6.1 Matrices & Determinants

Key Takeaways

  • An m×n matrix has m rows and n columns; addition needs equal order, while AB needs columns of A equal to rows of B.
  • det([a b; c d]) = ad − bc; a 2×2 matrix is invertible iff its determinant is nonzero.
  • For 3×3 matrices, expand along a row/column (cofactor expansion) or use the rule of Sarrus for a quick numeric check.
  • A⁻¹ = (1/det A) · adj A; for 2×2, swap diagonals and negate off-diagonals, then divide by det A.
  • Cramer’s rule solves Ax = b via xᵢ = det(Aᵢ)/det(A) when det(A) ≠ 0 and is efficient for small systems.
Last updated: July 2026

Why Matrices Matter in Technical Mathematics

Matrices and determinants are foundational pre-engineering mathematics. This review rewards fluent recall of order rules, determinant shortcuts, and whether an inverse exists before moving to longer elimination methods. It does not assert that PAF publishes a mathematics paper, count, timing, or topic allocation for this route.

Think of a matrix as a rectangular array of numbers used to store linear data compactly: coefficients of simultaneous equations, transformation rules, or tabulated engineering quantities. Useful exercises include addition, multiplication, determinants, invertibility checks, and Cramer’s rule for small systems.

Order, Types, and Notation

An m × n matrix has m rows and n columns. Element aᵢⱼ sits in row i, column j (1-based indexing in FSc texts).

TypeConditionRecognition cue
Row matrix1 × nSingle row of data
Column matrixm × 1Single column (often a solution vector)
Square matrixm = nOnly square matrices have determinants/inverses in the usual school sense
DiagonalSquare; aᵢⱼ = 0 for i ≠ jOff-diagonals zero
Identity Iₙ1s on diagonal, 0 elsewhereAI = IA = A
Zero / null OAll entries 0A + O = A
SymmetricAᵀ = AMirror across main diagonal
Skew-symmetricAᵀ = −ADiagonal must be zero

Transpose: (Aᵀ)ᵢⱼ = Aⱼᵢ. Useful identities: (Aᵀ)ᵀ = A, (A + B)ᵀ = Aᵀ + Bᵀ, (AB)ᵀ = BᵀAᵀ (order reverses).

Matrix Operations

Addition / subtraction. Same order required. Add or subtract corresponding entries. Commutative: A + B = B + A.

Scalar multiplication. kA multiplies every entry by k.

Multiplication. If A is m × n and B is n × p, then AB is m × p with

(AB)ᵢⱼ = Σₖ aᵢₖ bₖⱼ.

Multiplication is associative and distributive, but generally not commutative: AB may exist while BA does not, or both exist yet AB ≠ BA.

Worked — multiplication.

Let A = [[1, 2], [0, −1]] (2×2) and B = [[3, 0], [1, 4]] (2×2).

AB = [[1·3+2·1, 1·0+2·4], [0·3+(−1)·1, 0·0+(−1)·4]] = [[5, 8], [−1, −4]].

BA = [[3·1+0·0, 3·2+0·(−1)], [1·1+4·0, 1·2+4·(−1)]] = [[3, 6], [1, −2]].

So AB ≠ BA—classic MCQ trap if you assume commutativity.

Worked — order check. A is 2×3, B is 2×3 → A + B is fine, but AB is undefined (3 ≠ 2). BA would need B’s columns (3) equal to A’s rows (2)—also undefined. Always check inner dimensions first.

Determinants: 2×2 and 3×3

Only square matrices have a determinant (a scalar).

2×2 formula. For A = [[a, b], [c, d]],

det(A) = |A| = ad − bc.

Worked. |[[4, −2], [3, 5]]| = 4·5 − (−2)·3 = 20 + 6 = 26.

3×3 cofactor expansion along row 1:

For A = [[a₁₁, a₁₂, a₁₃], [a₂₁, a₂₂, a₂₃], [a₃₁, a₃₂, a₃₃]],

|A| = a₁₁(a₂₂a₃₃ − a₂₃a₃₂) − a₁₂(a₂₁a₃₃ − a₂₃a₃₁) + a₁₃(a₂₁a₃₂ − a₂₂a₃₁).

(Signs along row 1 are +, −, +.)

Worked — 3×3.

A = [[2, 1, 0], [−1, 3, 4], [0, 2, 1]].

|A| = 2(3·1 − 4·2) − 1((−1)·1 − 4·0) + 0(...) = 2(3 − 8) − 1(−1 − 0) + 0 = 2(−5) − (−1) = −10 + 1 = −9.

Properties that save time on MCQs:

PropertyConsequence
Swap two rows/columnsDeterminant changes sign
Two identical rows/columnsdet = 0
Factor k from one row/columndet multiplies by k
Row replaced by row + k·(another row)det unchanged
AB
Aᵀ
kA

If det(A) = 0, A is singular (not invertible). If det(A) ≠ 0, A is nonsingular (invertible).

Inverse of a Matrix (Introduction)

For square A, the inverse A⁻¹ satisfies A A⁻¹ = A⁻¹ A = I.

General formula: A⁻¹ = (1/|A|) · adj(A), where adj(A) is the adjoint (transpose of the cofactor matrix), provided |A| ≠ 0.

2×2 quick rule. If A = [[a, b], [c, d]] and Δ = ad − bc ≠ 0,

A⁻¹ = (1/Δ) [[d, −b], [−c, a]].

Swap the main-diagonal entries, negate the off-diagonal entries, divide by the determinant.

Worked — inverse.

A = [[3, 1], [2, 4]]. Δ = 12 − 2 = 10.

A⁻¹ = (1/10) [[4, −1], [−2, 3]] = [[0.4, −0.1], [−0.2, 0.3]].

Check: A A⁻¹ = I₂ (spot-check first row: 3·0.4 + 1·(−0.2) = 1.2 − 0.2 = 1; 3·(−0.1) + 1·0.3 = 0).

Common trap: If Δ = 0, stop—there is no inverse. Do not “force” the 2×2 swap formula.

Cramer’s Rule (Idea and 2×2 Practice)

For a linear system A x = b with square coefficient matrix A and |A| ≠ 0, Cramer’s rule says each unknown is a ratio of determinants:

xᵢ = |Aᵢ| / |A|,

where Aᵢ is A with column i replaced by the constant vector b.

Worked — 2×2 system.

Solve:

2x + y = 5
3x − y = 4

A = [[2, 1], [3, −1]], b = [[5], [4]].
|A| = −2 − 3 = −5.

A₁ (replace col 1): [[5, 1], [4, −1]], |A₁| = −5 − 4 = −9 → x = (−9)/(−5) = 9/5.
A₂ (replace col 2): [[2, 5], [3, 4]], |A₂| = 8 − 15 = −7 → y = (−7)/(−5) = 7/5.

Check: 2(9/5) + 7/5 = 18/5 + 7/5 = 25/5 = 5 ✓.

For 3×3 systems the same idea applies, but arithmetic grows. Build fluency with 2×2 Cramer exercises and single 3×3 determinants before attempting full three-unknown systems.

Efficient Practice Strategy

  1. Read the order before multiplying—undefined products are free marks if you catch them.
  2. Compute 2×2 dets in one line (ad − bc); expand 3×3 along a row with zeros when possible.
  3. Invertibility test is just “is det zero?”—do not build the full inverse unless asked.
  4. Cramer for 2×2: three tiny determinants beat substitution when options are fractions.
  5. Watch AB vs BA and |kA| = kⁿ|A| traps.

Mastering these mechanics builds a strong base for later linear-algebra ideas used in engineering coursework.

Test Your Knowledge

If A is a 2×3 matrix and B is a 3×2 matrix, which statement is correct?

A
B
C
D
Test Your Knowledge

What is the determinant of [[5, −3], [2, 4]]?

A
B
C
D
Test Your Knowledge

A square matrix A has det(A) = 0. Which conclusion follows?

A
B
C
D
Test Your Knowledge

Using Cramer’s rule on 3x + y = 7 and x − 2y = 0, the value of y is:

A
B
C
D