### Video Transcript

In this video, weβll learn how to
check whether a two-by-two matrix has an inverse and then find its inverse if
possible. For matrices, thereβs no such thing
as division. We can add, subtract, and multiply
them, but we canβt divide matrices. There is, however, a related
concept which is called inversion. And itβs extremely useful for
helping us to find the inverse of a matrix and to solve matrix equations. We say that an π-by-π matrix,
thatβs a square matrix, is invertible if there exists a second π-by-π square
matrix such that the product of the matrix and its inverse is πΌ, the identity
matrix. In the two-by-two case, thatβs the
matrix one, zero, zero, one.

Now, remember, multiplying matrices
is not commutative. It canβt be done in any order. However, when multiplying a matrix
by its inverse in any order, weβll always get the identity matrix. Note, though, that not all matrices
do have an inverse. Weβre going to look at the process
for finding the inverse of a two-by-two matrix and see what criteria our matrices
need to meet to ensure that they do have an inverse. We say that for a two-by-two matrix
π, π, π, π, its inverse is found by multiplying one over the determinant of π΄
by the two-by-two matrix π, negative π, negative π, π. Where the determinant of π΄ is the
product of the top-left and bottom-right elements minus the product of the top-right
and bottom-left elements.

Notice also that to get from the
two-by-two matrix π΄ to the two-by-two matrix the inverse of π΄, we swap these
elements and we change the sign of these. So, what does this latter part of
the formula mean for finding the inverse of a matrix π΄? Well, we know that one divided by
zero is undefined. We say then that if the determinant
of the matrix is zero, it is not invertible; it doesnβt have an inverse. Itβs also worth noting that only
square matrices are invertible. So, letβs have a look at some
examples.

Is the matrix three, one, negative
three, negative one invertible?

Remember, for a two-by-two matrix
π΄ given by π, π, π, π, the inverse of π΄ is one over the determinant of π΄
times π, negative π, negative π, π, where the determinant of π΄ is equal to ππ
minus ππ. We say that the determinant of the
matrix doesnβt exist. In other words, π΄ is not
invertible if its determinant is equal to zero. So, all we need to do to establish
whether a matrix is invertible is evaluate its determinant and see if thatβs equal
to zero. So, letβs find the determinant of
the matrix three, one, negative three, negative one. And we use these bars either side
of the matrix to represent its determinant.

The determinant is found by
subtracting the product of the top-right and bottom-left elements from the product
of the top left and bottom right. So here, π times π is three
multiplied by negative one. We then subtract π multiplied by
π. Thatβs one multiplied by negative
three. And so, our determinant is three
multiplied by negative one minus one multiplied by negative three. Thatβs negative three minus
negative three, which is negative three plus three, which is equal to zero. So, the determinant of our matrix
three, one, negative three, negative one is zero. We can therefore say that the
matrix doesnβt have an inverse. And therefore, the answer to this
question is no, itβs not invertible.

In our next example, weβll extend
this idea about the criteria for matrix invertibility.

Given that the matrix seven, one,
negative seven, π is invertible, what must be true of π?

Remember, for a square matrix π΄
equals π, π, π, π to be invertible, its determinant must not be equal to
zero. Now, for this matrix, its
determinant is ππ minus ππ. So, letβs find an expression for
the determinant of our matrix. Itβs the product of the top-left
and bottom-right elements, thatβs seven times π or seven π, minus the product of
the top-right and bottom-left elements. Thatβs one multiplied by negative
seven. So, the determinant of our matrix
is seven π minus negative seven, which we can write as seven π plus seven.

Weβre told that our matrix is
invertible, so its determinant cannot be equal to zero. In other words, seven π plus seven
cannot be equal to zero. We need to find the values of π
such that this expression is not equal to zero. So, weβre going to solve the
inequation. Weβll solve it like solving a
normal equation. But instead of our answer being π
is equal to some constant, we know π will not be equal to the result.

Letβs begin by subtracting seven
from both sides. When we do, we find that seven π
will not be equal to negative seven. Next, we divide through by seven,
and we find π cannot be equal to negative one. So, if π is equal to negative one,
the matrix does not have an inverse. So, for it to be invertible, π
cannot be equal to negative one.

In our next example, weβll look at
how to establish whether two matrices are multiplicative inverses of one
another.

Are the matrices one, two, three,
four and one, one-half, one-third, and one-quarter multiplicative inverses of each
other?

Letβs remind ourselves what we mean
by the phrase multiplicative inverses. We say that an π-by-π matrix,
thatβs a square matrix, is invertible if there exists a second π-by-π matrix such
that the product of the matrix and its inverse in any order is πΌ, where πΌ is the
identity matrix. In the case of two-by-two identity
matrices, thatβs one, zero, zero, one. So, we could answer this question
in one of two ways. We could find the inverse of each
matrix and check whether it matches the original of the other matrix. Alternatively, we could find the
product of the matrices and see if we get the identity matrix. Letβs look at the latter
method.

Weβre going to multiply one, two,
three, four by one, one-half, a third, a quarter. And we remind ourselves that to do
so, we begin by finding the dot product of the elements in the first row of the
first matrix by the elements in the first column of the second. So, thatβs the dot product of one,
two and one, one-third. Thatβs one multiplied by one plus
two multiplied by one-third. Thatβs five-thirds. Next, we find the dot product of
the elements in the first row of our first matrix and of the second column in our
second. So, thatβs one multiplied by a half
plus two multiplied by one-quarter, which is one.

Now, we find the dot product of the
elements in the second row of our first matrix and the first column of our
second. Thatβs three times one plus four
times one-third, which is 13 over three. Finally, the dot product of three,
four and a half, one-quarter. Thatβs three times a half plus four
times a quarter, which is five over two. And so, we see that when we
multiply the matrix one, two, three, four by the matrix one, one-half, a third, a
quarter, we get a two-by-two matrix of five-thirds, one, thirteen-thirds, and five
over two. This is quite clearly not equal to
the identity matrix one, zero, zero, one. And so, we can say no, the two
matrices are not multiplicative inverses of each other.

Weβll now calculate the
multiplicative inverse of a two-by-two matrix.

Find the multiplicative inverse of
the matrix π΄ equals negative four, negative 10, three, five, if possible.

Remember, to find the
multiplicative inverse or the inverse of a two-by-two matrix π΄ equals π, π, π,
π, we multiply one over the determinant of π΄ by the matrix π, negative π,
negative π, π. Remember that this means that if
the determinant of π΄ is equal to zero, weβre performing the calculation one divided
by zero, which is undefined. And this means the inverse does not
exist. So, letβs begin by calculating the
determinant of our matrix π΄ equals negative four, negative 10, three, five.

To find its determinant, we
multiply the element in the top left by the element in the bottom right. And then we subtract the product of
the elements in the top right by the bottom left. So, the determinant of our matrix
is negative four times five minus negative 10 times three. Thatβs negative 20 minus negative
30, which is negative 20 plus 30. And so we see that the determinant
of our matrix π΄ is 10. Thatβs quite clearly not equal to
zero. So, the inverse of our matrix π΄
does exist. The formula tells us to multiply
one over the determinant of π΄, so thatβs one over 10, by a two-by-two matrix which
is found by switching the elements in the top left and bottom right and then
changing the sign of the other two elements.

So, the inverse of π΄ is one-tenth
times five, 10, negative three, negative four. And we know that we can multiply a
matrix by a scalar by simply multiplying each of its individual elements. One-tenth times five is
five-tenths. One-tenth times 10 is
ten-tenths. One-tenth times negative three is
negative three-tenths. And one-tenth times negative four
is negative four-tenths. All thatβs left to evaluate the
multiplicative inverse of our matrix π΄ is to simplify each of these fractions. And so, we see that the inverse of
our matrix π΄ is one-half, one, negative three over 10, negative two-fifths.

In our next example, weβll consider
how we combine this process with another operation on matrices.

Consider the matrices π΄ and
π΅. Determine the inverse of π΄ plus
π΅. Here, π΄ is given as the matrix
negative three, negative two, negative five, negative seven. And π΅ is negative one, two, eight,
nine.

This expression here means to find
the inverse of the sum of these two matrices. So, letβs simply begin by
calculating the sum of π΄ and π΅. We do this by adding the individual
elements on each row and each column. So, the element in the first row
and first column is negative three plus negative one, which is negative four. The second element in the first row
is negative two plus two, which is zero. We then add negative five and eight
to get three. And finally, we add negative seven
and nine to get two. So, the sum of our two matrices is
the two-by-two matrix negative four, zero, three, two.

We now need to find the inverse of
this matrix. And so, we recall that for a
two-by-two matrix π, π, π, π, its inverse is one over the determinant of π΄
times π, negative π, negative π, π. Now, the determinant of π΄ is
actually ππ minus ππ. Itβs the product of the elements in
the top left and bottom right of our matrix minus the product of the elements in the
top right and bottom left. Note, of course, this means that if
the determinant is equal to zero, weβre performing the calculation one divided by
zero, which is undefined. And so, our matrix does not have an
inverse.

So, letβs begin by calculating the
determinant of our matrix π΄ plus π΅. Itβs negative four times two minus
zero times three, which is simply negative eight. So, we know the inverse of the sum
of our two matrices does exist, and weβre now ready to work it out. Itβs one over the determinant of
our matrix, so one over negative eight. And then we swap the element on the
top left with the element on the bottom right and change the sign of the other two
elements. So, the inverse of π΄ plus π΅ is
one over negative eight times two, zero, negative, three, negative four. Now, of course, one divided by
negative eight is the same as negative one-eighth. And we know that we can multiply a
matrix by a scalar by multiplying each of its individual elements.

Negative one-eighth multiplied by
two is negative one-quarter or negative 0.25. Negative one-eighth multiplied by
zero is, of course, zero. Then, we multiply negative
one-eighth by negative three. Well, a negative multiplied by a
negative is a positive. So, we get three-eighths which is
0.375. Finally, we multiply negative four
by negative one-eighth, and we get one-half or 0.5. And so, the inverse of our matrix
π΄ plus π΅ is negative 0.25, zero, 0.375, and 0.5. Note, of course, that we could
check our solution by ensuring that when we find the product of the matrix π΄ plus
π΅ and its inverse, we get the identity matrix. Thatβs the matrix one, zero, zero,
one.

In our final example, weβll look at
what we mean when we say a matrix is singular and how this definition can help us to
solve problems.

Find the set of real values of π₯
that make the two-by-two matrix π₯ minus three, eight, two, π₯ plus three
singular.

Letβs begin by defining the word
singular when it comes to matrices. We say that a matrix is singular if
itβs not invertible; it doesnβt have an inverse. We know that a matrix is invertible
if its determinant is not equal to zero, and the converse is also true. So, in other words, a matrix is
singular if its determinant is equal to zero. In this case then, we need to find
the set of real values of π₯ such that the determinant of our matrix is equal to
zero. The determinant of a two-by-two
matrix π, π, π, π is ππ minus ππ. We subtract the product of the
elements in the top right and bottom left from the product of those in the top left
and bottom right.

So, in this case, thatβs π₯ minus
three times π₯ plus three minus eight times two. If we distribute these parentheses,
we get π₯ times π₯, which is π₯ squared, plus three π₯ minus three π₯ minus three
times three, which is nine. That simplifies to π₯ squared minus
nine. And eight multiplied by two is
16. So the determinant of our matrix is
π₯ squared minus nine minus 16, which is π₯ squared minus 25. Weβre trying to find the set of
values of π₯ that make our matrix singular. In other words, which values of π₯
make the determinant zero? So, letβs set our expression for
the determinant equal to zero and solve for π₯. That is, π₯ squared minus 25 equals
zero.

Adding 25 to both sides of this
equation gives us π₯ squared equals 25. And then weβll take the square root
of both sides of our equations, remembering to take the positive and negative square
root of 25. That gives us π₯ is equal to
positive or negative five. We can use these squiggly brackets
to help us represent the set of values that make our matrix singular. They are negative five and
five. Note that at this stage, we could
check our solutions by substituting each value of π₯ into our original matrix and
then checking that the determinant is indeed equal to zero.

Weβll now summarize the key points
from this video. We saw that an π-by-π matrix, a
square matrix, is invertible if there exists a second matrix such that the product
of that matrix and its inverse is πΌ, the identity matrix. We saw that if a two-by-two matrix
is defined as π, π, π, π, then its inverse is one over the determinant of π
times the matrix π, negative π, negative π, π. And we, of course, calculate the
determinant of π΄ by finding the product of ππ and subtracting the product of
ππ. Finally, we saw that a matrix is
said to be singular if it doesnβt have an inverse. In other words, if the determinant
of that matrix is equal to zero.