### Video Transcript

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.