13.3 Matrices & Vectors

13.3 Matrices and Vectors for the PAF GD Pilot Initial Test

Quick Answer: This section tests 2x2 matrix operations (addition, multiplication, determinant, inverse) and vector basics (magnitude, direction, addition, dot product, unit vector). The two facts that show up repeatedly: the determinant of [[a,b],[c,d]] is ad - bc, and the dot product of perpendicular vectors is zero.

Matrix Operations

A matrix is a rectangular array of numbers. For 2x2 matrices the exam uses four operations:

Addition and subtraction (same dimensions only): element-wise.

[[a,b],[c,d]] + [[e,f],[g,h]] = [[a+e, b+f],[c+g, d+h]]

Scalar multiplication: multiply every entry by the scalar k.

Matrix multiplication: for A (2x2) and B (2x2), the product AB is

AB = [[a,b],[c,d]] * [[e,f],[g,h]] = [[ae + bg, af + bh],[ce + dg, cf + dh]]

Note that AB is generally not equal to BA: matrix multiplication is not commutative. It is, however, associative: A(BC) = (AB)C.

Determinant of a 2x2 matrix A = [[a,b],[c,d]]:

det(A) = |A| = ad - bc

A matrix is invertible if and only if its determinant is non-zero.

Inverse of a 2x2 matrix:

A^(-1) = (1/det(A)) * [[d, -b],[-c, a]]

The recipe: swap the main diagonal entries (a and d), negate the off-diagonal entries (b and c), then divide by the determinant.

Worked example. Find the inverse of A = [[2, 1],[5, 3]].

1. det(A) = 23 - 15 = 6 - 5 = 1.
2. Swap diagonal and negate off-diagonal: [[3, -1],[-5, 2]].
3. Divide by det = 1: A^(-1) = [[3, -1],[-5, 2]].
4. Check: A * A^(-1) = [[23 + 1(-5), 2*(-1) + 12],[53 + 3*(-5), 5*(-1) + 3*2]] = [[1, 0],[0, 1]] = I. Correct.

Vectors

A vector has both magnitude and direction. In 2D, write v = (x, y) or v = xi + yj where i, j are unit vectors along the axes.

• Magnitude: |v| = sqrt(x^2 + y^2).
• Direction: the angle with the positive x-axis is theta = tan^(-1)(y/x).
• Addition: (x1, y1) + (x2, y2) = (x1 + x2, y1 + y2) (component-wise, parallelogram rule).
• Scalar multiplication: k*(x, y) = (kx, ky).
• Unit vector in the direction of v: u = v / |v| = (x/|v|, y/|v|). It has magnitude 1.

Dot product of a = (a1, a2) and b = (b1, b2):

a . b = a1b1 + a2b2 = |a||b|cos(theta)

where theta is the angle between them. Key special cases:

• If a . b = 0, the vectors are perpendicular (theta = 90 degrees).
• If a . b > 0, the angle is acute; if a . b < 0, the angle is obtuse.
• a . a = |a|^2.

Worked example. Find the magnitude of v = (3, 4) and the unit vector in its direction.

1. |v| = sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5.
2. Unit vector u = (3/5, 4/5). Check: (3/5)^2 + (4/5)^2 = 9/25 + 16/25 = 1, so |u| = 1.

Matrix Rules Table

Vectors Quick Reference

Vectors and matrices appear in navigation, force resolution, and coordinate-transform problems—all directly relevant to a pilot's working knowledge, which is why the PAF includes them on the Initial Test.

Dot-Product Angle Worked Example

Find the angle between a = (1, 2) and b = (3, -1).

1. Dot product: a . b = 13 + 2(-1) = 3 - 2 = 1.
2. Magnitudes: |a| = sqrt(1^2 + 2^2) = sqrt(5); |b| = sqrt(3^2 + (-1)^2) = sqrt(10).
3. cos(theta) = (a . b) / (|a|*|b|) = 1 / (sqrt(5)sqrt(10)) = 1 / sqrt(50) = 1 / (5sqrt(2)) approx 0.1414.
4. theta = cos^(-1)(0.1414) approx 81.87 degrees.

Because the dot product is positive (1 > 0), the angle is acute — a quick sanity check before computing. Had a . b been zero, the vectors would be perpendicular; had it been negative, the angle would be obtuse.

Solving a 2x2 Linear System via the Inverse

To solve Ax = b, multiply both sides by A^(-1): x = A^(-1)b. For example, solve [[2, 1],[5, 3]] * [x, y] = [4, 9]. Using the inverse [[3, -1],[-5, 2]] found earlier: [x, y] = [[3, -1],[-5, 2]] * [4, 9] = [34 + (-1)9, (-5)4 + 29] = [12 - 9, -20 + 18] = [3, -2]. Verify: 23 + 1(-2) = 4 and 53 + 3(-2) = 9. This method fails when det(A) = 0 (no unique solution), which is why the determinant is checked first.

Common Traps

• Singular matrix: if det(A) = 0, the inverse does not exist and the system has either no solution or infinitely many. Do not write A^(-1) = (1/0)*... — stop and report the matrix as singular.
• Non-commutativity: AB is not equal to BA in general. When a question asks for AB, multiply in the order given; reversing the order changes the result. Only AI = IA = A and A*A^(-1) = A^(-1)*A = I are guaranteed to commute.
• Unit vector magnitude: v/|v| has magnitude 1 by construction, but forgetting to divide both components (only dividing one) leaves a non-unit vector. Always divide every component by |v|.