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Consistency of Linear Equations

A system of linear equations is consistent if it has at least one solution. It is inconsistent if it has no solution.

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Student-friendly explanation

For AX = B, determinant methods help test solution type. If |A| ≠ 0, the system has a unique solution. If |A| = 0, use adj(A)B to distinguish cases: when adj(A)B is a non-zero matrix, the system is inconsistent; when adj(A)B is a zero matrix, the system may have infinitely many solutions or needs further examination depending on the equations.

How to write this in exams

  1. 1

    Start with the exact idea

    A system of linear equations is consistent if it has at least one solution. It is inconsistent if it has no solution.

  2. 2

    Then show how to use it

    Write the system as AX = B. Compute |A|. If |A| ≠ 0, state unique solution. If |A| = 0, compare dependence of equations or compute adj(A)B. Then classify as inconsistent, infinitely many solutions, or needing further verification.

  3. 3

    Add one concrete example

    For equations x + y = 2 and 2x + 2y = 5, A = [[1,1],[2,2]], so |A| = 0. The left sides are proportional, but constants are not proportional; hence the system has no solution.

  4. 4

    Avoid this incomplete answer

    Saying the inverse exists when |A| = 0 and trying to solve by X = A^-1B.

Definition

A system of linear equations is consistent if it has at least one solution. It is inconsistent if it has no solution.

Example

For equations x + y = 2 and 2x + 2y = 5, A = [[1,1],[2,2]], so |A| = 0. The left sides are proportional, but constants are not proportional; hence the system has no solution.

Rule to remember

For AX = B: if |A| ≠ 0, unique solution. If |A| = 0 and adj(A)B is non-zero, no solution. If |A| = 0 and adj(A)B is zero, the system is consistent with infinitely many solutions or requires further equation-level verification.

Memory hook

Non-zero determinant gives one answer; zero determinant needs a consistency test.

Examples and method

Worked example

Test consistency of x + y = 3 and 2x + 2y = 6. A = [[1,1],[2,2]], B = [[3],[6]]. |A| = 1×2 - 1×2 = 0. The second equation is exactly twice the first, including the constant. Therefore the equations represent the same line and have infinitely many solutions; the system is consistent.

Method to apply

Write the system as AX = B. Compute |A|. If |A| ≠ 0, state unique solution. If |A| = 0, compare dependence of equations or compute adj(A)B. Then classify as inconsistent, infinitely many solutions, or needing further verification.

Diagram support

No diagram is required. For two-variable systems, line graphs can optionally show intersecting, parallel, or coincident lines.

How CBSE asks it

Students may be asked to check consistency, find values of a parameter for which a system has unique or non-unique solutions, or classify solution type.

Avoid common mistakes

Common confusion

Assuming |A| = 0 always means infinitely many solutions. It may also mean no solution.

Common wrong answer

Saying the inverse exists when |A| = 0 and trying to solve by X = A^-1B.

Exam tip

When |A| = 0, compare both coefficients and constants or use adj(A)B; do not declare the result from determinant alone.

Quick check

If |A| ≠ 0 for AX = B, what type of solution does the system have?

It has a unique solution.

Answer writing and exam use

1-mark answer

A system of linear equations is consistent if it has at least one solution. It is inconsistent if it has no solution.

2-mark answer

A system of linear equations is consistent if it has at least one solution. It is inconsistent if it has no solution. For AX = B: if |A| ≠ 0, unique solution. If |A| = 0 and adj(A)B is non-zero, no solution. If |A| = 0 and adj(A)B is zero, the system is consistent with infinitely many solutions or requires further equation-level verification. For equations x + y = 2 and 2x + 2y = 5, A = [[1,1],[2,2]], so |A| = 0. The left sides are proportional, but constants are not proportional; hence the system has no solution.

3-mark answer

For AX = B, determinant methods help test solution type. If |A| ≠ 0, the system has a unique solution. If |A| = 0, use adj(A)B to distinguish cases: when adj(A)B is a non-zero matrix, the system is inconsistent; when adj(A)B is a zero matrix, the system may have infinitely many solutions or needs further examination depending on the equations. For AX = B: if |A| ≠ 0, unique solution. If |A| = 0 and adj(A)B is non-zero, no solution. If |A| = 0 and adj(A)B is zero, the system is consistent with infinitely many solutions or requires further equation-level verification. Test consistency of x + y = 3 and 2x + 2y = 6. A = [[1,1],[2,2]], B = [[3],[6]]. |A| = 1×2 - 1×2 = 0. The second equation is exactly twice the first, including the constant. Therefore the equations represent the same line and have infinitely many solutions; the system is consistent. Students may be asked to check consistency, find values of a parameter for which a system has unique or non-unique solutions, or classify solution type. Saying the inverse exists when |A| = 0 and trying to solve by X = A^-1B.
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