This is a $2 \times 3$ matrix (2 rows, 3 columns).

Matrix Notation

• Dimension: $m \times n$ (m rows, n columns)
• Element: $a_{ij}$ is the element in row $i$, column $j$
• Square matrix: $m = n$ (same number of rows and columns)

Matrix Addition and Subtraction

Matrices can be added or subtracted only if they have the same dimensions.

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

Properties:

• Commutative: $A + B = B + A$
• Associative: $(A + B) + C = A + (B + C)$
• Identity: $A + O = A$ (where $O$ is the zero matrix)

Scalar Multiplication

Multiply each element by the scalar:

$cA = c\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} = \begin{bmatrix} ca_{11} & ca_{12} \\ ca_{21} & ca_{22} \end{bmatrix}$

Matrix Multiplication

For $A_{m \times n}$ and $B_{n \times p}$, the product $AB$ is an $m \times p$ matrix.

Note: The number of columns in $A$ must equal the number of rows in $B$.

The element in row $i$, column $j$ of $AB$ is: $(AB)_{ij} = \sum_{k=1}^{n} a_{ik}b_{kj}$

Example (2×2 matrices):

$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} e & f \\ g & h \end{bmatrix} = \begin{bmatrix} ae+bg & af+bh \\ ce+dg & cf+dh \end{bmatrix}$

Properties:

• NOT commutative: $AB \neq BA$ in general
• Associative: $(AB)C = A(BC)$
• Distributive: $A(B + C) = AB + AC$

Identity Matrix

The identity matrix $I_n$ is an $n \times n$ matrix with 1s on the diagonal and 0s elsewhere:

$I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Property: $AI = IA = A$

Determinant (2×2 matrices)

For a $2 \times 2$ matrix: $\det(A) = \det\begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - bc$

Determinant (3×3 matrices)

For a $3 \times 3$ matrix, use cofactor expansion: $\det\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} = a(ei-fh) - b(di-fg) + c(dh-eg)$

Properties:

• If $\det(A) = 0$, the matrix is singular (not invertible)
• If $\det(A) \neq 0$, the matrix is invertible
• $\det(AB) = \det(A) \cdot \det(B)$

Matrix Inverse (2×2)

For $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, if $\det(A) = ad - bc \neq 0$:

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

Property: $AA^{-1} = A^{-1}A = I$

Solving Systems with Matrices

A system of linear equations can be written as $AX = B$ where:

• $A$ is the coefficient matrix
• $X$ is the variable matrix
• $B$ is the constant matrix

Example:

$\begin{cases} 2x + 3y = 7 \\ 4x - y = 5 \end{cases}$

Can be written as: $\begin{bmatrix} 2 & 3 \\ 4 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 7 \\ 5 \end{bmatrix}$

Solution: $X = A^{-1}B$ (if $A$ is invertible)

Applications

1. Systems of equations: Solve $AX = B$
2. Transformations: Represent rotations, reflections, scaling
3. Cryptography: Encode and decode messages
4. Economics: Input-output models
5. Computer graphics: 3D transformations

Step 1: Calculate the determinant $\det(A) = (3)(4) - (2)(5) = 12 - 10 = 2$

Since $\det(A) = 2 \neq 0$, the matrix is invertible.

Step 2: Use the inverse formula for 2×2 matrices

For $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$: $A^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

Step 3: Apply the formula $A^{-1} = \frac{1}{2}\begin{bmatrix} 4 & -2 \\ -5 & 3 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ -\frac{5}{2} & \frac{3}{2} \end{bmatrix}$

Answer: $A^{-1} = \begin{bmatrix} 2 & -1 \\ -\frac{5}{2} & \frac{3}{2} \end{bmatrix}$

Verification: Check that $AA^{-1} = I$ $\begin{bmatrix} 3 & 2 \\ 5 & 4 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ -\frac{5}{2} & \frac{3}{2} \end{bmatrix} = \begin{bmatrix} 6-5 & -3+3 \\ 10-10 & -5+6 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ ✓

Where $A = \begin{bmatrix} 2 & 1 \\ 3 & -2 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, $B = \begin{bmatrix} 5 \\ 4 \end{bmatrix}$

Step 2: Find $A^{-1}$

Calculate determinant: $\det(A) = (2)(-2) - (1)(3) = -4 - 3 = -7$

Since $\det(A) \neq 0$, $A$ is invertible.

$A^{-1} = \frac{1}{-7}\begin{bmatrix} -2 & -1 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} \frac{2}{7} & \frac{1}{7} \\ \frac{3}{7} & -\frac{2}{7} \end{bmatrix}$

Step 3: Solve $X = A^{-1}B$

$X = \begin{bmatrix} \frac{2}{7} & \frac{1}{7} \\ \frac{3}{7} & -\frac{2}{7} \end{bmatrix} \begin{bmatrix} 5 \\ 4 \end{bmatrix}$

$X = \begin{bmatrix} \frac{10}{7} + \frac{4}{7} \\ \frac{15}{7} - \frac{8}{7} \end{bmatrix} = \begin{bmatrix} \frac{14}{7} \\ \frac{7}{7} \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$

Answer: $x = 2$, $y = 1$

Verification:

• $2(2) + 1 = 5$ ✓
• $3(2) - 2(1) = 6 - 2 = 4$ ✓