A rectangular array of numbers arranged in rows and columns, enclosed in brackets

\[ A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{bmatrix} \]

Dimension (Order):

Number of rows × number of columns (m × n)

Element (Entry):

\( a_{ij} \) = element in row i, column j

Square Matrix:

Matrix with same number of rows and columns (n × n)

Zero Matrix:

Matrix with all entries equal to 0

Identity Matrix (I):

Square matrix with 1's on main diagonal and 0's elsewhere

Example: \( I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)

Equal Matrices:

Same dimensions and corresponding elements are equal

Add or subtract corresponding elements

\[ A + B = \begin{bmatrix} a_{11} + b_{11} & a_{12} + b_{12} \\ a_{21} + b_{21} & a_{22} + b_{22} \end{bmatrix} \]

\[ A - B = \begin{bmatrix} a_{11} - b_{11} & a_{12} - b_{12} \\ a_{21} - b_{21} & a_{22} - b_{22} \end{bmatrix} \]

\( \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} + \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 5 \\ 7 & 9 \end{bmatrix} \)

Multiply every element by the scalar

\[ kA = \begin{bmatrix} ka_{11} & ka_{12} \\ ka_{21} & ka_{22} \end{bmatrix} \]

\( 3 \begin{bmatrix} 2 & 4 \\ 1 & 5 \end{bmatrix} = \begin{bmatrix} 6 & 12 \\ 3 & 15 \end{bmatrix} \)

Element in row i, column j of AB is the dot product of row i of A with column j of B

\[ (AB)_{ij} = \sum_{k=1}^{n} a_{ik}b_{kj} \]

\( \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix} = \begin{bmatrix} 19 & 22 \\ 43 & 50 \end{bmatrix} \)

First element: (1)(5) + (2)(7) = 5 + 14 = 19

\[ \det(A) = |A| = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc \]

\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = a\begin{vmatrix} e & f \\ h & i \end{vmatrix} - b\begin{vmatrix} d & f \\ g & i \end{vmatrix} + c\begin{vmatrix} d & e \\ g & h \end{vmatrix} \]

= a(ei - fh) - b(di - fg) + c(dh - eg)

\( \begin{vmatrix} 3 & 2 \\ 1 & 4 \end{vmatrix} = (3)(4) - (2)(1) = 12 - 2 = 10 \)

The inverse \( A^{-1} \) of matrix A satisfies:

\[ AA^{-1} = A^{-1}A = I \]

\[ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \]

\[ A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \]

where ad - bc ≠ 0

Find inverse of \( A = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} \)

To solve AX = B for matrix X:

\[ X = A^{-1}B \]

Steps:

1. Find \( A^{-1} \)

2. Multiply both sides by \( A^{-1} \) on the left

3. Simplify: \( A^{-1}AX = A^{-1}B \)

4. Result: \( IX = A^{-1}B \), so \( X = A^{-1}B \)

Reflection over x-axis:

\( \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \)

Reflection over y-axis:

\( \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} \)

Rotation 90° counterclockwise:

\( \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \)

Rotation 180°:

\( \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \)

Dilation by factor k:

\( \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix} \)

To transform a shape, organize vertices as columns in a matrix, then multiply by transformation matrix

\( \text{Image} = \text{Transformation Matrix} \times \text{Vertex Matrix} \)

Combine coefficient matrix with constants, separated by vertical bar

System: \( \begin{cases} 2x + 3y = 8 \\ x - y = 1 \end{cases} \)

\[ \left[\begin{array}{cc|c} 2 & 3 & 8 \\ 1 & -1 & 1 \end{array}\right] \]

1. Swap two rows

2. Multiply a row by a nonzero constant

3. Add a multiple of one row to another row

Goal: Reduce to row echelon form (REF) or reduced row echelon form (RREF)

• Leading entry (first nonzero) in each row is 1

• Leading 1's move to the right in successive rows

• All entries below leading 1 are 0