plus five multiplied by five. Simplifying each of our elements
gives us the matrix 31, negative one, five, 20. If matrix π΄ is equal to negative
six, negative one, negative five, five, then π΄ squared is equal to 31, negative
one, five, 20. We notice that the order of the
matrix stays the same. And this is true when we raise any
square matrix to any power.

We will now consider what happens
when we need to square and cube a three-by-three matrix.

Consider the matrix π΄ which is
equal to one, one, two, one, zero, one, two, one, zero. There are two parts to this
question. Firstly, find π΄ squared, and
secondly, find π΄ cubed.

We recall that in order to square
any matrix, we multiply it by itself. Also for the matrix π΄ squared to
be well defined, matrix π΄ must be a square matrix. In this case, it is a
three-by-three matrix. This means that π΄ squared is equal
to one, one, two, one, zero, one, two, one, zero multiplied by one, one, two, one,
zero, one, two, one, zero. When multiplying two matrices, we
begin by multiplying the elements in the first row of the first matrix by the first
column of the second matrix. The element in the top-left corner
will be equal to one times one plus one times one plus two times two. This is equal to six.

Next, we multiply the elements of
the first row of the first matrix by the second column of the second matrix. This gives us one multiplied by one
plus one multiplied by zero plus two multiplied by one, which is equal to three. Moving on to the third column of
the second matrix, we also get an answer of three, as one multiplied by two plus one
multiplied by one plus two multiplied by zero equals three. We then repeat this process for the
second row of the first matrix. This gives us answers of three,
two, and two. Finally, multiplying the third row
of the first matrix by each of the columns in the second matrix gives us values
three, two, and five. π΄ squared is the three-by-three
matrix six, three, three, three, two, two, three, two, five.

The second part of our question
asks us to find the matrix π΄ cubed. We can do this by multiplying
matrix π΄ by the matrix π΄ squared or by multiplying the matrix π΄ squared by the
matrix π΄. In this question, we will use the
first method. π΄ cubed is equal to one, one, two,
one, zero, one, two, one, zero multiplied by six, three, three, three, two, two,
three, two, five. We use the same method as the first
part of the question. We begin by multiplying the
elements of the first row of the first matrix by the first column of the second
matrix. One multiplied by six plus one
multiplied by three plus two multiplied by three is equal to 15.

Multiplying the first row of the
first matrix by the second column of the second matrix gives us nine. And multiplying by the third column
of the second matrix gives us 15. The second row of the matrix π΄
cubed has elements nine, five, and eight. Finally, the third row is equal to
15, eight, and eight. The matrix π΄ cubed is equal to 15,
nine, 15, nine, five, eight, 15, eight, eight. If weβre given any square matrix
π΄, we can calculate π΄ squared by multiplying the matrix by itself and π΄ cubed by
multiplying matrix π΄ by the matrix π΄ squared.

In our next question, we will
simplify an expression by using squared and cubed matrices.

Given the matrix π΄, which is equal
to four, zero, negative three, seven, calculate π΄ cubed minus three π΄ squared.

We recall that when given any
square matrix π΄, we can calculate the matrix π΄ squared by multiplying π΄ by
itself. In this question, π΄ squared is
equal to four, zero, negative three, seven multiplied by four, zero, negative three,
seven. When multiplying matrices, we begin
by multiplying the elements in the first row of the first matrix by the elements in
the first column of the second matrix. Four multiplied by four plus zero
multiplied by negative three is equal to 16. We then repeat this for the second
column of the second matrix. Four multiplied by zero plus zero
multiplied by seven is equal to zero.

Next, we multiply the second row of
the first matrix by each of the columns in the second matrix. This gives us negative 33 and
49. π΄ squared is equal to the
two-by-two matrix 16, zero, negative 33, 49. In our expression, we want three π΄
squared. This means we need to multiply the
matrix π΄ squared by the scalar or constant three. We multiply each of the elements by
three, giving us 48, zero, negative 99, 147. Next, we need to calculate π΄
cubed, and we know this is equal to π΄ multiplied by π΄ squared. π΄ cubed is therefore equal to
four, zero, negative three, seven multiplied by 16, zero, negative 33, 49. We multiply these two matrices in
the same way as we calculated π΄ squared, giving us π΄ cubed is equal to 64, zero,
negative 279, 343.

We now have a matrix for π΄ cubed
and also for three π΄ squared. We are asked in the question to
subtract these. We have 64, zero, negative 279, 343
minus 48, zero, negative 99, 147. When subtracting matrices, we
subtract the individual components or elements. This means that we begin by
subtracting 48 from 64. This is equal to 16. Subtracting the top-right elements
gives us zero. Negative 279 minus negative 99 is
the same as adding 99 to negative 279. This is equal to negative 180. Finally, 343 minus 147 is equal to
196. If the matrix π΄ is equal to four,
zero, negative three, seven, then π΄ cubed minus three π΄ squared is equal to 16,
zero, negative 180, 196.

In our final question, we will
solve a system of linear equations by applying operations on matrices.

Given that the matrix π is equal
to five, six, negative five, negative four, find the values of π₯ and π¦ that
satisfy π squared plus π₯π plus π¦πΌ is equal to π, where π is the zero matrix
of order two by two and πΌ is the unit matrix of order two by two.

We recall that any zero matrix has
all elements equal to zero. Therefore, the matrix π is equal
to zero, zero, zero, zero. The unit or identity matrix has
ones on its main or leading diagonal and zeros elsewhere. Therefore, πΌ is equal to one,
zero, zero, one. When multiplying any matrix by a
constant, in this question π₯ and π¦, we simply multiply each of the elements in the
matrix by the constant. This means that the matrix π¦πΌ is
equal to π¦, zero, zero, π¦. As π is equal to five, six,
negative five, negative four, then π₯π is equal to five π₯, six π₯, negative five
π₯, negative four π₯. Finally, we need to calculate the
matrix π squared. This is equal to matrix π
multiplied by itself. We need to multiply five, six,
negative five, negative four by five, six, negative five, negative four.

When multiplying matrices, we
multiply each of the rows in the first matrix by each of the columns in the second
matrix. Five multiplied by five plus six
multiplied by negative five is equal to negative five. Multiplying the elements of the
first row in the first matrix by the second column in the second matrix gives us
six. Repeating this for the second row
of the first matrix, we get the elements negative five and negative 14. Substituting the four matrices into
our equation, we have negative five, six, negative five, negative 14 plus five π₯,
six π₯, negative five π₯, negative four π₯ plus π¦, zero, zero, π¦ is equal to zero,
zero, zero, zero. We can now set up four equations
comparing the corresponding components or elements.

Comparing the top-left elements, we
have negative five plus five π₯ plus π¦ is equal to zero. We will call this equation one. The top-right elements give us six
plus six π₯ plus zero is equal to zero. As there was only one unknown in
this equation, we can solve it. Subtracting six from both sides
gives us six π₯ is equal to negative six. Dividing both sides of this
equation by six gives us π₯ is equal to negative one. We can then substitute this value
back into equation one to calculate the value of π¦. Simplifying the left-hand side, we
have negative 10 plus π¦ is equal to zero. This means that π¦ is equal to
10. The values of π₯ and π¦ are
negative one and 10, respectively.

We do need to check, however, that
these values hold for the bottom row of our matrices. Here, we have the two equations:
negative five plus negative five π₯ plus zero is equal to zero and negative 14 plus
negative four π₯ plus π¦ is equal to zero. Adding five π₯ to both sides of the
first equation gives us five π₯ is equal to negative five. Dividing both sides by five, we see
that π₯ once again is equal to negative one. In the second equation,
substituting π₯ is negative one gives us negative 14 plus four plus π¦ is equal to
zero. Once again, this confirms that π¦
is equal to 10. If the matrix π equals five, six,
negative five, negative four, then the values of π₯ and π¦ that satisfy the equation
π squared plus π₯π plus π¦πΌ equals π are π₯ is equal to negative one and π¦ is
equal to 10.

We will now summarize the key
points from this video. For a square matrix π΄ and positive
integer π, π΄ to the πth power is equal to π΄ multiplied by π΄ multiplied by π΄,
and so on, multiplied by π΄, where there are π instances of π΄. The power of a matrix is only well
defined if the matrix is square. In this video, we dealt with
two-by-two and three-by-three square matrices. However, this extends to any
π-by-π square matrix. In this video, we used the general
rule to square and cube matrices, where π΄ squared is equal to π΄ multiplied by π΄
and π΄ cubed is equal to π΄ squared multiplied by π΄ or π΄ multiplied by π΄
squared. This method can also be extended
when dealing with higher powers.