Linear Algebra and Matrix Operations

Linear algebra is the toolbox for simultaneous equations, which appear everywhere on the FE: truss joint equilibrium, mesh-current circuit analysis, and force resolution. Vectors (dot and cross products) drive statics moments and work calculations. The Handbook gives the formulas; you supply speed and accuracy.

Matrix Operations

An m×n matrix has m rows, n columns. Add/subtract element-by-element (same size required). Matrix multiplication of A (m×n) by B (n×p) gives C (m×p) with Cᵢⱼ = Σₖ Aᵢₖ Bₖⱼ — the inner dimensions must match.

Key rule: Matrix multiplication is associative, (AB)C = A(BC), but not commutative: AB ≠ BA in general. Order matters.

Determinants

2×2: det[a b; c d] = ad − bc.

3×3 (cofactor expansion along row 1): det(A) = a₁₁(a₂₂a₃₃ − a₂₃a₃₂) − a₁₂(a₂₁a₃₃ − a₂₃a₃₁) + a₁₃(a₂₁a₃₂ − a₂₂a₃₁). Watch the alternating + − + sign pattern.

Properties: det(AB) = det(A)det(B); det(Aᵀ) = det(A); swapping two rows flips the sign; identical rows give det = 0; det(A) = 0 ⟺ A is singular (non-invertible).

Worked example — 2×2 determinant. det[3 2; 1 4] = (3)(4) − (2)(1) = 12 − 2 = 10. The trap is computing the off-diagonal product as additive; it is subtracted.

Worked example — 3×3 determinant. For A = [1 2 3; 0 4 5; 1 0 6], expand along row 1: 1·(4·6 − 5·0) − 2·(0·6 − 5·1) + 3·(0·0 − 4·1) = 1·24 − 2·(−5) + 3·(−4) = 24 + 10 − 12 = 22. Because det ≠ 0, this matrix is non-singular and any system Ax = b built from it has a unique solution. The most common arithmetic slip is the middle sign: the cofactor of a₁₂ carries a leading minus.

Solving Systems Ax = b

Cramer's Rule

For det(A) ≠ 0, xᵢ = det(Aᵢ)/det(A), where Aᵢ is A with its i-th column replaced by b. Best for 2×2 and 3×3 systems on the FE.

Worked example — Cramer's Rule. Solve 2x + 3y = 8 and x − y = 1. Here A = [2 3; 1 −1], b = [8; 1]. det(A) = 2(−1) − 3(1) = −5. For x, replace column 1: det[8 3; 1 −1] = 8(−1) − 3(1) = −11, so x = −11/−5 = 11/5 = 2.2. For y, replace column 2: det[2 8; 1 1] = 2(1) − 8(1) = −6, so y = −6/−5 = 6/5 = 1.2. Check: 2(2.2) + 3(1.2) = 4.4 + 3.6 = 8 ✓.

Matrix Inverse

For a 2×2 matrix, A⁻¹ = (1/(ad−bc))·[d −b; −c a], existing only when det(A) ≠ 0. Then x = A⁻¹b. Properties: AA⁻¹ = I; (AB)⁻¹ = B⁻¹A⁻¹ (the order reverses); (Aᵀ)⁻¹ = (A⁻¹)ᵀ.

Singular Systems

If det(A) = 0, the system has either no solution (inconsistent) or infinitely many (dependent) — never a unique one. Recognizing this from a zero determinant is a frequent FE question.

Eigenvalues and Eigenvectors

For a square matrix A, the eigenvalue λ and eigenvector v satisfy Av = λv. Find eigenvalues from the characteristic equation det(A − λI) = 0.

Worked example — eigenvalues. For A = [4 1; 2 3], det(A − λI) = (4−λ)(3−λ) − (1)(2) = λ² − 7λ + 12 − 2 = λ² − 7λ + 10 = (λ−5)(λ−2) = 0, so λ = 5 and λ = 2. Eigenvalues appear as natural frequencies of vibrating systems, as stability indicators in controls (negative real parts ⇒ stable), and as principal stresses (eigenvalues of the stress tensor).

Vectors — Dot and Cross Products

The FE Other Disciplines exam explicitly tests vector operations, which feed directly into statics (moments) and work calculations.

For vectors A = (Aₓ, Aᵧ, A_z) and B = (Bₓ, Bᵧ, B_z):

The angle between vectors comes from cos θ = (A·B)/(|A||B|). Two vectors are perpendicular when A·B = 0 and parallel when A×B = 0.

Worked example — dot product and angle. Let A = (1, 2, 2) and B = (2, 0, 1). Then A·B = (1)(2) + (2)(0) + (2)(1) = 4. The magnitudes are |A| = √(1+4+4) = 3 and |B| = √(4+0+1) = √5 ≈ 2.236. So cos θ = 4/(3·2.236) = 0.596, giving θ ≈ 53.4°.

Worked example — cross product. For A = (1, 0, 0) and B = (0, 1, 0), A×B = (0·0 − 0·1, 0·0 − 1·0, 1·1 − 0·0) = (0, 0, 1). The result is perpendicular to both inputs (the +z axis), illustrating the right-hand rule: x̂ × ŷ = ẑ. Reversing the order gives B×A = (0, 0, −1) — the cross product is anti-commutative, a detail the FE likes to test.

Engineering use — moment of a force. The moment of a force F about a point is M = r × F, where r is the position vector from the point to the force's line of action. The magnitude |M| = |r||F|sin θ is exactly the familiar "force times perpendicular distance," and the direction (by the right-hand rule) tells you the axis of rotation. Recognizing that statics moment problems ARE cross products lets you reuse one formula across both the Mathematics and Statics portions of the exam, and the dot product W = F·d similarly computes the work done by a constant force over a displacement d.