# Lesson Video: Rank of a Matrix: Determinants Mathematics

In this video, 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.

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### 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 𝐴 is the number of rows or columns 𝑛 of the largest square 𝑛-by-𝑛 submatrix of 𝐴 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 𝑝-by-𝑞 matrix 𝐴, that is, 𝑝 rows and 𝑞 columns, has lower and upper bounds of zero and the minimum of 𝑝 and 𝑞, respectively. That is, zero is less than or equal to the rank of 𝐴, which is less than or equal to the minimum of 𝑝 and 𝑞. 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 𝐴 is equal to zero if, and only if, 𝐴 is the zero matrix, that is, a matrix with elements only equal to zero. This is very easily shown by considering the general 𝑛-by-𝑛 matrix 𝐴 with general elements. If any one of these elements, say 𝑥 𝑖𝑗, 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 𝐴 must be greater than or equal to one and not equal to zero. Conversely, if we consider first the 𝑛-by-𝑛 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 𝐴 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 𝑎, 𝑏, 𝑐, 𝑑 not equal to the zero matrix. The only possible two-by-two submatrix of 𝐴 is just 𝐴 itself. Therefore, if the determinant of 𝐴 is zero, there is no two-by-two submatrix with a nonzero determinant. And therefore, the rank of 𝐴 cannot be equal to two. Since 𝐴 is also not equal to the zero matrix, then by the previous result, the rank of 𝐴 also cannot be zero. Therefore, the only remaining possibility is that the rank of 𝐴 is equal to one.

Conversely, if the rank of 𝐴 is equal to one, there cannot be a two-by-two submatrix of 𝐴 with a nonzero determinant. Therefore, the only two-by-two submatrix of 𝐴, 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 𝐴, if 𝐴 is the zero matrix, which should be obvious, we can immediately conclude that its rank is zero. If 𝐴 is not the zero matrix, we can check the determinant of 𝐴, and if it is equal to zero, then the rank of 𝐴 is one. And if not, the rank of 𝐴 is equal to two. Let’s 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 𝐴 is the number of rows or columns 𝑛 of the largest square 𝑛-by-𝑛 submatrix of 𝐴 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 𝐴 is equal to zero if and only if 𝐴 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 𝐴 is itself, there is no two-by-two submatrix with a nonzero determinant. Therefore, the rank of 𝐴 cannot be two. And this leaves only one option remaining: the rank of 𝐴 must be equal to one.

Now let’s 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 𝐴 is the number of rows or columns of the largest square 𝑛-by-𝑛 submatrix of 𝐴 with a nonzero determinant. Recall also that the rank of the matrix is between zero and the minimum of 𝑝 and 𝑞, where 𝑝 is the number of rows of 𝐴 and 𝑞 is the number of columns of 𝐴. This matrix has two rows and three columns. Therefore, the rank of 𝐴 must be less than or equal to the smaller of these numbers, which is two. Recall also that the rank of 𝐴 is equal to zero if and only if 𝐴 is the zero matrix. This matrix clearly isn’t 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 let’s 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 𝐴, calculate the determinant of this submatrix, and if the determinant is nonzero, then the rank of 𝐴 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 𝐴 is equal to the number of rows or columns in this submatrix. Now let’s 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 𝐴 is the number of rows or columns in the largest square submatrix of 𝐴 with a nonzero determinant. Recall also that the rank of 𝐴 is greater than or equal to zero and less than or equal to the minimum of 𝑝 and 𝑞, where 𝑝 is the number of rows in 𝐴 and 𝑞 is the number of columns in 𝐴. This is a three-by-three matrix. The rank of 𝐴 must therefore be between zero and three. And finally, recall that the rank of 𝐴 is equal to zero if and only if 𝐴 is the zero matrix. This matrix clearly isn’t the zero matrix. Therefore, its rank cannot be zero.

The largest possible square submatrix of 𝐴 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 𝐴, in this case itself, with a nonzero determinant. Therefore, the rank of 𝐴 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 matrix’s 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 𝐴 is the number of rows or columns in the largest square submatrix of 𝐴 with a nonzero determinant. Recall also that the rank of 𝐴 is greater than or equal to zero and less than or equal to the minimum of 𝑝 and 𝑞, where 𝑝 is the number of rows in 𝐴 and 𝑞 is the number of columns in 𝐴. Since 𝐴 in this case is a three-by-three matrix, the rank of 𝐴 is between zero and three. Recall also that the rank of 𝐴 is equal to zero if and only if 𝐴 is the zero matrix. This matrix clearly isn’t 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 𝐴, and it has a determinant of zero. Therefore, the rank of 𝐴 cannot be three. We now seek a two-by-two submatrix of 𝐴 with a nonzero determinant. This is a problem because there are nine possible two-by-two submatrices of 𝐴, 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 𝐴 with a nonzero determinant. However, if we select a two-by-two submatrix that doesn’t 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 𝐴 with a nonzero determinant. Therefore, the rank of 𝐴 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 𝐴 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 𝐴 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 𝐴 is two. We can verify this directly by taking the determinant of 𝐴 and showing it to be zero. This is the only three-by-three submatrix of 𝐴. Therefore, its rank must not be three. And taking a two-by-two submatrix of 𝐴, being careful to include the middle column, we get a nonzero determinant. Therefore, the rank of 𝐴 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 𝐴 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 𝑎, 𝑏, 𝑐 and two more rows that are scalar multiples 𝜆 and 𝜇 of the first row. By extension, all three columns are also scalar multiples of each other. Taking the determinant of 𝐴 by expanding along the top row, the terms in each of the brackets will all cancel each other out. Therefore, the determinant of 𝐴 is zero, and the rank of 𝐴 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 𝐴 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 𝐴 will all be zero. Therefore, the rank of 𝐴 cannot be two either. 𝐴 is not the zero matrix. Therefore, the only remaining possibility is the rank of 𝐴 is equal to one.

This statement is even more powerful than the last, since it is an “if and only if” 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 𝐴 contains no rows or columns that are scalar multiples of each other and the determinant of 𝐴 is equal to zero, then the rank of 𝐴 is two. This is easily shown because the determinant of 𝐴 being equal to zero means that the rank of 𝐴 cannot be three. And secondly, the matrix contains no rows or columns that are scalar multiples of each other, which first implies that 𝐴 is not the zero matrix. Therefore, its rank is not zero. And by the previous result, the rank of 𝐴 is not equal to one. Therefore, the only remaining option is that the rank of 𝐴 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 𝐴, first we check if 𝐴 is the zero matrix. If it is, then the rank of 𝐴 is zero. If not, we check the matrix to see if any of the rows or columns are scalar multiples of each other. If 𝐴 has exactly two rows or columns that are scalar multiples of each other, then the rank of 𝐴 is two. If all three rows and columns are scalar multiples of each other, then the rank of 𝐴 is one. If none of the rows or columns are scalar multiples of each other, we check the determinant of 𝐴. If the determinant is zero, then the rank of 𝐴 is two. And if not, the rank of 𝐴 is equal to three.

Let’s apply this process to an example matrix. Firstly, clearly 𝐴 is not the zero matrix. On closer inspection, 𝐴 contains no rows or columns that are scalar multiples of each other. And finally, taking the determinant of 𝐴 by expanding along the top row, we get a result of zero. Therefore, the rank of 𝐴 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 Rouché–Capelli theorem states that a system of linear equations with 𝑛 variables has solutions if and only if the rank of its coefficient matrix 𝐴 is equal to the rank of its augmented matrix 𝐴 stroke 𝑏.

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

Let’s 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 Rouché–Capelli theorem states that solutions to the system of linear equations exist if, and only if, the rank of its coefficient matrix 𝐴 is equal to the rank of its augmented matrix 𝐴 stroke 𝑏. Consider the coefficient matrix 𝐴. 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 𝐴 by expanding along the top row, we get a result of 1516. We have therefore found a three-by-three submatrix of 𝐴 with a nonzero determinant. Therefore, the rank of 𝐴 is three.

Now consider the augmented matrix 𝐴 stroke 𝑏. 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, 𝐴 stroke 𝑏 contains 𝐴 as a submatrix, which we have just shown to have a nonzero determinant. Therefore, the rank of 𝐴 stroke 𝑏 must be at least three and at most three. And therefore, the rank of 𝐴 stroke 𝑏 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 𝑛, the number of variables in the system of linear equations, which is three. By the Rouché–Capelli 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.

Let’s finish this video by recapping some key points. The rank of a matrix 𝐴 is given by the number of rows or columns of the largest square submatrix of 𝐴 with a nonzero determinant. For a matrix 𝐴 with 𝑝 rows and 𝑞 columns, the rank of 𝐴 is between zero and the minimum of 𝑝 and 𝑞. And finally, the Rouché–Capelli theorem states that a system of linear equations has solutions if and only if the rank of its coefficient matrix 𝐴 is equal to the rank of its augmented matrix 𝐴 stroke 𝑏.

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