### Video Transcript

In this video, we will learn how to
find the rank of a matrix using determinants and how we can use this to find the
number of solutions to a system of linear equations.

The rank of a matrix RK of π΄ is
the number of rows or columns π of the largest square π-by-π submatrix of π΄
whose determinant is nonzero. Consider the following matrix as an
example. This is a four-by-two matrix, and
the largest possible square submatrix we can take of it is a two-by-two matrix. We can choose from several possible
two-by-two matrices by deleting any two rows. Sticking with this two-by-two
submatrix by deleting the top two rows, taking the determinant gives one times nine
minus negative one time seven, which is equal to 16, which is nonzero.

We have therefore found a
two-by-two submatrix of the original matrix with a nonzero determinant. Therefore, the rank of the original
matrix is equal to the number of rows or columns of this submatrix, which is
two. This hints at an important
result. The rank of π-by-π matrix π΄,
that is, π rows and π columns, has lower and upper bounds of zero and the minimum
of π and π, respectively. That is, zero is less than or equal
to the rank of π΄, which is less than or equal to the minimum of π and π. This makes sense because for the
lower bound zero, clearly we cannot have a submatrix with fewer than zero rows or
columns. And for the upper bound, clearly
the upper bound of a square matrix inside the original matrix will be whichever is
lower of the number of rows or columns.

This leads to another important
result. The rank of a matrix π΄ is equal to
zero if, and only if, π΄ is the zero matrix, that is, a matrix with elements only
equal to zero. This is very easily shown by
considering the general π-by-π matrix π΄ with general elements. If any one of these elements, say
π₯ ππ, is not equal to zero, then we can find a one-by-one matrix, that is, a
single element, with a nonzero determinant. Therefore, the rank of π΄ must be
greater than or equal to one and not equal to zero. Conversely, if we consider first
the π-by-π zero matrix, any square submatrix that we take of this matrix will also
be the zero matrix, which will have a determinant of zero. Therefore, we cannot find a
submatrix with the determinant not equal to zero. Therefore, the rank of the zero
matrix is zero.

We can use these results to draw a
helpful conclusion about two-by-two matrices specifically. A two-by-two matrix π΄ not equal to
the zero matrix has rank equal to one if and only if it has a determinant of
zero. To show this, consider the general
two-by-two matrix π, π, π, π not equal to the zero matrix. The only possible two-by-two
submatrix of π΄ is just π΄ itself. Therefore, if the determinant of π΄
is zero, there is no two-by-two submatrix with a nonzero determinant. And therefore, the rank of π΄
cannot be equal to two. Since π΄ is also not equal to the
zero matrix, then by the previous result, the rank of π΄ also cannot be zero. Therefore, the only remaining
possibility is that the rank of π΄ is equal to one.

Conversely, if the rank of π΄ is
equal to one, there cannot be a two-by-two submatrix of π΄ with a nonzero
determinant. Therefore, the only two-by-two
submatrix of π΄, itself, has a determinant of zero. This result is very significant
since it means we can instantly find the rank of a two-by-two matrix simply by
taking its determinant. Starting with any two-by-two matrix
π΄, if π΄ is the zero matrix, which should be obvious, we can immediately conclude
that its rank is zero. If π΄ is not the zero matrix, we
can check the determinant of π΄, and if it is equal to zero, then the rank of π΄ is
one. And if not, the rank of π΄ is equal
to two. Letβs look at an example of how to
use this process to quickly find the rank of a two-by-two matrix.

Find the rank of the matrix two,
24, four, 48.

Recall that the rank of a matrix π΄
is the number of rows or columns π of the largest square π-by-π submatrix of π΄
with a nonzero determinant. This implies that for a two-by-two
matrix like the one we have, the rank is between zero and two. Recall also that the rank of π΄ is
equal to zero if and only if π΄ is the zero matrix. Clearly, this matrix is not the
zero matrix. Therefore, its rank is not equal to
zero. Taking the determinant of the
original matrix, we get two times 48 minus 24 times four, which is equal to
zero. Since the only two-by-two submatrix
of π΄ is itself, there is no two-by-two submatrix with a nonzero determinant. Therefore, the rank of π΄ cannot be
two. And this leaves only one option
remaining: the rank of π΄ must be equal to one.

Now letβs take a look at an example
of using this approach to find the rank of larger matrices.

Find the rank of the following
matrix using determinants: seven, six, eight, negative eight, three, eight.

Recall that the rank of a matrix π΄
is the number of rows or columns of the largest square π-by-π submatrix of π΄ with
a nonzero determinant. Recall also that the rank of the
matrix is between zero and the minimum of π and π, where π is the number of rows
of π΄ and π is the number of columns of π΄. This matrix has two rows and three
columns. Therefore, the rank of π΄ must be
less than or equal to the smaller of these numbers, which is two. Recall also that the rank of π΄ is
equal to zero if and only if π΄ is the zero matrix. This matrix clearly isnβt the zero
matrix. Therefore, its rank cannot be
zero.

We now seek the largest square
submatrix of the original matrix with a nonzero determinant. The largest possible square
submatrix of the original matrix will be a two by two. So letβs choose the two-by-two
matrix formed from deleting the right-most column. Taking the determinant of this
submatrix, we get seven times three minus six times negative eight, which is equal
to 21 plus 48, which is equal to 69, which is not equal to zero. We have therefore found a
two-by-two submatrix of the original matrix with a nonzero determinant. Therefore, the rank of the original
matrix is two.

The techniques shown so far can be
boiled down to a three-step method to find the rank of any matrix. The first step is to find the
largest possible square submatrix of π΄, calculate the determinant of this
submatrix, and if the determinant is nonzero, then the rank of π΄ is equal to the
number of rows or columns in the submatrix. If the determinant of the submatrix
is zero, we repeat step one for other possible square submatrices of the same
size. And finally, if a square submatrix
with a nonzero determinant has not yet been found, we repeat steps one and two for
submatrices one row and column smaller until a submatrix with a nonzero determinant
is found. Then the rank of π΄ is equal to the
number of rows or columns in this submatrix. Now letβs take a look at how we
apply this to even larger matrices.

Find the rank of the matrix
negative 16, negative 11, negative 14, 17, 19, negative 24, three, negative six,
negative 24.

Recall that the rank of a matrix π΄
is the number of rows or columns in the largest square submatrix of π΄ with a
nonzero determinant. Recall also that the rank of π΄ is
greater than or equal to zero and less than or equal to the minimum of π and π,
where π is the number of rows in π΄ and π is the number of columns in π΄. This is a three-by-three
matrix. The rank of π΄ must therefore be
between zero and three. And finally, recall that the rank
of π΄ is equal to zero if and only if π΄ is the zero matrix. This matrix clearly isnβt the zero
matrix. Therefore, its rank cannot be
zero.

The largest possible square
submatrix of π΄ is just itself, a three-by-three matrix. Taking the determinant of the
matrix by expanding along the top row, we get a result of 8130, which is not equal
to zero. We have therefore found a
three-by-three submatrix of π΄, in this case itself, with a nonzero determinant. Therefore, the rank of π΄ is
three.

You may be wondering why we did not
simply start with taking the determinant of the matrix since this is all that was
needed to verify that its rank was three. The reason is that we did not know
the matrixβs rank was three and its determinant could well have been equal to zero,
in which case these facts are easier to verify first. Sometimes we need to be a little
more strategic when selecting a submatrix of the original matrix.

Find the rank of the following
matrix.

Recall that the rank of a matrix π΄
is the number of rows or columns in the largest square submatrix of π΄ with a
nonzero determinant. Recall also that the rank of π΄ is
greater than or equal to zero and less than or equal to the minimum of π and π,
where π is the number of rows in π΄ and π is the number of columns in π΄. Since π΄ in this case is a
three-by-three matrix, the rank of π΄ is between zero and three. Recall also that the rank of π΄ is
equal to zero if and only if π΄ is the zero matrix. This matrix clearly isnβt the zero
matrix. Therefore, its rank cannot be
zero.

The largest possible square
submatrix of this matrix is just itself, a three-by-three matrix. Taking the determinant of the
matrix by expanding along the top row, we get a result of zero. This is the only possible
three-by-three submatrix of π΄, and it has a determinant of zero. Therefore, the rank of π΄ cannot be
three. We now seek a two-by-two submatrix
of π΄ with a nonzero determinant. This is a problem because there are
nine possible two-by-two submatrices of π΄, and we may need to check every single
one of them.

Consider, for instance, if the
original matrix had looked like this. The only two-by-two submatrix we
can select that might have a nonzero determinant is this one. In this hypothetical example, the
choice is clear, but it might not be for our question. If we look at the original matrix,
we can see that the bottom row is an exact scalar multiple of the top row. Any two-by-two submatrix selected
from these two rows will have determinant of zero. And we might suppose from this that
this means that there are no two-by-two submatrices with a nonzero determinant.

We might suppose from this that
there are no two-by-two submatrices of π΄ with a nonzero determinant. However, if we select a two-by-two
submatrix that doesnβt come from just these two rows that are scalar multiples of
each other, for example, by removing the bottom row and the right-most column, we
get a nonzero determinant. We have therefore found a
two-by-two submatrix of π΄ with a nonzero determinant. Therefore, the rank of π΄ is
two.

In this example, we saw how we can
use the fact that two rows of the matrix are scalar multiples of each other to find
the rank of the matrix more quickly. This leads to the following
result. If a three-by-three matrix π΄
containing no zero rows or columns contains two rows or columns that are scalar
multiples of each other and a third row or column that is not a scalar multiple of
the other two, then the rank of π΄ is equal to two.

Consider the following matrix as an
example. It has no rows that are scalar
multiples of each other, but it does have two columns that are scalar multiples of
each other. The right-hand column is exactly
two times the left-hand column, and the middle column is not a scalar multiple of
either. We can therefore immediately
conclude that the rank of π΄ is two. We can verify this directly by
taking the determinant of π΄ and showing it to be zero. This is the only three-by-three
submatrix of π΄. Therefore, its rank must not be
three. And taking a two-by-two submatrix
of π΄, being careful to include the middle column, we get a nonzero determinant. Therefore, the rank of π΄ is
two.

In some cases, we may encounter a
matrix where all three rows and columns are scalar multiples of each other. In this case, we have the following
result. A three-by-three matrix π΄ not
equal to the zero matrix has rank equal to one if and only if it contains three rows
or columns that are scalar multiples of each other. Such matrix will look like
this. We have a nonzero row π, π, π
and two more rows that are scalar multiples π and π of the first row. By extension, all three columns are
also scalar multiples of each other. Taking the determinant of π΄ by
expanding along the top row, the terms in each of the brackets will all cancel each
other out. Therefore, the determinant of π΄ is
zero, and the rank of π΄ cannot be three. This will be the result no matter
how we move around the rows and columns, as long as they are scalar multiples of
each other.

To verify that the rank of π΄ is
not two, we need not take all nine determinants of all nine two-by-two
submatrices. Instead, we can just use the
general result that the determinant of a two-by-two matrix whose rows and columns
are scalar multiples of each other will always be equal to zero. Therefore, the determinants of all
nine two-by-two submatrices of π΄ will all be zero. Therefore, the rank of π΄ cannot be
two either. π΄ is not the zero matrix. Therefore, the only remaining
possibility is the rank of π΄ is equal to one.

This statement is even more
powerful than the last, since it is an βif and only ifβ statement. This means that if we have a
three-by-three matrix that does not meet this criteria, its rank cannot be one. This leads to one final result. If a three-by-three matrix π΄
contains no rows or columns that are scalar multiples of each other and the
determinant of π΄ is equal to zero, then the rank of π΄ is two. This is easily shown because the
determinant of π΄ being equal to zero means that the rank of π΄ cannot be three. And secondly, the matrix contains
no rows or columns that are scalar multiples of each other, which first implies that
π΄ is not the zero matrix. Therefore, its rank is not
zero. And by the previous result, the
rank of π΄ is not equal to one. Therefore, the only remaining
option is that the rank of π΄ is two.

All of these results together allow
us to follow a general procedure to find the rank of a three-by-three matrix. Starting with a three-by-three
matrix π΄, first we check if π΄ is the zero matrix. If it is, then the rank of π΄ is
zero. If not, we check the matrix to see
if any of the rows or columns are scalar multiples of each other. If π΄ has exactly two rows or
columns that are scalar multiples of each other, then the rank of π΄ is two. If all three rows and columns are
scalar multiples of each other, then the rank of π΄ is one. If none of the rows or columns are
scalar multiples of each other, we check the determinant of π΄. If the determinant is zero, then
the rank of π΄ is two. And if not, the rank of π΄ is equal
to three.

Letβs apply this process to an
example matrix. Firstly, clearly π΄ is not the zero
matrix. On closer inspection, π΄ contains
no rows or columns that are scalar multiples of each other. And finally, taking the determinant
of π΄ by expanding along the top row, we get a result of zero. Therefore, the rank of π΄ is
two.

One of the most important
implications of the rank of a matrix is the number of solutions to the system of
linear equations it represents. The RouchΓ©βCapelli theorem states
that a system of linear equations with π variables has solutions if and only if the
rank of its coefficient matrix π΄ is equal to the rank of its augmented matrix π΄
stroke π.

More specifically, if the rank of
π΄ is not equal to the rank of π΄ stroke π, the system has no solutions. If the rank of π΄ is equal to the
rank of π΄ stroke π which is equal to π, the number of variables in the system,
then the system has one unique solution. And finally, if the rank of π΄ is
equal to the rank of π΄ stroke π but not equal to π, the number of variables in
the system, then the system has infinitely many solutions.

Letβs look at an example of how to
apply this theorem to quickly find the number of solutions to a system of linear
equations.

Find the number of solutions for
the following system of linear equations.

Recall that the RouchΓ©βCapelli
theorem states that solutions to the system of linear equations exist if, and only
if, the rank of its coefficient matrix π΄ is equal to the rank of its augmented
matrix π΄ stroke π. Consider the coefficient matrix
π΄. It is clearly not the zero matrix,
and it contains no rows or columns that are scalar multiples of each other. Taking the determinant of π΄ by
expanding along the top row, we get a result of 1516. We have therefore found a
three-by-three submatrix of π΄ with a nonzero determinant. Therefore, the rank of π΄ is
three.

Now consider the augmented matrix
π΄ stroke π. This is a three-by-four matrix. And recall that the rank of a
matrix cannot exceed the minimum of either the number of rows or columns. Therefore, the rank of this matrix
must be at most three. By construction, π΄ stroke π
contains π΄ as a submatrix, which we have just shown to have a nonzero
determinant. Therefore, the rank of π΄ stroke π
must be at least three and at most three. And therefore, the rank of π΄
stroke π is equal to three.

So we have shown that the rank of
the augmented matrix is equal to the rank of the coefficient matrix and both are
equal to π, the number of variables in the system of linear equations, which is
three. By the RouchΓ©βCapelli theorem, this
system, therefore, has solutions. And since the rank of the
coefficient matrix and augmented matrix are both equal to the number of variables in
the system, the system has one unique solution.

Letβs finish this video by
recapping some key points. The rank of a matrix π΄ is given by
the number of rows or columns of the largest square submatrix of π΄ with a nonzero
determinant. For a matrix π΄ with π rows and π
columns, the rank of π΄ is between zero and the minimum of π and π. And finally, the RouchΓ©βCapelli
theorem states that a system of linear equations has solutions if and only if the
rank of its coefficient matrix π΄ is equal to the rank of its augmented matrix π΄
stroke π.

More specifically, if the rank of
π΄ is not equal to the rank of π΄ stroke π, the system has no solutions. If the rank of π΄ is equal to the
rank of π΄ stroke π and both are equal to π, the number of variables in the
system, then the system has one unique solution. And if the rank of π΄ is equal to
the rank of π΄ stroke π and not equal to the number of variables in the system,
then the system has infinitely many solutions.