Video Transcript
Inverse of a Matrix: The Adjoint
Method
In this video, weβre going to
discuss how to find the inverse of any square matrix with nonzero determinant by
using the adjoint method. Weβll go through the method in
detail, explaining how to find the matrix minors, how to construct the cofactor
matrix, and then how to construct the adjoints of our matrix. Weβll then go through some examples
explaining how we can apply this method to find inverses.
Before we discuss this new method
of finding the inverse of a square matrix, weβre going to need to go through a few
new concepts first. Weβre going to start with defining
a matrix minor. If we have a matrix π΄ which is of
order π times π, then the matrix minor, which we represent with capital π΄ ππ,
is the matrix π΄ where we remove the πth row and the πth column from matrix
π΄. So our matrix minor π΄ ππ is
exactly the same as matrix π΄. However, we removed the πth row
and πth column. And we know if we remove one row
and one column from matrix π΄, then our new order will be π minus one by π minus
one. And the easiest way to understand
this concept is through an example.
Letβs start with the matrix π΄,
which is a three-by-four matrix given as follows. Now, letβs see how we would
construct the matrix minor π΄ two three. From our definition of a matrix
minor, we can see we need to remove the πth row and πth column from our matrix
π΄. First, our value of π is two. So we need to remove the second row
from our matrix π΄. Next, we can see the value of π is
equal to three. So we need to remove the third
column from matrix π΄. Then our matrix minor π΄ two three
will be all the remaining elements we didnβt remove. π΄ two three will be the following
matrix.
However, this is of course not the
only matrix minor we couldβve constructed. Now, letβs construct the matrix
minor π΄ three one. This time, we can see the value of
π is equal to three. So we need to remove the third row
from our matrix π΄. And we can see the value of π is
one. So we need to remove the first
column from matrix π΄. And then we can construct our
matrix π΄ three one by using all of the remaining elements, the ones we didnβt
remove from π΄. This gives us π΄ three one is the
following matrix.
The methods we go through in this
video will help us determine if we can find the inverse of any π-by-π matrix. However, we will need to find the
determinant of any matrix we want to find the inverse of. And weβve seen previously itβs
difficult to calculate the determinant for large square matrices. So weβll focus mainly on
three-by-three matrices.
And before we do this, weβre going
to need to recall how we find the determinant of a two-by-two matrix. Let π΄ be equal to the two-by-two
matrix π, π, π, π. Then the determinant of π΄ is equal
to ππ minus ππ. Weβre now ready to discuss the key
part which will help us find the inverse of any square matrix with nonzero
determinant. We need to discuss what the
cofactor matrix of a matrix π΄ is.
First, because weβre using this to
find the inverse of a matrix, we need our matrix π΄ to be a square matrix. Letβs say its order is π by
π. Next, to construct our cofactor
matrix, weβre going to need to find all of the matrix minors of π΄. Weβll call these π΄ ππ. Weβre now ready to construct the
cofactor matrix of π΄. Weβll call this matrix πΆ. And weβll define it element by
element. The entry in row π and column π
of our cofactor matrix will be given by negative one raised to the power of π plus
π multiplied by the determinant of the matrix minor π΄ ππ. And remember, our matrix π΄ is a
square matrix of order π by π. So our values of π will be running
through the rows of our matrix π΄. And the values of π will be
running through the columns of our matrix π΄. So our values of π range from one
to π, and our values of π range from one to π. This means our cofactor matrix will
be a square matrix. It will be of order π by π.
Itβs also worth pointing out
sometimes youβll see this written out in its entire matrix form. And if we were to do this, we would
get the following matrix representation of our cofactor matrix. And itβs worth pointing out all
weβve done here is create an π by π square matrix where in row π and column π
weβve used our formula to construct the entry. But usually, itβs a lot easier to
work with our definition for each entry individually. Before we discuss how weβre going
to use the cofactor matrix to find the inverse of a matrix, letβs go through some
examples.
Given that π΄ is equal to the
three-by-three matrix negative five, eight, negative seven, six, zero, one,
five, negative four, negative eight, determine the value of negative one to the
power of one plus two multiplied by the determinant of the matrix minor π΄ one
two.
Weβre given a three-by-three
square matrix. And weβre asked to determine
the value of negative one to the power of one plus two times the determinant of
the matrix minor π΄ one two. And while not necessary to
answering this question, itβs worth pointing out this will be the entry in row
one and column two of our cofactor matrix.
The first step to answering
this question is to remember what we mean by the matrix π΄ one two. We call this a matrix
minor. The matrix minor π΄ ππ means
we remove row π and column π from our matrix π΄. In our case, we can see the
value of π is equal to one and π is equal to two. We can then write this into our
definition for the matrix minor. We see that the matrix minor π΄
one two means we remove row one and column two from matrix π΄.
So to find our matrix minor π΄
one two, we need to start with our matrix π΄ and then remove row one. This means we remove the
following three entries. Then, we also need to remove
column two. This means we need to remove
the entire second column from matrix π΄. And we can see this leaves us
with only four elements. Then, we can construct our
matrix minor π΄ one two by constructing a matrix with the four remaining
elements. This gives us π΄ one two is the
two-by-two matrix six, one, five, negative eight.
But the question is not just
asking us to find the matrix minor π΄ one two. We also need to find its
determinant. And since π΄ one two is a
two-by-two matrix, we can do this by recalling the formula for the determinant
of a two-by-two matrix. We recall the determinant of
the square matrix π, π, π, π is equal to ππ minus ππ. In our case, we can see that π
is equal to six and π is equal to negative eight. And we can also see that our
value of π is equal to one and π is equal to five. So by using this formula, we
have the determinant of π΄ one two is equal to six times negative eight minus
one times five. And if we calculate this
expression, we get negative 53.
Weβre now ready to find the
value of the expression given to us in the question. We have negative one to the
power of one plus two multiplied by the determinant of π΄ one two is equal to
negative one cubed, since one plus two is equal to three, and negative 53, since
we already found the value of this determinant. And we can simplify this to
give us 53. Therefore, we were able to show
for the square matrix π΄ given to us in the question, the value of negative one
to the power of one plus two times the determinant of the matrix minor π΄ one
two is equal to 53.
Letβs now go through an example
of finding a cofactor matrix of a three-by-three square matrix. Weβll start with the
three-by-three square matrix π΄ is equal to three, zero, negative three,
negative two, negative three, negative six, seven, three, negative five. And now we recall the element
in row π and column π of our cofactor matrix will be negative one to the power
of π plus π multiplied by the determinant of the matrix minor π΄ ππ.
So to find our cofactor matrix,
weβre first going to need to find all of our matrix minors. Letβs start with the matrix
minor π΄ one one. Remember, this means weβre
going to need to remove the first row and the first column from matrix π΄. This gives us the following
two-by-two matrix. This gives us π΄ one one is
negative three, negative six, three, negative five. To find our cofactor matrix,
weβre going to need to find all of our matrix minors.
Letβs now find the matrix minor
π΄ one two. This means we need to remove
row one and column two from our matrix π΄. And doing this leaves us with
the following four elements. So the matrix minor π΄ one two
is negative two, negative six, seven, negative five. Because our matrix π΄ is a
three-by-three matrix, our values of π and π will range from one to three. So weβll have nine total matrix
minors to calculate. And we can find all nine of
these by using the same method. We remove row π and column π
from our matrix π΄. This gives us the following
nine matrix minors.
We can now see from our
definition of the cofactor matrix weβre now going to need to find the
determinant of all of these matrix minors. And since all of these are
two-by-two matrices, we can do this by recalling the determinant of the
two-by-two matrix π, π, π, π is equal to ππ minus ππ. So letβs start by finding the
determinant of the matrix minor π΄ one one. This is equal to negative three
times negative five minus negative six multiplied by three. And if we calculate this
expression, itβs equal to 33.
We can then do exactly the same
to find the determinant of our matrix minor π΄ one two. Itβs equal to negative two
times negative five minus negative six multiplied by seven. And if we calculate this
expression, we see itβs equal to 52. Using the exact same method, we
could find the determinants of all of our matrix minors. We would get the following
values.
Now that we found the
determinants of all of our matrix minors, we can find all of the elements of our
cofactor matrix. Weβll start with the entry in
row one and column one of our cofactor matrix. It will be equal to negative
one to the power of one plus one multiplied by the determinant of our matrix
minor π΄ one one. Well, we know negative one to
the power of one plus one is equal to one. And weβve already shown that
the determinant of π΄ one one is equal to 33. So πΆ one one will be equal to
33. So weβve shown that πΆ one one
is equal to 33. And in fact, we can add this
into our cofactor matrix in row one, column one.
We can do the same to find the
entry in row one, column two. Itβs equal to negative one to
the power of one plus two times the determinant of π΄ one two. And since the determinant of π΄
one two is 52, this simplifies to give us negative 52. And we can then add this into
our cofactor matrix in row one, column two. And we could then do exactly
the same to find all the remaining entries of our cofactor matrix. Doing this would give us the
following values. And just as we did before, we
can then add these into our cofactor matrix πΆ.
There is one thing worth
pointing out here. When we calculate πΆ ππ, we
multiply the determinant of our matrix minor by negative one to the power of π
plus π. This means in row one, column
one, weβll always multiply by positive one. And then in row one and column
two, weβll always multiply by negative one. And this pattern will
continue. And some people prefer to use
this rather than multiplying by negative one to the power of π plus π. We just need to remember to
multiply our determinant by the value.
Now that weβve constructed our
cofactor matrix, we can discuss how we can use this to find the inverse of our
matrix. First, we need one last
definition. The adjoint matrix of a matrix
π΄, denoted adjoint π΄, is equal to the transpose of our matrix πΆ, where πΆ is
the cofactor matrix of π΄. And itβs also worth remembering
when we take the transpose of a matrix, we switch the rows and columns
around.
And now weβre finally ready to
determine how to find the inverse of a square matrix. We have if π΄ is a square
matrix and the determinant of π΄ is not equal to zero, then the inverse of π΄
will be equal to one divided by the determinant of π΄ multiplied by the adjoint
of π΄. So to find the inverse of any
square matrix, there are two parts. First, we need to find the
determinant of π΄, and then we need to find the adjoint of π΄. And remember, if the
determinant of π΄ is equal to zero, then the matrix does not have an
inverse. So normally, we check this
first.
Letβs now go through some examples
of using the adjoint method to find the inverse of some matrices.
Consider the matrix one, zero,
three, one, zero, one, three, one, zero. Determine whether the matrix has an
inverse by finding whether the determinant is nonzero. If the determinant is nonzero, find
the inverse using the formula for the inverse which involves the cofactor
matrix.
Weβre given a three-by-three
matrix. And the first thing weβre asked to
do is determine whether this matrix has an inverse by first finding the determinant
of our matrix and checking whether this is equal to zero. Remember, if the determinant of a
matrix is equal to zero, then that matrix cannot be invertible. And for a square matrix, if its
determinant is not equal to zero, then it is invertible. So we need to start by finding the
determinant of our matrix. Weβll call this matrix π΄.
Thereβs a lot of different ways of
finding the determinant of a matrix. The easiest way is to find which
row or column contains the most number of zeros. For our matrix π΄, we can see this
is column two. It has two zeros. Then, we need to recall our formula
for the determinant of a three-by-three matrix where we choose a column π. This gives us the following
expression, where lowercase π one π, lowercase π two π, and lowercase π three
π are the entries in row one, column π; row two, column π; and row three, column
π of our matrix π΄. And capital π΄ one π, capital π΄
two π, and capital π΄ three π are our matrix minors.
Since we chose the second column,
weβll set our value of π equal to two. So by using π is equal to two, we
get the following expression. And we can simplify this. First, negative one to the power of
one plus two is equal to negative one. Negative one to the power of two
plus two is equal to one. And negative one to the power of
three plus two is equal to negative one. So we can simplify our expression
to give us the following.
We can then use our definition of
the matrix π΄ to find some of these values. First, lowercase π of one two is
the entry in row one, column two of our matrix. We can see this is zero. Next, lowercase π two two is the
entry in row two, column two. We can see this is also zero. Finally, lowercase π three two is
the entry in row three, column two. We can see this is equal to
one. So our first two terms have a
factor of zero and, therefore, are equal to zero. So this entire expression
simplifies to give us negative one times the determinant of the matrix minor π
three two. And we can find our matrix minor π΄
three two from our definition of π΄.
Remember, we need to remove row
three and column two from our matrix π΄. And this leaves us with only four
elements: one, three, one, one. So our adjoint matrix π΄ three two
is the two-by-two matrix one, three, one, one. So the determinant of π΄ is
negative one times the determinant of the two-by-two matrix one, three, one,
one. And we know how to find the
determinant of a two-by-two matrix. The determinant of a two-by-two
matrix π, π, π, π is equal to ππ minus ππ. And by using this, we can show the
determinant of the two-by-two matrix one, three, one, one is equal to one times one
minus three times one. And of course, we still need to
multiply this by negative one. And if we simplify this expression,
we see we get two. And therefore, because the
determinant of our square matrix is not equal to zero, we can conclude it must have
an inverse.
The next part of our question asked
us to find the inverse of our matrix by using the formula which involves the
cofactor matrix. So letβs clear some space and then
recall how we would do this for our matrix π΄. We recall we can find the inverse
of a matrix by following five steps.
First, we need to calculate the
determinant of our matrix π΄. We do this first because if this is
equal to zero, then we canβt find an inverse. And we already did this in the
first part of our question. We found that the determinant of
our matrix π΄ was equal to two.
The second thing we need to do is
find all of our matrix minors. Remember, the matrix minor π΄ ππ
is our matrix π΄ where we remove row π and column π. Letβs start by finding π΄ one
one. This means we need to remove the
first row and first column from our matrix π΄. So we remove the first row and the
first column from our matrix π΄. And by doing this, weβre left with
only four elements. So the matrix minor π΄ one one is
equal to the two-by-two matrix zero, one, one, zero. We need to find all of our matrix
minors. We now need to find π΄ one two. That means we remove the first row
and the second column from our matrix π΄. By doing this, we can see weβre
left with four entries: one, one, three, and zero. So the matrix minor π΄ one two is
the two-by-two matrix one, one, three, zero. And we can use exactly the same
method to find all of our matrix minors. We get the following nine
matrices.
The third thing we need to do is
construct our cofactor matrix. And remember, the entry in row π,
column π of our cofactor matrix is equal to negative one to the power of π plus π
multiplied by the determinant of the matrix minor π΄ ππ. This means we need to find the
determinants of all nine of our matrix minors. And we know how to find the
determinant of two-by-two matrices. For example, the determinant of π΄
one one will be equal to zero times zero minus one times one. And we can evaluate this
expression. Itβs equal to negative one.
We can do exactly the same to find
the determinant of our second cofactor matrix. We get itβs equal to one times zero
minus one times three, which we can calculate is equal to negative three. And we can do the exact same thing
to find the determinant of all the rest of our matrix minors. We get the following values. But remember, we need to multiply
these by negative one to the power of π plus π. In other words, we multiply the
determinant of π΄ one one by one. We then multiply the determinant of
π΄ one two by negative one, this gives us positive three, and multiply the
determinant of π΄ one three by one. We multiply the determinant of π΄
two one by negative one. This gives us positive three. And we can continue this on for all
of our matrix minors. This gives us the following
values.
And remember, each of these values
is an entry in our cofactor matrix. So by filling row π, column π
with each of these values, we get our cofactor matrix πΆ is the following
three-by-three matrix. So letβs clear our working and move
on to the fourth step. We now need to find our adjoint
matrix. Thatβs the transpose of our
cofactor matrix.
Remember, the transpose of a matrix
means we need to switch the rows with the columns. So when we transpose our cofactor
matrix, our first row will be negative one, three, zero. So we write in the first row
negative one, three, zero. And our second row will be three,
negative nine, two. And our third row will be one,
negative one, zero. And this is the adjoint of our
matrix π΄. All thatβs left to do now is use
our formula to find π΄ inverse.
Using our formula for π΄ inverse,
we get that π΄ inverse is equal to the following expression. And then we can simplify that
matrix to find the three-by-three matrix which is the inverse of π΄.
Letβs now go over the key points of
this video. First, the matrix minor π΄ ππ of
a matrix π΄ is obtained by removing row π and column π from π΄. We also know for a square matrix,
its cofactor matrix will have the same order. And the entries of our cofactor
matrix are generated by using the following formula. We know the adjoint of π΄ is the
transpose of the cofactor matrix. And finally, for a square matrix
whose determinant is nonzero, we found the inverse of π΄ is equal to one over the
determinant of π΄ times the adjoint of π΄.