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