Mathematics for Data Science II (BSMA1003) formula sheet

$\text{vector} = \text{a list of numbers (row or column)}$

$c \cdot \begin{pmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{pmatrix} = \begin{pmatrix} c \cdot v_1 \\ c \cdot v_2 \\ \vdots \\ c \cdot v_n \end{pmatrix}$

$A_{ij}$

$(AB)_{ij} = \sum\limits_{k=1}^{n} A_{ik} B_{kj}$

$a_1 x_1 + a_2 x_2 + \cdots + a_n x_n = b$

$\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc$

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

$\det(A) = \sum_{j=1}^{n} a_{1,j} C_{1,j}$

$M_{ij}$

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

$\det(A^n) = (\det A)^n$

$\det(A^{-1}) = \frac{1}{\det A}$

$\det(A^{\mathrm{T}}) = \det A$

$\det(tA) = t^n \det A$

$x_1 = \frac{\det(A_{x_1})}{\det(A)}, \quad x_2 = \frac{\det(A_{x_2})}{\det(A)}$

$x_i = \frac{\det(A_{x_i})}{\det(A)},\quad i = 1,2,\ldots,n$

$\text{adj}(A) = C^T$

$\text{1. Every non-zero row has its first non-zero entry (called the **leading entry**) equal to 1.}$

$\text{2. Each leading entry is in a column to the **right** of the leading entry in the row above it.}$

$\text{3. Any rows that are entirely zero must lie at the **bottom** of the matrix.}$

$\text{4. Every leading 1 must be the **only** non-zero entry in its column.}$

$\text{If any row of the coefficient matrix is all zeros **but** the corresponding entry in the augmented column is non-zero, the system has **no solution**.}$

$R_i \leftrightarrow R_j$

$R_i \leftarrow t \cdot R_i$

$R_i \leftarrow R_i + c \cdot R_j$

$A \mid b = \begin{pmatrix} 3 & 2 & \vert & 6 \\ 1 & 0 & \vert & 2 \end{pmatrix}$

$\text{If } R \mid c \text{ has a row } [0 \ \cdots \ 0 \mid d] \text{ with } d \neq 0, \text{ then no solution.}$

$2 \cdot (1, 2) + 1 \cdot (2, 1) = (2\cdot1 + 1\cdot2,\ 2\cdot2 + 1\cdot1) = (4, 5)$

$a_1 v_1 + a_2 v_2 + \cdots + a_n v_n = 0$

$a_1 \mathbf{v}_1 + a_2 \mathbf{v}_2 + \cdots + a_n \mathbf{v}_n = \mathbf{0}$

$\text{If } a_1 \mathbf{v}_1 + \cdots + a_n \mathbf{v}_n = \mathbf{0}, \text{ then } a_1 = \cdots = a_n = 0.$

$V = \begin{pmatrix} | & | & & | \\ \mathbf{v}_1 & \mathbf{v}_2 & \cdots & \mathbf{v}_n \\ | & | & & | \end{pmatrix}, \quad V\mathbf{x} = \mathbf{0}$

$\text{span}(S) = \left\{ \sum_{i=1}^k a_i \mathbf{v}_i \mid k \in \mathbb{N}, \mathbf{v}_i \in S, a_i \in \mathbb{R} \right\}$

$\text{span}(S) = \{ a_1 v_1 + \dots + a_k v_k \mid v_i \in S, a_i \in \mathbb{R} \}$

$S \text{ is linearly independent if } a_1 v_1 + \dots + a_k v_k = 0 \text{ implies } a_1 = \dots = a_k = 0$

$a_1 \mathbf{v}_1 + \cdots + a_k \mathbf{v}_k = \mathbf{0} \implies a_1 = \cdots = a_k = 0$

$\text{span}(S) = \{ c_1 \mathbf{v}_1 + \cdots + c_k \mathbf{v}_k \mid c_i \in \mathbb{R}, \mathbf{v}_i \in S \}$

$\dim(V)$

$\text{rank}(A)$

$\text{dim}(W) = \text{number of non-zero rows in row echelon form}$

$\text{pivot columns of } R \text{ indicate which original columns form a basis}$