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Solving Linear Equations Using Matrix Inverse

A system of linear equations can be written as AX = B. If |A| ≠ 0, the unique solution is X = A^-1B.

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Student-friendly explanation

In the matrix method, the coefficient matrix A must match the fixed order of variables in X and the equation order used in B. Once the system is written as AX = B, the inverse method is valid only when A is square and |A| ≠ 0. Multiplying both sides on the left by A^-1 gives A^-1AX = A^-1B, so IX = A^-1B and therefore X = A^-1B. This left multiplication is important because matrix multiplication is not generally commutative; B A^-1 is either undefined or incorrect in most exam setups. If |A| = 0, the system must be tested for consistency instead of using the inverse formula for a unique answer.

How to write this in exams

  1. 1

    Start with the exact idea

    A system of linear equations can be written as AX = B. If |A| ≠ 0, the unique solution is X = A^-1B.

  2. 2

    Then show how to use it

    Arrange equations in a fixed variable order. Form A, X, and B. Compute |A|. If non-zero, find A^-1. Multiply A^-1B in the correct order. State the values of variables and, if time permits, substitute back to check.

  3. 3

    Add one concrete example

    The equations 2x + y = 5 and x - y = 1 give A = [[2,1],[1,-1]], X = [[x],[y]], B = [[5],[1]]. Since |A| = -3 ≠ 0, X = A^-1B gives x = 2 and y = 1.

  4. 4

    Avoid this incomplete answer

    Multiplying BA^-1 or changing the order of variables midway.

Definition

A system of linear equations can be written as AX = B. If |A| ≠ 0, the unique solution is X = A^-1B.

Example

The equations 2x + y = 5 and x - y = 1 give A = [[2,1],[1,-1]], X = [[x],[y]], B = [[5],[1]]. Since |A| = -3 ≠ 0, X = A^-1B gives x = 2 and y = 1.

Rule to remember

Write the system as AX = B. If |A| ≠ 0, then A^-1 exists and X = A^-1B. If |A| = 0, do not apply the inverse formula for a unique solution.

Memory hook

Coefficients times variables equals constants: AX = B.

Examples and method

Worked example

Solve 2x + y = 5 and x - y = 1. A = [[2,1],[1,-1]], B = [[5],[1]]. |A| = 2(-1) - 1(1) = -3. A^-1 = 1/(-3)[[-1,-1],[-1,2]] = [[1/3,1/3],[1/3,-2/3]]. X = A^-1B = [[1/3,1/3],[1/3,-2/3]][[5],[1]] = [[2],[1]]. Hence x = 2, y = 1.

Method to apply

Arrange equations in a fixed variable order. Form A, X, and B. Compute |A|. If non-zero, find A^-1. Multiply A^-1B in the correct order. State the values of variables and, if time permits, substitute back to check.

Diagram support

No diagram is required. A block layout showing A, X, and B can help students organize the system.

How CBSE asks it

Questions may ask students to solve two or three simultaneous linear equations using the matrix method, or to use a real-life case converted into linear equations.

Avoid common mistakes

Common confusion

Writing the constant matrix B incorrectly by changing the order of equations or constants.

Common wrong answer

Multiplying BA^-1 or changing the order of variables midway.

Exam tip

Always define A, X, and B clearly before applying X = A^-1B; this earns method credit and reduces arrangement errors.

Quick check

For AX = B, what is X if |A| ≠ 0?

If |A| ≠ 0, A inverse exists, so multiply AX = B on the left by A^-1 to get X = A^-1B.

Answer writing and exam use

1-mark answer

A system of linear equations can be written as AX = B. If |A| ≠ 0, the unique solution is X = A^-1B.

2-mark answer

A system of linear equations can be written as AX = B. If |A| ≠ 0, the unique solution is X = A^-1B. Write the system as AX = B. If |A| ≠ 0, then A^-1 exists and X = A^-1B. If |A| = 0, do not apply the inverse formula for a unique solution. The equations 2x + y = 5 and x - y = 1 give A = [[2,1],[1,-1]], X = [[x],[y]], B = [[5],[1]]. Since |A| = -3 ≠ 0, X = A^-1B gives x = 2 and y = 1.

3-mark answer

In the matrix method, the coefficient matrix A must match the fixed order of variables in X and the equation order used in B. Once the system is written as AX = B, the inverse method is valid only when A is square and |A| ≠ 0. Multiplying both sides on the left by A^-1 gives A^-1AX = A^-1B, so IX = A^-1B and therefore X = A^-1B. This left multiplication is important because matrix multiplication is not generally commutative; B A^-1 is either undefined or incorrect in most exam setups. If |A| = 0, the system must be tested for consistency instead of using the inverse formula for a unique answer. Write the system as AX = B. If |A| ≠ 0, then A^-1 exists and X = A^-1B. If |A| = 0, do not apply the inverse formula for a unique solution. Solve 2x + y = 5 and x - y = 1. A = [[2,1],[1,-1]], B = [[5],[1]]. |A| = 2(-1) - 1(1) = -3. A^-1 = 1/(-3)[[-1,-1],[-1,2]] = [[1/3,1/3],[1/3,-2/3]]. X = A^-1B = [[1/3,1/3],[1/3,-2/3]][[5],[1]] = [[2],[1]]. Hence x = 2, y = 1. Questions may ask students to solve two or three simultaneous linear equations using the matrix method, or to use a real-life case converted into linear equations. Multiplying BA^-1 or changing the order of variables midway.
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