Difficulty: Intermediate | Prerequisites: Matrix multiplication, elementary row operations, 2x2 and 3x3 determinants
Matrix inverses sit at the heart of solving linear systems. If you can invert the coefficient matrix, you can solve Ax = b in one step. This topic covers the two main routes to finding an inverse – row reduction and the adjoint method – and how to tell when a matrix has no inverse at all. You should already be comfortable multiplying matrices and performing row operations before diving in.
The inverse of a square matrix A is the matrix A⁻¹ such that AA⁻¹ = A⁻¹A = I. You can find it by augmenting [A | I] and row-reducing until the left side becomes I, at which point the right side is A⁻¹. Alternatively, for any n×n matrix, compute A⁻¹ = (1/det(A)) · adj(A). A matrix is invertible only when its determinant is non-zero.
Inverse matrix (A⁻¹)
The unique matrix such that AA⁻¹ = A⁻¹A = I (the identity matrix). Think of it as the "undo" matrix: if A transforms a vector, A⁻¹ transforms it back.
Identity matrix (I)
A square matrix with 1s on the main diagonal and 0s everywhere else. In simple terms, it is the matrix equivalent of multiplying by 1.
Invertible (non-singular) matrix
A square matrix whose determinant is not zero, meaning an inverse exists. If det(A) = 0, the matrix is singular and has no inverse.
Row reduction (Gauss-Jordan elimination)
A sequence of elementary row operations (scaling a row, swapping rows, adding a multiple of one row to another) used to transform a matrix into reduced row echelon form. In simple terms, it is the systematic method of simplifying a matrix one row operation at a time.
Cofactor (Cᵢⱼ)
The signed minor of entry aᵢⱼ. You delete row i and column j, take the determinant of what remains, then multiply by (−1)^(i+j). Think of it as the "weight" each entry carries when computing the determinant.
Cofactor matrix
The matrix formed by replacing every entry of A with its cofactor.
Adjoint / adjugate (adj(A))
The transpose of the cofactor matrix. This is the matrix you multiply by 1/det(A) to get the inverse via the adjoint method.
Reduced row echelon form (RREF)
The final form after Gauss-Jordan elimination, where each leading entry is 1 and is the only non-zero entry in its column. If the left half of [A | I] reaches RREF as the identity, the right half is A⁻¹.
This is the workhorse method. It works for any size of square matrix and is typically the most efficient approach on exams.
Form the augmented matrix [A | I], where I is the identity matrix of the same size.
Apply elementary row operations to the left side, aiming to turn it into I.
Every operation you perform on the left is simultaneously performed on the right.
When the left side reaches I, the right side is A⁻¹.
If at any point you produce a row of all zeros on the left side, stop: the matrix is not invertible.
2×2 example (from 9.1.1a)
Start with A = [5, 0; 0, 3]. Augment: [5, 0 | 1, 0; 0, 3 | 0, 1].
R₁/5 → R₁ gives [1, 0 | 1/5, 0; 0, 3 | 0, 1]
R₂/3 → R₂ gives [1, 0 | 1/5, 0; 0, 1 | 0, 1/3]
The left side is I, so A⁻¹ = [1/5, 0; 0, 1/3].
2×2 example (from 9.1.1b)
A = [3, 2; −1, 1]. Augment and reduce:
R₁/3 → R₁ gives [1, 2/3 | 1/3, 0; −1, 1 | 0, 1]
R₂ + R₁ → R₂ gives [1, 2/3 | 1/3, 0; 0, 5/3 | 1/3, 1]
R₂ · (3/5) → R₂ gives [1, 2/3 | 1/3, 0; 0, 1 | 1/5, 3/5]
R₁ – (2/3)R₂ → R₁ gives [1, 0 | 1/5, −2/5; 0, 1 | 1/5, 3/5]
So A⁻¹ = (1/5)[1, −2; 1, 3].
When the matrix is not invertible (from 9.1.1c)
A = [1, 2; 2, 4]. After R₂ – 2R₁ → R₂, the augmented matrix becomes [1, 2 | 1, 0; 0, 0 | −2, 1]. The left side is in row echelon form but is not the identity. A row of zeros on the left means A is singular, so no inverse exists.
3×3 example (from 9.3.2a)
A = [1, 2, −1; 0, −1, 2; 0, 1, 0]. Augment [A | I] and row-reduce. After a sequence of operations the result is A⁻¹ = [1, 1/2, −3/2; 0, 0, 1; 0, 1/2, 1/2].
The steps are the same as for 2×2, just longer. Keep your arithmetic tidy and work one row operation at a time.
The formula is:
A⁻¹ = (1 / det(A)) · adj(A)
where adj(A) is the transpose of the cofactor matrix.
Steps:
Compute det(A). If det(A) = 0, stop: A is not invertible.
Build the cofactor matrix: for each entry aᵢⱼ, delete row i and column j, compute the determinant of the remaining (n−1)×(n−1) matrix, and multiply by (−1)^(i+j).
Transpose the cofactor matrix to get adj(A).
Multiply by 1/det(A).
3×3 example (from 9.3.1a)
A = [1, 2, 3; 0, 1, 2; 0, 0, 1]. det(A) = 1 (product of diagonal entries for a triangular matrix).
Cofactors: C₁₁ = 1, C₁₂ = 0, C₁₃ = 0, C₂₁ = −2, C₂₂ = 1, C₂₃ = 0, C₃₁ = 1, C₃₂ = −2, C₃₃ = 1.
adj(A) = [1, −2, 1; 0, 1, −2; 0, 0, 1] (transpose of the cofactor matrix).
A⁻¹ = (1/1) · adj(A) = [1, −2, 1; 0, 1, −2; 0, 0, 1].
3×3 example (from 9.3.1b)
A = [1, 2, −1; 0, 1, 0; −5, 2, 3]. det(A) = −2.
After computing all nine cofactors and transposing, adj(A) = [3, −8, 1; 0, −2, 0; 5, −12, 1].
A⁻¹ = (−1/2) · adj(A) = [−3/2, 4, −1/2; 0, 1, 0; −5/2, 6, −1/2].
2×2 shortcut formula
For A = [a, b; c, d], A⁻¹ = (1/(ad – bc)) · [d, –b; –c, a]. Swap the diagonal entries, negate the off-diagonal entries, divide by the determinant.
General formula
A⁻¹ = (1/det(A)) · adj(A)
Determinant of an inverse
det(A⁻¹) = 1/det(A). This is used in problem AP 1, where the inverse is found by row reduction and its determinant is computed to recover det(A).
Matrix inverses turn up whenever a system of linear equations needs to be solved repeatedly with different right-hand sides but the same coefficient matrix. In engineering, this happens in circuit analysis (solving for currents given different voltage sources) and in computer graphics (applying and reversing transformations to 3D objects).
Students often think any square matrix has an inverse. It does not. A matrix is invertible only if its determinant is non-zero. Always check det(A) first when using the adjoint method.
Students sometimes forget to transpose the cofactor matrix when computing the adjoint. The adjoint is the transpose of the cofactor matrix, not the cofactor matrix itself.
A common sign error: the cofactor Cᵢⱼ includes the sign factor (−1)^(i+j). Forgetting the sign or miscounting i+j is one of the most frequent mistakes on exams.
Students occasionally stop at row echelon form during row reduction instead of continuing to reduced row echelon form. You need the full identity on the left side, not just an upper triangular matrix.
⚠️ You will almost certainly be asked to find the inverse of a 2×2 or 3×3 matrix using row reduction. Be fluent in both methods.
⚠️ Expect a question where the matrix is not invertible. Recognise a zero row on the left side of the augmented matrix, or det(A) = 0, as the signal to stop and state that no inverse exists.
⚠️ The relationship det(A⁻¹) = 1/det(A) is a common short-answer or proof question.
True or false: Every square matrix has an inverse.
False. Only matrices with non-zero determinant are invertible.
True or false: The adjoint of A is the cofactor matrix.
False. The adjoint is the transpose of the cofactor matrix.
Fill in the blank: If det(A) = 5, then det(A⁻¹) = ___.
1/5.
True or false: If row reduction of [A | I] produces a row of zeros on the left side, A has no inverse.
True.
Fill in the blank: For a 2×2 matrix [a, b; c, d], the determinant used in the inverse formula is ___.
ad – bc.
Q: Find the inverse of A = [3, 2; −1, 1] using row reduction.
A: Augment [A | I] and row-reduce. A⁻¹ = (1/5)[1, −2; 1, 3] = [1/5, −2/5; 1/5, 3/5].
Q: Using the adjoint method, find A⁻¹ for A = [3, 0, 0; 0, −1, 0; 0, 0, 3].
A: det(A) = 3(−1)(3) = −9. Compute all cofactors, transpose, and divide by −9. A⁻¹ = (−1/9)[−3, 0, 0; 0, 9, 0; 0, 0, −3] = [1/3, 0, 0; 0, −1, 0; 0, 0, 1/3].
Q: Explain why the matrix [1, 2; 2, 4] is not invertible.
A: det(A) = (1)(4) – (2)(2) = 0. Since the determinant is zero, A is singular and has no inverse. Equivalently, row 2 is twice row 1, so the rows are linearly dependent.
Q: If A is invertible and det(A⁻¹) = 10/3, what is det(A)?
A: det(A) = 1/det(A⁻¹) = 3/10.
Q: State the formula for the inverse of a 2×2 matrix [a, b; c, d].
A: A⁻¹ = (1/(ad – bc)) · [d, –b; –c, a], provided ad – bc ≠ 0.
This connects to determinants because an inverse exists only when det(A) ≠ 0, and the adjoint method uses the determinant directly in the formula.
This connects to Cramer's rule because Cramer's rule is essentially using the inverse (via determinants) to solve Ax = b one variable at a time.
This connects to linear systems more broadly: if A is invertible, the unique solution to Ax = b is x = A⁻¹b.
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