TCX2101 | CT3 Helpsheet (A4 Double-Sided)

Sun, 05 Apr 2026 00:00:00 +0800

RREF & Linear Systems (Q1)

RREF Rules (3 conditions)

Each leading entry is 1
Each leading 1 is the only nonzero in its column (zeros above AND below)
Leading 1s move right as you go down

REF = zeros below only. RREF = zeros above AND below.

Pivot vs Free Variables

Scan each column of RREF:

Has leading 1 → pivot variable (equation decides)
No leading 1 → free variable (you choose = $s, t, \ldots$)

$\text{#free vars} = \text{#variables} - \text{#pivots}$

Vector Form from RREF

$\mathbf{x}_p$: Set all free vars = 0, read the RHS column (insert 0 at free var positions).

Direction vectors: For each free variable column:

Read the column entries at each pivot row
Move to other side of $=$ (flip sign) → fills pivot var slots
Free var itself → 1
Other free vars → 0

$$\mathbf{x} = \mathbf{x}_p + s\mathbf{v}_1 + t\mathbf{v}_2$$

Free vars	Solution type
0	Unique ($\mathbf{x}_p$ only, RREF left = $I$)
1	Line ($\mathbf{x}_p + s\mathbf{v}$)
2	Plane ($\mathbf{x}_p + s\mathbf{v}_1 + t\mathbf{v}_2$)

Inconsistent: Row $[0\ 0\ \cdots\ 0\ |\ b]$ with $b \neq 0$ → no solution.

Determinants & Properties (Q2)

Cofactor Expansion

Pick the row/column with most zeros. Sign pattern (checkerboard):

$$C_{ij} = (-1)^{i+j} M_{ij}$$

2×2: $\det = ad - bc$ (minus, not plus!)

4×4 → 3×3 → 2×2 (same method, just nesting)

Determinant Properties

Formula	Result
$\det(kA)$	$= k^n \cdot \det(A)$ where $n$ = matrix size
$\det(AB)$	$= \det(A) \cdot \det(B)$
$\det(A^{-1})$	$= 1/\det(A)$
$\det(A^n)$	$= \det(A)^n$
$\det(A^T)$	$= \det(A)$
$\det(I)$	$= 1$

$\det(kA) = k^n \det(A)$, NOT $k \cdot \det(A)$. Each of $n$ rows gets multiplied by $k$.

Invertibility

$A$ is invertible $\iff \det(A) \neq 0$

All equivalent:

$\det(A) \neq 0$
RREF of $A = I_n$
$A$ has $n$ pivots
$A\mathbf{x} = \mathbf{b}$ has unique solution for all $\mathbf{b}$
$A\mathbf{x} = \mathbf{0}$ has only $\mathbf{x} = \mathbf{0}$
Columns of $A$ are linearly independent
Columns of $A$ span $\mathbb{R}^n$

Zero entries in the matrix does NOT mean not invertible. Only $\det = 0$ matters.

Span, Subspace, Basis (Q3)

Subspace Proof Template (3 conditions)

Given $V = {\ \mathbf{x} \in \mathbb{R}^n : [\text{equation} = 0]\ }$

Zero vector: Plug in $\mathbf{0}$ → does equation hold?
Closed under $+$: Let $\mathbf{u}, \mathbf{v} \in V$. Show $\mathbf{u}+\mathbf{v}$ satisfies equation.
Closed under scalar $\times$: Let $\mathbf{u} \in V$, $c \in \mathbb{R}$. Show $c\mathbf{u}$ satisfies equation.

Homogeneous $(= 0)$: always works (sum of zeros = zero, scalar $\times$ zero = zero). Non-homogeneous $(= 5, 1, \ldots)$: NEVER a subspace ($\mathbf{0}$ fails).

Span Test

Is $\mathbf{w}$ in $\text{span}{\mathbf{v}_1, \mathbf{v}_2, \ldots}$?

Set up $[\mathbf{v}_1\ |\ \mathbf{v}_2\ |\ \cdots\ |\ \mathbf{w}]$ and row reduce.

Consistent → yes (read coefficients)
Contradiction row → no

Basis from Equation

Solve equation for one variable
Remaining variables = free parameters ($s, t, \ldots$)
Write as linear combination → coefficient vectors = basis

Dimension

$\dim(V) = $ number of vectors in any basis

$\dim(\mathbb{R}^n) = n$
$\dim({\mathbf{0}}) = 0$ (basis = empty set $\emptyset$)
$\dim(\text{Nul}(A)) = $ #free vars (Rank-Nullity)
More vectors than $\dim$ → dependent
Equal count + independent → basis
Fewer + independent → can extend to basis

Linear Independence Check

Put vectors as columns, row reduce, count pivots.

All columns have pivots → independent
A column without pivot → dependent

Building as I prep — last updated: 2026-04-05

Linear-Algebra on ENKR's Blog | Jing Hui PANG