Exam Practice

A matrix form where all non-zero rows are above zero rows, each leading entry is strictly to the right of the leading entry above it, and all entries below pivots are zero.

An REF matrix where every leading entry (pivot) is $1$, and each pivot is the only non-zero entry in its entire column. Every matrix has a **unique** RREF.

A linear system is consistent iff the rightmost augmented column has no pivot ($[0 \; \cdots \; 0 \mid b]$ with $b \ne 0$). It has a unique solution iff there are no free variables.

The set of all linear combinations $c_1\mathbf{v}_1 + \cdots + c_p\mathbf{v}_p$ of the vectors $\mathbf{v}_1, \dots, \mathbf{v}_p$.

The columns of an $m \times n$ matrix $A$ span $\mathbb{R}^m$ if and only if $A$ has a pivot position in every one of its $m$ rows.

The homogeneous equation $A\mathbf{x} = \mathbf{0}$ has a non-trivial solution if and only if the equation has at least one free variable.

Expressing the general solution of $A\mathbf{x} = \mathbf{b}$ as $\mathbf{x} = \mathbf{p} + t_1\mathbf{v}_1 + \cdots + t_k\mathbf{v}_k$, where $\mathbf{p}$ is a particular solution and $\mathbf{v}_i$ span the solution space of $A\mathbf{x} = \mathbf{0}$.

The columns of $A$ are linearly independent iff $A\mathbf{x} = \mathbf{0}$ has only the trivial solution $\mathbf{x} = \mathbf{0}$, which means $A$ has a pivot in every column.

For any linear transformation $T: \mathbb{R}^n \to \mathbb{R}^m$, $T(\mathbf{x}) = A\mathbf{x}$ where $A = [T(\mathbf{e}_1) \; T(\mathbf{e}_2) \; \cdots \; T(\mathbf{e}_n)]$.

$T$ is one-to-one iff $A$ has a pivot in every column (no free variables). $T$ is onto $\mathbb{R}^m$ iff $A$ has a pivot in every row (spans $\mathbb{R}^m$).

$(AB)^T = B^T A^T$. The factor order reverses when taking the transpose of a matrix product.

For $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$, provided $\det A = ad - bc \ne 0$.

Row reduce $[A \mid I_n]$. If $A$ is row equivalent to $I_n$, then $[A \mid I_n] \sim [I_n \mid A^{-1}]$.

For square $n \times n$ matrix $A$: $A$ is invertible $\iff n$ pivots $\iff A\mathbf{x} = \mathbf{0}$ has only trivial solution $\iff$ columns independent $\iff$ columns span $\mathbb{R}^n \iff T$ is one-to-one and onto.

Writing $A = LU$ where $L$ is unit lower-triangular ($1$s on diagonal) and $U$ is row echelon form. Solves $A\mathbf{x} = \mathbf{b}$ in two steps: $L\mathbf{y} = \mathbf{b}$ then $U\mathbf{x} = \mathbf{y}$.