Matrix Algebra Reference

A matrix is a rectangular array of numbers arranged in rows and columns.

An m × n matrix has m rows and n columns:

Entry aᵢⱼ = row i, column j (row index FIRST).   Example: a₂₃ = element in row 2, column 3.

Two matrices A and B are equal if and only if:

• They have the same size (same m and n), AND
• Every corresponding entry is equal: aᵢⱼ = bᵢⱼ for all i, j

Size rule: A is m×n and B is n×p → AB is m×p. Inner dimensions must match.

/* (m × n)(n × p) = (m × p) — inner n's cancel */

Entry (AB)ᵢⱼ = (row i of A) · (col j of B) = aᵢ₁b₁ⱼ + aᵢ₂b₂ⱼ + ··· + aᵢₙbₙⱼ

(AB)₁₁ = 1·5 + 2·7 = 19   (AB)₁₂ = 1·6 + 2·8 = 22
(AB)₂₁ = 3·5 + 4·7 = 43   (AB)₂₂ = 3·6 + 4·8 = 50

Row 1: [1·7+2·10   1·8+2·11   1·9+2·12] = [27   30   33]
Row 2: [3·7+4·10   3·8+4·11   3·9+4·12] = [61   68   75]
Row 3: [5·7+6·10   5·8+6·11   5·9+6·12] = [95   106   117]

Where: r₀ = position vector of a known point; d = direction vector; t ∈ ℝ is the free parameter.

Given two points P₁ = (x₁, y₁, z₁) and P₂ = (x₂, y₂, z₂):

d = P₂ − P₁ = (x₂−x₁,  y₂−y₁,  z₂−z₁)

/* Any scalar multiple of d is also a valid direction vector */
/* The parametric form is NOT unique — any point on the line */
/* can serve as r₀, and any non-zero multiple of d is valid  */

Step 1: Direction vector
d = P₂ − P₁ = (10−2, 3−(−1)) = (8, 4)
  (can simplify: d = (2, 1) by dividing by 4)

Step 2: Vector form  using P₁ as r₀
r = (2, −1) + t(2, 1)

Step 3: Parametric form
x = 2 + 2t
y = −1 + t

Step 4: General form
From parametric: t = y + 1, so x = 2 + 2(y+1) = 2y + 4
→  x − 2y = 4

Direction vector:
d = P₂ − P₁ = (−6−(−9),  −15−3,  4−4) = (3, −18, 0)
Simplify: d = (1, −6, 0)

Vector form  (using P₁):
r = (−9, 3, 4) + t(1, −6, 0)

Parametric form:
x = −9 + t
y =  3 − 6t
z =  4     ← z is constant (line is horizontal in z)

Step 1: Direction vectors
u = P₂ − P₁ = (6−5, 1−0, 0−0) = (1, 1, 0)
v = P₃ − P₁ = (3−5, 0−0, 1−0) = (−2, 0, 1)

Step 2: Normal vector  n = (a, b, c)  such that n·u=0 and n·v=0
n · u = 0:   a + b       = 0    → b = −a
n · v = 0:  −2a     + c = 0    → c = 2a

Set a=1:  n = (1, −1, 2)

Step 3: General form  using P₁=(5,0,0):
1(x−5) + (−1)(y−0) + 2(z−0) = 0
x − y + 2z = 5

Step 4: Vector / parametric form  using r₀=P₁, directions u and v:
r = (5,0,0) + s(1,1,0) + t(−2,0,1)

x = 5 +  s − 2t
y =     s
z =         t

A linear system is converted to an augmented matrix [A|b]:

The vertical bar separates the coefficient matrix A from the constants b.

A matrix is in Reduced Row Echelon Form if ALL four hold:

Leading variables (have pivot): x₁ (col 1), x₃ (col 3)
Free variable (no pivot): x₂ (col 2) → let x₂ = t

A vector v is a linear combination of vectors v₁, v₂, …, vₖ if there exist scalars c₁, c₂, …, cₖ such that:

v = c₁v₁ + c₂v₂ + ··· + cₖvₖ

Example: Is [5, 7] a linear combination of [1,2] and [3,1]?
Set up: c₁[1,2] + c₂[3,1] = [5,7]  →  c₁ + 3c₂ = 5  and  2c₁ + c₂ = 7

Consistent — so [5,7] = (16/5)[1,2] + (3/5)[3,1] ✓

Span{v₁, v₂, …, vₖ} = the set of ALL possible linear combinations of v₁, …, vₖ.

The span is always a subspace — it contains the origin, is closed under addition, and closed under scalar multiplication.

Form the matrix with the vectors as columns and row-reduce. The number of pivots = rank = dimension of the span.

When: Three linearly independent vectors in ℝ³ (no vector is in the span of the other two).

Adding a third vector that is already in the span of the first two does NOT increase the span. The rank stays the same.

A set spans all of ℝⁿ iff you can reach every vector in ℝⁿ as a linear combination. Test:

b = 2v₁ + v₂ = 2[1,0,1] + [1,1,2] = [3,1,4] ✓

A set of vectors {v₁, v₂, …, vₖ} is linearly independent if the only solution to:

3 vectors in ℝ² → immediately DEPENDENT (k > n quick rule)

Verify: Form A = [v₁ | v₂ | v₃] — already in RREF:

x₁ = −x₃,   x₂ = −x₃,   x₃ = t (free)

Non-trivial solution: c₁ = −1, c₂ = −1, c₃ = 1
Check: (−1)[1,0] + (−1)[0,1] + 1[1,1] = [0,0] ✓

If the set is dependent, the free variable solution tells you how to express one vector in terms of the others:

(a) 2A + B:

(b) A·B:

(1·0+2·2   1·(−1)+2·5) = (4   9)
(3·0+4·2   3·(−1)+4·5) = (8   17)

(c) B·A:

(0·1+(−1)·3   0·2+(−1)·4) = (−3   −4)
(2·1+5·3    2·2+5·4) = (17   24)

AB:  (3×4)(4×2) → inner dims match (4=4) → defined, result is 3×2
BA:  (4×2)(3×4) → inner: 2 ≠ 3 → UNDEFINED
AC:  (3×4)(3×2) → inner: 4 ≠ 3 → UNDEFINED
BC:  (4×2)(3×2) → inner: 2 ≠ 3 → UNDEFINED
AB+C:  AB is 3×2,  C is 3×2 → same size → defined, result is 3×2

Approach:  d = Q − P,  then r = P + td.

d = (4−1,  0−2,  −1−3) = (3, −2, −4)

Vector form:
r = (1, 2, 3) + t(3, −2, −4)

Parametric form:
x = 1 + 3t
y = 2 − 2t
z = 3 − 4t

Approach:  u = B−A,  v = C−A,  find n⊥u and n⊥v.

u = (0−1, 2−0, 0−0) = (−1, 2, 0)
v = (0−1, 0−0, 3−0) = (−1, 0, 3)

Find n=(a,b,c) s.t. n·u=0 and n·v=0:
n·u: −a + 2b     = 0  →  a = 2b
n·v: −a     + 3c = 0  →  a = 3c

Set b=3:  a=6, c=2.   n = (6, 3, 2)

General form using point A=(1,0,0):
6(x−1) + 3(y−0) + 2(z−0) = 0
6x + 3y + 2z = 6

Free variable: x₃ = t

Approach: Form A = [v₁ | v₂ | v₃] and row reduce.

Approach: Set up [v₁ | v₂ | v₃ | b] and row reduce.