1. RREF & Linear Systems (Q1)
| Concept | Details |
|---|---|
| RREF | REF + leading 1s + zeros above pivots too |
| Pivot / Free | Pivot col = leading 1 (forced). Free col = no leading 1 (choose s,t,…). #free = #vars − #pivots |
| ⚠️ | RREF ≠ identity when free vars exist — leading 1s don’t need diagonal |
| ⚠️ | Solve for PIVOT vars, not free vars |
| 1. xₚ | Set free vars = 0, read RHS at pivot rows (0 at free positions) |
| 2. Dir. vectors | Per free var: pivot-row entries, flip sign → pivot slots. That free = 1, others = 0 |
| 3. Combine | x = xₚ + sv₁ + tv₂ + … |
| Result | 0 free = Unique · 1 free = Line · 2 free = Plane · Zero row but RHS ≠ 0 → no solution |
2. Determinants & Invertibility (Q2)
Method Choice by Size
- 2×2: $\det = ad - bc$ (minus, not plus!)
- 3×3: cofactor expansion — pick row/col with most zeros (sign: checkerboard +-+-)
- 4×4: row reduce to triangular (track sign flips from swaps, scalar pulls from scaling)
Determinant Properties
| Formula | Result |
|---|---|
| $\det(kA)$ | $k^n \det(A)$ — NOT $k\det(A)$! Each of $n$ rows gets $\times k$ |
| $\det(AB)$ | $\det(A)\det(B)$ |
| $\det(A^{-1})$ | $1/\det(A)$ |
| $\det(A^n)$ | $\det(A)^n$ |
| $\det(A^T)$ | $\det(A)$ |
| $\det(-M)$ | $(-1)^n\det(M)$ — even $n$ (4×4): sign stays! |
| $\det(A^T A^{-1})$ | $= 1$ always (for invertible $A$) |
| ⚠️ Triangular | det = diagonal product. Only triangular! |
- Simplify first: $M^2 M^{-1} = M$, then $\det = \det(M)$
- ⚠️ det is NOT linear: $\det(A+B) \neq \det(A) + \det(B)$. Try $A=B=I$: $\det(2I)=4 \neq 2$
- ⚠️ $\det(A)=0,\ \det(B)=0 \not\Rightarrow \det(A+B)=0$. Try $A=[[1,0],[0,0]],\ B=[[0,0],[0,1]] \Rightarrow A+B=I$
2×2 Inverse
$A = [[a,b],[c,d]] \Rightarrow A^{-1} = \frac{1}{ad-bc} \cdot [[d,-b],[-c,a]]$ — swap diagonal, negate off-diagonal
Invertibility — Package Deal (flip one, all flip)
| ✅ Invertible | ❌ Not invertible |
|---|---|
| $\det(A) \neq 0$ | $\det(A) = 0$ (regardless of zero entries) |
| RREF $= I_n$ | RREF has zero row |
| All $n$ columns have pivots | A column has no pivot |
| No free variables | Free variables exist |
| $A\mathbf{x}=\mathbf{0}$ only $\mathbf{x}=\mathbf{0}$ | $A\mathbf{x}=\mathbf{0}$ has other/inf many solutions |
| Columns linearly independent | Columns dependent |
Linear Independence Test
Put vectors as columns, row reduce, count pivots.
- Every column has pivot → independent
- Any column without pivot → dependent (that vector = combo of pivot columns)
- Contains 0 = (0,…,0) → always dependent
- ⚠️ Pairwise non-scalar ≠ independent. Always use pivot-count, never pairwise check.
3. Definitions (Vector Spaces)
| Term | Meaning |
|---|---|
| Subspace | Flat through origin, closed under + and scalar × |
| Span | All combinations $c_1 v_1 + \cdots + c_k v_k$ |
| Basis | Independent set that spans $V$ (no redundant vectors) |
| Dimension | Number of vectors in a basis |
4. Subspace Proof — Worked Example: $V = {(x,y,z) : 2x - y + z = 0}$
Homogeneous $(= 0)$: always a subspace. Non-homogeneous $(\neq 0)$: never ($\mathbf{0}$ fails). For 2+ defining equations: check ALL equations in each closure step.
(i) Zero vector: $2(0) - 0 + 0 = 0\ \checkmark$. So $\mathbf{0} \in V$.
(ii) Closed under $+$: Let $\mathbf{u}, \mathbf{v} \in V$. Then $2u_1 - u_2 + u_3 = 0\ \ldots(1), \quad 2v_1 - v_2 + v_3 = 0\ \ldots(2)$. Consider $\mathbf{u}+\mathbf{v}$: $$2(u_1+v_1) - (u_2+v_2) + (u_3+v_3)$$ $$= (2u_1 - u_2 + u_3) + (2v_1 - v_2 + v_3) \quad [\text{regroup}]$$ $$= 0 + 0 \quad [\text{by }(1),(2)]$$ $$= 0\ \checkmark \quad \therefore\ \mathbf{u}+\mathbf{v} \in V.$$
(iii) Closed under scalar $\times$: Let $\mathbf{u} \in V$, $c \in \mathbb{R}$. Then $2u_1 - u_2 + u_3 = 0\ \ldots(3)$. Consider $c\mathbf{u}$: $$2(cu_1) - (cu_2) + (cu_3)$$ $$= c(2u_1 - u_2 + u_3) \quad [\text{factor } c]$$ $$= c \cdot 0 \quad [\text{by }(3)]$$ $$= 0\ \checkmark \quad \therefore\ c\mathbf{u} \in V.$$
$\therefore V$ is a subspace of $\mathbb{R}^3$.
5. Basis from Equation
🎯 Variables not in the equation are automatically FREE. dim(V) = n − (#independent equations). From RREF → pivot/free already decided. From raw equation → isolate any variable with nonzero coefficient.
Example ($\mathbb{R}^4$, 1 eq — $w$ is free since not in equation): $V = {(x,y,z,w) \in \mathbb{R}^4 : x - 2y + z = 0}$. Solve for $x$: $x = 2y - z$. Let $y = s,\ z = t,\ w = r$ (3 free → $\dim = 3$).
$$(x,y,z,w) = (2s - t,\ s,\ t,\ r) = s(2,1,0,0) + t(-1,0,1,0) + r(0,0,0,1)$$
Basis $= {(2,1,0,0),\ (-1,0,1,0),\ (0,0,0,1)},\ \dim(V) = 3$
6. Span Test — Is $\mathbf{w} \in \text{span}{\mathbf{v}_1, \mathbf{v}_2}$?
Stack as augmented $[\mathbf{v}_1 \mid \mathbf{v}_2 \mid \mathbf{w}]$, row reduce. Ex: $\mathbf{v}_1=(1,3,2),\ \mathbf{v}_2=(2,-1,4),\ \mathbf{w}=(2,1,5)$:
| $\mathbf{v}_1$ | $\mathbf{v}_2$ | $\mathbf{w}$ | After row reduce | $\mathbf{v}_1$ | $\mathbf{v}_2$ | $\mathbf{w}$ | |
|---|---|---|---|---|---|---|---|
| 1 | 2 | 2 | → | 1 | 2 | 2 | |
| 3 | -1 | 1 | 0 | -7 | -5 | ||
| 2 | 4 | 5 | 0 | 0 | 1 |
Row $[0\ 0\ |\ 1]$ = contradiction → $\mathbf{w} \notin \text{span}$. No contradiction → $\mathbf{w} \in \text{span}$.
7. Rank-Nullity
$\text{rank}(A) + \text{nullity}(A) = n$ where $n$ = number of columns of $A$. Rank = #pivot cols, nullity = #free cols.
Null space basis: same as §1 vector form but RHS = 0. Coefficient matrix from equations: each equation = 1 row, each variable = 1 column.
⚠️ Verify: plug each basis vector back into $A\mathbf{x} = \mathbf{0}$. Every row must $= 0$. If not, you flipped a sign.