Rank of a matrix and the consistency of linear systems form the first half of the linear algebra module in VTU Engineering Mathematics 1 (BMATS101 under the 2022 scheme, 18MAT11 under 2018). The method is row reduction, and the same procedure answers two questions: how many independent equations a system contains, and whether it has a solution. Students who learn the row operations carefully find this module straightforward. Students who guess at the reduced form make errors that are hard to spot.
What the Rank of a Matrix Means
The rank of a matrix A, written rank(A), is the number of linearly independent rows (or equivalently, columns) it contains. It equals the number of non-zero rows after the matrix is reduced to row echelon form. Rank can never exceed the smaller of the number of rows and columns.
In practical terms, rank tells you how much new information the rows give. Two identical rows contribute only once. A row that is a combination of others adds nothing new.
Elementary Row Operations
You may perform three operations without changing the rank:
- Interchange two rows.
- Multiply a row by a non-zero number.
- Add a multiple of one row to another row.
The goal is to reach row echelon form, where each row has more leading zeros than the row above it, and any zero rows are at the bottom. The number of non-zero rows in that form is the rank.
Worked Example 1: Rank of a Singular 3 × 3 Matrix
Find the rank of A =
[ 1 2 3 ]
[ 2 4 6 ]
[ 1 0 1 ]
Apply R₂ → R₂ − 2R₁: the second row becomes [0 0 0].
Apply R₃ → R₃ − R₁: the third row becomes [0 −2 −2].
The matrix is now
[ 1 2 3 ]
[ 0 0 0 ]
[ 0 −2 −2 ]
Interchange R₂ and R₃ to get a triangular form:
[ 1 2 3 ]
[ 0 −2 −2 ]
[ 0 0 0 ]
There are two non-zero rows, so rank(A) = 2.
Notice that the second row of the original matrix is exactly twice the first, which is why one row disappeared. The determinant of A is zero, consistent with the rank being less than 3.
Worked Example 2: Rank of a Non-Singular Matrix
Find the rank of B =
[ 1 2 1 ]
[ 0 1 3 ]
[ 2 5 4 ]
Apply R₃ → R₃ − 2R₁: the third row becomes [0 1 2].
Apply R₃ → R₃ − R₂: the third row becomes [0 0 −1].
The matrix is now upper triangular with three non-zero rows, so rank(B) = 3. Check: the determinant of B is 1(4 − 15) − 2(0 − 6) + 1(0 − 2) = −11 + 12 − 2 = −1, which is non-zero, consistent with full rank.
Consistency of a Linear System
Consider the system AX = B, where A is the coefficient matrix and B is the column of right-hand sides. Form the augmented matrix [A | B] by adding B as an extra column. Then the system is:
- Consistent with a unique solution if rank(A) = rank([A | B]) = number of unknowns.
- Consistent with infinitely many solutions if rank(A) = rank([A | B]) < number of unknowns.
- Inconsistent (no solution) if rank(A) ≠ rank([A | B]).
The test is always the same: compare the rank of the coefficient matrix with the rank of the augmented matrix.
Worked Example 3: Testing Consistency
Test the system x + 2y + 3z = 6, 2x + 4y + 6z = 12, x + z = 4.
Write the augmented matrix and reduce:
[ 1 2 3 | 6 ]
[ 2 4 6 | 12 ]
[ 1 0 1 | 4 ]
R₂ → R₂ − 2R₁ gives [0 0 0 | 0]. R₃ → R₃ − R₁ gives [0 −2 −2 | −2].
The reduced augmented matrix has two non-zero rows, so rank([A | B]) = 2.
The coefficient matrix A is the matrix from Worked Example 1, with rank 2. Since rank(A) = rank([A | B]) = 2, the system is consistent. With 3 unknowns and rank 2, there are infinitely many solutions, with one free parameter.
To solve, take the reduced equations: x + 2y + 3z = 6 and −2y − 2z = −2, so y + z = 1. Let z = t. Then y = 1 − t, and x = 6 − 2(1 − t) − 3t = 4 − t. The general solution is x = 4 − t, y = 1 − t, z = t.
Check with t = 0: (4, 1, 0). Substitute into x + 2y + 3z = 4 + 2 + 0 = 6 ✓, and x + z = 4 ✓.
Worked Example 4: An Inconsistent System
Test x + y + z = 1, x + y + z = 2.
The coefficient matrix has rank 1 (both rows are the same). The augmented matrix is
[ 1 1 1 | 1 ]
[ 1 1 1 | 2 ]
Subtract the first row from the second: [0 0 0 | 1]. This row says 0 = 1, which is impossible. So rank([A | B]) = 2, but rank(A) = 1, and the system is inconsistent. No solution exists.
How to Structure a Rank Question
- Write the matrix or augmented matrix clearly. Separate the coefficient part from the constants with a bar if it helps.
- Use the pivot in the top-left corner. Eliminate the entries below it using row operations.
- Move to the next column and repeat. Continue until the matrix is in echelon form.
- Count the non-zero rows. That is the rank.
- Compare ranks for consistency. Write the comparison explicitly in the answer.
Common Mistakes That Cost Marks
Mistake 1: Interchanging rows without tracking the operation. Row interchanges do not change the rank, but they can make the arithmetic harder to check. Write each operation next to the matrix.
Mistake 2: Dividing by a zero pivot. If the leading entry is zero, interchange rows first. Do not divide by zero.
Mistake 3: Stopping too early. A matrix may look reduced but still contain a row that can be eliminated. Continue until the echelon form is clear.
Mistake 4: Comparing ranks incorrectly for consistency. Compare rank(A) with rank([A | B]), not with the number of equations. The consistency test is about the ranks of these two matrices.
Practice Questions
- Find the rank of [1 2 3; 4 5 6; 7 8 9].
- Test the consistency of x + y = 2, 2x + 2y = 4, x − y = 0.
- Solve x + 2y − z = 3, 2x + y + z = 4, 3x + 3y = 7 if consistent.
Checks: (1) R₂ − 4R₁ gives [0 −3 −6], R₃ − 7R₁ gives [0 −6 −12]. The second is twice the first, so after one more elimination there are two non-zero rows. Rank = 2. (2) The augmented matrix reduces to [1 1 | 2], [0 0 | 0], [0 −2 | −2]. Rank of A is 2 and of the augmented matrix is 2, so the system is consistent with a unique solution, x = 1, y = 1. Check: x + y = 2, 2x + 2y = 4 and x − y = 0 all hold. (3) Reducing gives [1 2 −1 | 3], [0 −3 3 | −2], and a zero row. Rank of A and of the augmented matrix are both 2, with 3 unknowns, so there are infinitely many solutions: x = 5/3 − t, y = 2/3 + t, z = t for any t. Check with t = 0: (5/3, 2/3, 0) gives x + 2y − z = 3 and 3x + 3y = 7.
A Matrix with a Parameter
Find the rank of A = [[1, 2, 3], [2, k, 6], [3, 6, 9]] as a function of k.
Apply R₂ → R₂ − 2R₁: the second row becomes [0, k − 4, 0].
Apply R₃ → R₃ − 3R₁: the third row becomes [0, 0, 0].
The matrix now has rank 1 if k − 4 = 0, and rank 2 otherwise, because the only remaining non-zero row is the first row together with the second row when k ≠ 4.
Answer: rank(A) = 1 when k = 4, and rank(A) = 2 when k ≠ 4.
Check k = 4: the matrix is [[1, 2, 3], [2, 4, 6], [3, 6, 9]], where every row is a multiple of [1, 2, 3], so the rank is 1 ✓. Check k = 0: [[1, 2, 3], [2, 0, 6], [3, 6, 9]] has a non-zero 2 × 2 minor, for example 1·0 − 2·2 = −4, so the rank is at least 2. Since the third row is 3 times the first, the rank is exactly 2 ✓.
This kind of question asks you to find the values of a parameter where the rank changes. Reduce the matrix in general form, keep the parameter in the pivot position, and identify the values that make a pivot zero. Those values are the special cases.
Connecting This Topic to the Full Module
Rank and the consistency of linear systems open the linear algebra module of VTU Engineering Mathematics 1, and they lead directly to eigenvalues. For the full module outline, the topics most often tested, and the places where marks are usually lost, see our VTU Engineering Maths 1 Notes — Complete Module-Wise Guide (BMATS101).
Frequently Asked Questions
Q: Is rank the same as the number of pivots?
Yes. In echelon form, each non-zero row has exactly one pivot, so the number of pivots equals the rank.
Q: Can a non-square matrix have full rank?
Yes. A 2 × 3 matrix has rank at most 2. Full rank means the rank equals the smaller dimension.
Q: Where does this fit in the module?
It comes before eigenvalues and Cayley-Hamilton. The full module outline is in our VTU Engineering Maths 1 notes guide.
For fully worked linear algebra problems, see the Eigenvalues and Eigenvectors Solved Examples page and the Engineering Math Survival Kit. To discuss batch options, use the contact page.