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Invertible Matrix and Inverse by Elementary Operations

A square matrix A is invertible if there exists a matrix A^-1 such that AA^-1 = A^-1A = I. A square matrix is invertible only when its determinant is non-zero.

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Student-friendly explanation

The inverse can be found using elementary row or column operations. For row operations, write A = IA and apply the same row operations to A and I until the left side becomes I; the right side then becomes A^-1. Row and column operations should not be mixed in one method.

How to write this in exams

  1. 1

    Start with the exact idea

    A square matrix A is invertible if there exists a matrix A^-1 such that AA^-1 = A^-1A = I. A square matrix is invertible only when its determinant is non-zero.

  2. 2

    Then show how to use it

    Check that A is square. Compute determinant if practical. If determinant is zero, state inverse does not exist. If using row operations, form [A | I]. Use only row operations to convert A into I. Read the transformed right side as A^-1. Verify by multiplying AA^-1 if needed.

  3. 3

    Add one concrete example

    For A = [[1, 2], [3, 7]], determinant = 1 x 7 - 2 x 3 = 1, so A is invertible.

  4. 4

    Avoid this incomplete answer

    Dividing by determinant zero in the 2 x 2 inverse formula, which falsely creates an inverse for a singular matrix.

Definition

A square matrix A is invertible if there exists a matrix A^-1 such that AA^-1 = A^-1A = I. A square matrix is invertible only when its determinant is non-zero.

Example

For A = [[1, 2], [3, 7]], determinant = 1 x 7 - 2 x 3 = 1, so A is invertible.

Rule to remember

Invertibility condition: A must be square and |A| != 0. For elementary row operations, transform [A | I] into [I | A^-1]. For 2 x 2 matrix [[a, b], [c, d]], inverse exists when ad - bc != 0 and A^-1 = 1/(ad - bc) [[d, -b], [-c, a]].

Memory hook

Inverse turns A into identity; row operations must carry identity along.

Examples and method

Worked example

Find the inverse of A = [[1, 2], [3, 7]] by row operations. Write [A | I] = [[1, 2 | 1, 0], [3, 7 | 0, 1]]. Apply R2 → R2 - 3R1: [[1, 2 | 1, 0], [0, 1 | -3, 1]]. Apply R1 → R1 - 2R2: [[1, 0 | 7, -2], [0, 1 | -3, 1]]. Hence A^-1 = [[7, -2], [-3, 1]]. Check: determinant = 1, so the inverse is valid.

Method to apply

Check that A is square. Compute determinant if practical. If determinant is zero, state inverse does not exist. If using row operations, form [A | I]. Use only row operations to convert A into I. Read the transformed right side as A^-1. Verify by multiplying AA^-1 if needed.

Diagram support

No geometric diagram is required. The augmented matrix layout [A | I] is the useful visual form for row-operation questions.

How CBSE asks it

Asked as finding inverse using elementary row or column operations, checking invertibility, solving matrix equations, or using inverse properties in short and long answers.

Avoid common mistakes

Common confusion

A frequent error is mixing row and column operations while finding the inverse. This breaks the elementary operation method.

Common wrong answer

Dividing by determinant zero in the 2 x 2 inverse formula, which falsely creates an inverse for a singular matrix.

Exam tip

Start by checking determinant for a 2 x 2 or 3 x 3 square matrix. If determinant is zero, inverse does not exist and row-reduction work is unnecessary.

Quick check

Is A = [[2, 4], [1, 2]] invertible?

No. Its determinant is 2 x 2 - 4 x 1 = 0, so A is not invertible.

Answer writing and exam use

1-mark answer

A square matrix A is invertible if there exists a matrix A^-1 such that AA^-1 = A^-1A = I. A square matrix is invertible only when its determinant is non-zero.

2-mark answer

A square matrix A is invertible if there exists a matrix A^-1 such that AA^-1 = A^-1A = I. A square matrix is invertible only when its determinant is non-zero. Invertibility condition: A must be square and |A| != 0. For elementary row operations, transform [A | I] into [I | A^-1]. For 2 x 2 matrix [[a, b], [c, d]], inverse exists when ad - bc != 0 and A^-1 = 1/(ad - bc) [[d, -b], [-c, a]]. For A = [[1, 2], [3, 7]], determinant = 1 x 7 - 2 x 3 = 1, so A is invertible.

3-mark answer

The inverse can be found using elementary row or column operations. For row operations, write A = IA and apply the same row operations to A and I until the left side becomes I; the right side then becomes A^-1. Row and column operations should not be mixed in one method. Invertibility condition: A must be square and |A| != 0. For elementary row operations, transform [A | I] into [I | A^-1]. For 2 x 2 matrix [[a, b], [c, d]], inverse exists when ad - bc != 0 and A^-1 = 1/(ad - bc) [[d, -b], [-c, a]]. Find the inverse of A = [[1, 2], [3, 7]] by row operations. Write [A | I] = [[1, 2 | 1, 0], [3, 7 | 0, 1]]. Apply R2 → R2 - 3R1: [[1, 2 | 1, 0], [0, 1 | -3, 1]]. Apply R1 → R1 - 2R2: [[1, 0 | 7, -2], [0, 1 | -3, 1]]. Hence A^-1 = [[7, -2], [-3, 1]]. Check: determinant = 1, so the inverse is valid. Asked as finding inverse using elementary row or column operations, checking invertibility, solving matrix equations, or using inverse properties in short and long answers. Dividing by determinant zero in the 2 x 2 inverse formula, which falsely creates an inverse for a singular matrix.
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