Lesson Explainer: Rank of a Matrix: Determinants | Nagwa Lesson Explainer: Rank of a Matrix: Determinants | Nagwa

# Lesson Explainer: Rank of a Matrix: Determinants Mathematics

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

### Definition: Rank of a Matrix

The “rank” of a matrix , , is the number of rows or columns, , of the largest square submatrix of for which the determinant is nonzero.

The largest possible square submatrix of a general matrix will be whichever of or is smaller. For example, for the matrix the largest possible square submatrix is . And likewise, for a matrix the largest possible square submatrix is .

The upper bound for the rank of a matrix is therefore the minimum (i.e., whichever is smallest) of the number of rows or columns.

The lower bound for the rank of a matrix is 0, but this can only be the case if we cannot find a matrix with a nonzero determinant, that is, if the matrix has no nonzero elements.

### Theorem: Upper and Lower Bounds for the Rank of a Matrix

The rank of an matrix , , has lower and upper bounds given by

### Theorem: The Rank of the Zero Matrix

if and only if  is the zero matrix, .

This means that the rank of a matrix can be found simply by computing its determinant.

### Corollary: The Rank of a 2 × 2 Matrix

A matrix , where , has rank  if and only if it has a determinant .

We can prove this corollary as follows.

The only submatrix of is itself. If , then the largest possible square submatrix with a nonzero determinant is a matrix. Since , there is at least one submatrix of with a nonzero determinant; therefore, .

Conversely, if , any submatrix of must have a determinant of 0. The only submatrix of is itself; therefore, .

The rank of a matrix can therefore be found by the following process:

Let’s look at an example of how to find the rank of a matrix using determinants.

### Example 1: Finding the Rank of a Matrix

Find the rank of the matrix .

Recall that the rank of a matrix is equal to the number of rows/columns of the largest square submatrix of that has a nonzero determinant.

Since the matrix is a square matrix, the largest possible square submatrix is the original matrix itself. Its rank must therefore be between 0 and 2 inclusive. We can see immediately that the matrix is not the zero matrix, and therefore its rank cannot be zero. Taking the determinant of the matrix,

Since the determinant of the matrix is zero, its rank cannot be equal to the number of rows/columns, 2.

The only remaining possibility is that the rank of the matrix is 1, which we do not need to verify by taking any further determinants.

Therefore, the rank of the matrix is 1.

We will now take a look at how this approach of using the determinant to find the rank extends to larger matrices.

### Example 2: Finding the Rank of a Matrix

Find the rank of the following matrix using determinants:

Recall that the rank of a matrix is equal to the number of rows/columns of the largest square submatrix of that has a nonzero determinant.

Since this is a matrix, the largest square submatrix we can take is , and therefore its rank must be between 0 and 2. The matrix is not the zero matrix; therefore, its rank cannot be 0.

Consider the submatrix obtained by “deleting” the right column:

Taking the determinant of this submatrix,

We have found a submatrix with a nonzero determinant; therefore, the rank of the original matrix must be 2.

The techniques shown so far can be boiled down to a three-step method.

### How To: Finding the Rank of a Matrix 𝐴

1. Consider the largest possible square submatrix of . Calculate the determinant of this submatrix. If the determinant is nonzero, the rank of the original matrix is given by the number of rows of the submatrix.
2. If the determinant of the submatrix is zero, repeat step 1 for other possible square submatrices of the same size.
3. If a submatrix with a nonzero determinant has not been found, repeat steps 1 and 2 for submatrices 1 row and column smaller until a submatrix with a nonzero determinant is found. The rank of the original matrix is equal to the number of rows/columns of this submatrix with a nonzero determinant.

matrices are among the most common matrices, particularly due to their occurrence in problems in 3D space.

Quickly finding the rank of a matrix is therefore a very useful tool.

Consider the matrix

We first need to consider the largest possible square submatrix of . Since is already a square matrix, the largest possible square submatrix of is just itself.

Once we have chosen a submatrix, we can take the determinant of the matrix. If the determinant is nonzero, then the rank of the original matrix is equal to the number of rows/columns of the submatrix.

If we take the determinant of the example matrix, , above by expanding along the top row, we find that

We have found that the determinant of this submatrix (in this case, itself) is zero. There are no other submatrices of ; therefore, by the definition of rank, cannot be 3.

The next step is to consider smaller square submatrices, in this case submatrices. If we take the submatrix obtained by “deleting” the bottom row and right column of , let’s call this submatrix , we see

Taking the determinant of ,

So, we have found a submatrix of for which the determinant is nonzero; therefore, .

Let’s look at a simple example of how to find the rank of another matrix.

### Example 3: Finding the Rank of a Given Matrix

Find the rank of the matrix

Recall that the rank of a matrix is equal to the number of rows/columns of the largest square submatrix of that has a nonzero determinant.

Since this is a matrix, its rank must be between 0 and 3. Also, since it is not the zero matrix, its rank cannot be 0.

The largest possible square submatrix of this matrix is just itself, so let’s expand the determinant of this matrix along the top row:

The determinant of the matrix is nonzero; therefore, its rank must be 3.

We sometimes need to take care when selecting a submatrix of the original matrix. It is possible to find one submatrix with a determinant , but another with a determinant . Consider, for example, the matrix

Taking the determinant of itself by expanding along the top row,

Therefore, . If we proceed by taking the submatrix obtained by “deleting” the bottom row and right column of , we find that

Taking the determinant of ,

So, we might conclude that either. However, if we had instead chosen the submatrix obtained by “deleting” the top row and right column of , we would have found

Taking the determinant of this matrix gives

So, we have found a submatrix of for which the determinant is nonzero; therefore, .

This property could mean that we may need to check multiple different submatrices of the same size before concluding the rank of the matrix. Fortunately, we do not need to do this, as we can skip this process by noticing something about the original matrix:

Notice that the second row of is a scalar multiple of the first. Specifically, each element in the second row is the element above it.

### Lemma: The Determinant of a 2 × 2 Matrix with Scalar Multiple Rows/Columns

A matrix has determinant  if and only if the rows/columns of are scalar multiples of each other.

### Corollary: Row/Column Redundancy and Determinants

A square matrix that contains a row/column that is a scalar multiple of another row/column will have a determinant of zero, and any submatrix of taken from those two rows/columns will also have a determinant of zero.

Let’s look at an example of how to use the techniques covered so far to find the rank of a matrix.

### Example 4: The Rank of a 3 × 3 Matrix

Find the rank of the following matrix:

Recall that the rank of a matrix is equal to the number of rows/columns of the largest square submatrix of that has a nonzero determinant.

Since this is a matrix, its rank must be between 0 and 3. Also, since it is not the zero matrix, its rank cannot be 0.

The largest possible submatrix is the original matrix itself. Taking the determinant of the original matrix by expanding along the top row, we get

This is the only possible submatrix, and its determinant is zero; therefore, the rank of the original matrix cannot be 3.

On inspecting the original matrix, we can see that the bottom row is an exact nonzero scalar multiple of the top row. This means that the determinant of any submatrix selected from these two rows will have a determinant of zero. We can verify this directly:

However, this does not mean that there is no submatrix of the original matrix with a nonzero determinant. The middle row of the original matrix is not a scalar multiple of the other two, so any determinant of a submatrix including the middle row will have a nonzero determinant.

Taking the matrix obtained by “deleting” the bottom row and right-hand column,

Taking the determinant of this submatrix,

We have found a submatrix of the original matrix that has a nonzero determinant; therefore, the rank of the original matrix is 2.

In the previous example, we saw that we need to be more selective with our choice of submatrix in seeking a matrix that contains rows/columns that are nonzero scalar multiples of each other.

In fact, however, for a matrix or smaller, if we notice that at least one row/column is a scalar multiple of another, we do not need to select a submatrix at all. Instead, we can immediately conclude the rank of the matrix from the number of rows/columns that are scalar multiples of each other.

### Theorem: The Rank of a 3 × 3 Matrix with Two Scalar Multiple Rows/Columns

If a matrix , containing no zero rows/columns, contains two rows/columns that are scalar multiples of each other and a third row/column that is not a scalar multiple of the other two, then .

In some cases, it may be that all three rows of a matrix are scalar multiples of each other.

### Theorem: The Rank of a 3 × 3 Matrix with Three Scalar Multiple Rows/Columns

A matrix , where , has rank  if and only if it contains three rows/columns that are scalar multiples of each other.

### Corollary: The Rank of a 3 × 3 Matrix with No Scalar Multiple Rows/Columns

If a matrix contains no rows/columns that are scalar multiples of each other and , then .

With these properties in mind, the rank of any matrix can be found much more quickly by the following process:

Let’s consider an example of how to use this method to quickly find the ranks of multiple matrices.

Consider the matrices

None of the matrices are the zero matrix, so they must all have a rank between 1 and 3.

Starting with , all three rows (or columns) are scalar multiples of each other; therefore, .

Next, does not have any rows that are scalar multiples of each other, but columns 1 and 3 are scalar multiples of each other, and column 2 is not; therefore, .

Next, does not have any rows or columns that are scalar multiples of each other. Taking the determinant of by expanding along the top row, we get

has a nonzero determinant; therefore, its rank is 3.

And finally, also has no rows/columns that are scalar multiples of each other. Taking the determinant of by expanding along the top row, we get

has a determinant of zero; therefore, its rank must be 2.

In the next example, we will see how we can extend this general technique for finding the rank of a matrix to solving algebraic problems.

### Example 5: The Rank of a Matrix

What value can not take if the rank of the matrix is 3?

The rank of an matrix can only be equal to if the determinant of the matrix is nonzero. Therefore, if the rank of the matrix above is 3 (rank ), then . This being the case, by finding the determinant of the matrix above and setting it equal to zero, solving for will give us the value that cannot take.

Taking the determinant of the matrix by expanding along the middle column, we find

If we now set , we have

Solving this equation for gives

So, if the rank of the matrix is 3, the value of cannot be 1.

One of the most significant implications of the rank of a matrix is the number of solutions to the system of linear equations it represents.

### Theorem: Rouché–Capelli Theorem

A system of linear equations with variables has solution(s) if and only if the rank of its coefficient matrix, , is equal to the rank of its augmented matrix, .

If , the system of equations has no solutions.

If , the system has one unique solution.

If , the system has infinitely many solutions.

Consider the following system of linear equations:

This system of equations may be represented by the matrix equation

Therefore, the coefficient matrix is given by and the augmented matrix is given by

Taking the determinant of by expanding along the top row,

Therefore, contains a submatrix (in this case, itself) with a nonzero determinant; therefore, its rank is 3. Since  contains the submatrix , its rank must also be 3.

Therefore, , so the system of equations has solution(s). Since also , the number of variables in the system, the system has one unique solution.

It is worth noting here that we can save time in finding the rank of the augmented matrix if we can show that it is at most equal to the rank of the coefficient matrix.

### Theorem: Rank of the Augmented Matrix

The rank of the augmented matrix, , of a system of linear equations is equal to or greater than the rank of the coefficient matrix, . That is,

This is easily shown. Since the coefficient matrix, , is itself a submatrix of the augmented matrix, , any submatrix of is also a submatrix of ; therefore, any square submatrix of with a nonzero determinant must also be a submatrix of . Therefore, is at least equal to .

Let’s look at an example of how to find the number of solutions to a system of linear equations by finding the determinants of the coefficient and augmented matrices, and how this strategy of finding their ranks using determinants greatly speeds up the process of finding the number of solutions to a system of linear equations.

### Example 6: Finding 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 Rouché–Capelli theorem states that a system of linear equations has solution(s) if and only if the rank of its coefficient matrix is equal to the rank of its augmented matrix.

In our case, the coefficient matrix, , is the matrix on the left of the equation:

The augmented matrix, , is formed by “appending” the solution matrix to the right-hand side of the coefficient matrix:

Recall that the rank of a matrix is equal to the number of rows/columns of the largest square submatrix of that has a nonzero determinant.

The only submatrix of the coefficient matrix is itself. Taking the determinant of by expanding along the top row,

We have found a submatrix of the coefficient matrix, (in this case, itself) with a nonzero determinant. Therefore, .

Since the augmented matrix, , is a matrix, its rank is at most 3, and since the augmented matrix has rank at least equal to the coefficient matrix, , its rank is also at least 3. Therefore, the .

Hence, we have , the number of variables in the system of equations. Therefore, the system of equations has one, unique solution.

In the next example, we will look at a system of equations for which the rank of the coefficient matrix is not equal to the rank of the augmented matrix.

### Example 7: Finding 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 Rouché–Capelli theorem states that a system of linear equations has solution(s) if and only if the rank of its coefficient matrix is equal to the rank of its augmented matrix.

In our case, the coefficient matrix, , is the matrix on the left of the equation:

The augmented matrix, , is formed by “appending” the solution matrix to the right-hand side of the coefficient matrix:

Recall that the rank of a matrix is equal to the number of rows/columns of the largest square submatrix of that has a nonzero determinant.

The only submatrix of the coefficient matrix, , is itself. Taking the determinant of by expanding along the top row,

This is the only possible submatrix of , which has a determinant of 0; therefore, its rank cannot be 3. also has no rows or columns that are scalar multiples of each other; therefore, its rank must be 2.

Next, we need to find the rank of the augmented matrix, . Since it is a matrix, its rank can be at most the minimum (i.e., whichever is smallest) of 3 and 4; therefore, .

We, therefore, seek a submatrix of with a nonzero determinant. One submatrix of is of course the coefficient matrix, , but we have already shown this to have a determinant of zero, so we must look for a different submatrix.

Consider then the submatrix, which we will call , formed from “deleting” the left-hand column of :

Taking the determinant of by expanding along the top row, we get

We have found a submatrix of with a nonzero determinant; therefore, .

Hence, , and by the Rouché–Capelli theorem, the system of equations has no solutions.

In the previous example, the rank of the augmented matrix was greater than the rank of the coefficient matrix, and we only needed to calculate one determinant of the augmented matrix to confirm its rank was 3.

However, it may be the case that the rank of an augmented matrix cannot be verified by calculating one determinant. For a augmented matrix, like in the previous example, in the worst case scenario we may need to take three determinants, in addition to the determinant of the coefficient matrix.

To avoid this time-consuming process, we can use one more theorem.

### Theorem: Rank of a Matrix with Linearly Dependent Rows/Columns

If an matrix contains a row/column that may be formed from a linear combination of any other rows/columns, then the rank of is strictly less than the minimum of and . That is,

Let’s look at one final example of how to find the number of solutions to a system of linear equations by using this time-saving theorem.

### Example 8: Finding the Number of Solutions to a System of Linear Equations

Find the number of solutions of the following system of linear equations:

Recall that the Rouché–Capelli theorem states that a system of linear equations has solution(s) if and only if the rank of its coefficient matrix is equal to the rank of its augmented matrix.

In our case, the coefficient matrix, , is the matrix on the left of the equation:

The augmented matrix, , is formed by “appending” the solution matrix to the right-hand side of the coefficient matrix:

Recall that the rank of a matrix is equal to the number of rows/columns of the largest of its square submatrices that has a nonzero determinant.

Notice also that our matrix, , has one row that may be formed from a linear combination of the other two. Specifically, row 2 is equal to the sum of row 1 and row 3:

Since one row is a linear combination of the other two, the rank of must be less than 3. We can verify this directly. The only submatrix of the coefficient matrix is itself. Taking the determinant of by expanding along the top row, we get

This is the only possible submatrix of , which has a determinant of 0; therefore, the rank of cannot be 3. also has no rows or columns that are scalar multiples of each other; therefore, its rank must be 2.

Next, we need to find the rank of the augmented matrix, . Since it is a matrix, its rank can be at most the minimum (i.e., whichever is smallest) of 3 and 4; therefore, .

We found that for the coefficient matrix, , row 2 was equal to the sum of row 1 and row 3. If the same still applies to the augmented matrix, , its rank also must be less than 3.

Looking at the sum of row 1 and row 3 in , we can see that this is indeed the case:

Therefore, the rank of the augmented matrix, , cannot be 3. Although unnecessary and time-consuming, this can be verified directly by finding the determinant of every other submatrix of and showing them all to be 0. Indeed, we find

Since the rank of the augmented matrix must be equal to or greater than the rank of the coefficient matrix, we must have that .

Therefore, the rank of the coefficient matrix is equal to the rank of the augmented matrix; hence, the system of linear equations has solution(s). Since the rank of the coefficient matrix, , is less than the number of variables in the system, , the system has infinitely many solutions.

We complete this explainer with some of the key points associated with the determinant of a matrix and its rank.

### Key Points

• For an matrix , its rank, , is given by the number of rows/columns of the largest square submatrix of (which may be itself) that has a nonzero determinant.
• .
•  if and only if  is the zero matrix, .
• A matrix , where , has rank  if and only if .
• The rank of any matrix can be found by the following process:
• Consider the largest possible square submatrix of . Calculate the determinant of this submatrix. If the determinant is nonzero, the rank of the original matrix is given by the number of rows of the submatrix.
• If the determinant of the submatrix is zero, repeat step 1 for other possible square submatrices of the same size.
• If a submatrix with a nonzero determinant has not been found, repeat steps 1 and 2 for submatrices 1 row and column smaller.
• The rank of a matrix can be found by the following process:
• The rank of a matrix can be found by the following process:
• The Rouché–Capelli theorem states that a system of linear equations with variables has solution(s) if and only if the rank of its coefficient matrix, , is equal to the rank of its augmented matrix, . If , the system of equations has no solutions. If , the system has one unique solution. If , the system has infinitely many solutions.
• The rank of the augmented matrix, , of a system of linear equations is equal to or greater than the rank of the coefficient matrix, . That is, .