Practice Maths

Determinants and Inverses

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Key Terms

Determinant (2×2)
det([[a,b],[c,d]]) = ad − bc.
Inverse formula (2×2)
A−1 = (1/det A) [[d, −b], [−c, a]]: swap the diagonal, negate the off-diagonal, divide by det.
Singular matrix
det(A) = 0 → A has no inverse; the system AX = B has no unique solution.
Non-singular (invertible)
det(A) ≠ 0 → A has a unique inverse.
Verification
AA−1 = A−1A = I (the identity matrix).
Product determinant
det(AB) = det(A) × det(B).

Determinants and Inverses — Key Facts

Determinant of a 2×2 matrix: det([[a,b],[c,d]]) = ad − bc

Inverse formula: A−1 = (1/det(A)) [[d,−b],[−c,a]]

Singular matrix: A matrix with det(A) = 0 has no inverse. It is called singular.

Invertible: A matrix with det(A) ≠ 0 is called invertible (or non-singular).

Verification: AA−1 = A−1A = I

Product determinant: det(AB) = det(A) × det(B)

Inverse of product: (AB)−1 = B−1A−1

Worked Example 1 — Finding A−1

Find A−1 for A = [[4, 1], [3, 2]].

Step 1: det(A) = 4(2) − 1(3) = 8 − 3 = 5

Step 2: Since det(A) = 5 ≠ 0, the inverse exists.

Step 3: Swap a and d, negate b and c, divide by det:

A−1 = (1/5) [[2, −1], [−3, 4]] = [[2/5, −1/5], [−3/5, 4/5]]

Step 4: Verify: AA−1 = [[4,1],[3,2]] × (1/5)[[2,−1],[−3,4]]

(1/5)[[4(2)+1(−3), 4(−1)+1(4)],[3(2)+2(−3), 3(−1)+2(4)]] = (1/5)[[5,0],[0,5]] = [[1,0],[0,1]] = I ✓

Worked Example 2 — Identifying a Singular Matrix

Show that B = [[2, 6], [1, 3]] is singular.

det(B) = 2(3) − 6(1) = 6 − 6 = 0

Since det(B) = 0, B is singular and has no inverse. Geometrically, this matrix maps all points in the plane onto a single line (it collapses the plane).

Hot Tip: When applying the inverse formula, students often forget to swap the a and d entries AND negate the b and c entries. A common error is to just divide each entry by the determinant, which gives the wrong answer. Always remember: swap the diagonal, negate the off-diagonal, divide by det.

What is the Determinant?

The determinant of a 2×2 matrix A = [[a,b],[c,d]] is the scalar value det(A) = ad − bc. This single number encodes enormous information about the matrix. Geometrically, |det(A)| represents the area scale factor: when A is applied as a linear transformation, it stretches all areas in the plane by a factor of |det(A)|. The sign indicates orientation: positive means orientation is preserved, negative means it is reversed (a reflection is involved).

For the 2×2 matrix, the formula det = ad − bc is easy to remember as the product of the main diagonal minus the product of the anti-diagonal. For larger matrices, the calculation is more complex, but the interpretation of det as a volume-scale factor remains valid.

The Inverse Matrix

The inverse of A, written A−1, satisfies AA−1 = A−1A = I. It is the matrix that “undoes” the transformation A. Just as dividing by a number a means multiplying by 1/a, solving the matrix equation AX = B involves multiplying by A−1: X = A−1B.

For a 2×2 matrix A = [[a,b],[c,d]] with det(A) ≠ 0, the inverse is:

A−1 = (1/(ad−bc)) [[d, −b], [−c, a]]

The process: swap the main diagonal entries (a and d exchange places), negate the off-diagonal entries (b becomes −b, c becomes −c), then divide the entire matrix by the determinant.

Singular Matrices

When det(A) = 0, the formula for A−1 involves division by zero — the inverse does not exist. Such a matrix is called singular. Geometrically, a singular matrix collapses the plane onto a lower-dimensional space: all points are mapped onto a single line (or even a single point). Once you crush space like this, you cannot undo it — hence no inverse exists.

A matrix is singular if and only if its rows (or columns) are linearly dependent — one row is a scalar multiple of the other. For example, [[2,6],[1,3]] is singular because row 1 = 2 × row 2.

The Multiplicative Property: det(AB) = det(A) det(B)

One of the most useful properties of the determinant is that it is multiplicative: det(AB) = det(A) × det(B). This tells us that if we compose two transformations, the combined area scale factor is the product of the individual scale factors. It also means that if either A or B is singular (determinant zero), then AB is also singular.

A direct consequence is the formula for the determinant of A−1: since AA−1 = I and det(I) = 1, we get det(A) × det(A−1) = 1, so det(A−1) = 1/det(A). The inverse “undoes” the scaling.

The Inverse of a Product

For invertible matrices A and B, the inverse of the product satisfies (AB)−1 = B−1A−1. The order reverses! This is the same reversal we saw for transposes: (AB)T = BTAT. The intuition is the same — to undo “first do A then do B,” you must “first undo B then undo A.”

This rule matters in exam questions where you are given det(A) and det(B) separately and asked for det(AB) or det(A−1B). Always use the multiplicative property rather than computing the full product matrix.

Solving AX = B Using the Inverse

The matrix equation AX = B represents a system of linear equations. If A is invertible, the unique solution is X = A−1B. Be careful: because matrix multiplication is not commutative, you must multiply on the left by A−1 to get X = A−1B, not XA−1 = BA−1. The order of multiplication must be preserved. If A is singular (det = 0), the system either has no solution or infinitely many solutions — the inverse method breaks down entirely.

Mastery Practice

  1. Fluency

    Q1 — Computing Determinants

    Find det(A) for each matrix: (a) A = [[5, 2], [3, 4]]    (b) A = [[−1, 3], [2, −6]]    (c) A = [[7, 0], [4, −2]]

  2. Fluency

    Q2 — Finding an Inverse

    Find A−1 for A = [[3, 1], [5, 2]]. Verify by computing AA−1.

  3. Fluency

    Q3 — Identifying Singular Matrices

    Determine which matrices are singular: (a) [[4, 2], [6, 3]]    (b) [[1, 0], [0, 5]]    (c) [[3, −6], [−1, 2]]

  4. Fluency

    Q4 — Determinant of a Product

    Given det(A) = 4 and det(B) = −3, find: (a) det(AB)    (b) det(A−1)    (c) det(A²)

  5. Understanding

    Q5 — Finding k for a Singular Matrix

    Find the value(s) of k for which the matrix [[k, 2], [3, k]] is singular.

  6. Understanding

    Q6 — Using A−1 to Solve a System

    Use the inverse matrix to solve the system: 2x + 3y = 8, x + 2y = 5.

  7. Understanding

    Q7 — Inverse of a Product

    Given A = [[2, 1], [1, 1]] and B = [[3, 0], [1, 2]], find (AB)−1 using two methods: (i) compute AB then invert; (ii) use (AB)−1 = B−1A−1.

  8. Understanding

    Q8 — Finding a Matrix from its Inverse

    The inverse of matrix M is M−1 = [[3, −1], [−5, 2]]. Find M.

  9. Problem Solving

    Q9 — Determining Solution Type Using the Determinant

    For each system, use det(A) to determine whether the system has a unique solution, no solution, or infinitely many solutions. Do not solve. (a) 3x − y = 2, 6x − 2y = 4    (b) 2x + y = 5, x − y = 1    (c) 4x + 6y = 3, 6x + 9y = 7

  10. Problem Solving

    Q10 — Proof: If det(A) ≠ 0, then AX = 0 has only the trivial solution

    Prove that if A is an invertible 2×2 matrix, then AX = 0 implies X = 0 (the zero vector).

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