Matrix Operations and Applications

Perform matrix addition, multiplication, find determinants and inverses, and solve systems using matrices.

Matrix Operations and Applications

Introduction to Matrices

A matrix is a rectangular array of numbers arranged in rows and columns.

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

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

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

📚 Practice Problems

1Problem 1easy

❓ Question:

Given $A = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 5 \\ -2 & 3 \end{bmatrix}$ , find $AB$ .

💡 Show Solution

Solution:

Given: $A = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 5 \\ -2 & 3 \end{bmatrix}$

Matrix multiplication formula: For element $(i,j)$ in $AB$ : multiply row $i$ of $A$ by column $j$ of $B$ .

Calculate each element:

First row, first column: $(2)(1) + (-1)(-2) = 2 + 2 = 4$

First row, second column: $(2)(5) + (-1)(3) = 10 - 3 = 7$

Second row, first column: $(3)(1) + (4)(-2) = 3 - 8 = -5$

Second row, second column: $(3)(5) + (4)(3) = 15 + 12 = 27$

Answer: $AB = \begin{bmatrix} 4 & 7 \\ -5 & 27 \end{bmatrix}$

Note: Matrix multiplication is not commutative. $BA$ would give a different result!

2Problem 2medium

❓ Question:

Find the inverse of $A = \begin{bmatrix} 3 & 2 \\ 5 & 4 \end{bmatrix}$ .

💡 Show Solution

Solution:

Given: $A = \begin{bmatrix} 3 & 2 \\ 5 & 4 \end{bmatrix}$

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

3Problem 3hard

❓ Question:

Use matrices to solve the system: $\begin{cases} 2x + y = 5 \\ 3x - 2y = 4 \end{cases}$

💡 Show Solution

Solution:

Step 1: Write in matrix form $AX = B$

$\begin{bmatrix} 2 & 1 \\ 3 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 5 \\ 4 \end{bmatrix}$

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:

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

4Problem 4medium

❓ Question:

Find the product AB where A = [2 1; 3 4] and B = [5 6; 7 8].

💡 Show Solution

Step 1: Check dimensions: A is 2×2, B is 2×2 Product AB will be 2×2

Step 2: Calculate element (1,1): Row 1 of A × Column 1 of B: 2(5) + 1(7) = 10 + 7 = 17

Step 3: Calculate element (1,2): Row 1 of A × Column 2 of B: 2(6) + 1(8) = 12 + 8 = 20

Step 4: Calculate element (2,1): Row 2 of A × Column 1 of B: 3(5) + 4(7) = 15 + 28 = 43

Step 5: Calculate element (2,2): Row 2 of A × Column 2 of B: 3(6) + 4(8) = 18 + 32 = 50

Step 6: Write result matrix: AB = [17 20] [43 50]

Answer: [17 20] [43 50]

5Problem 5hard

❓ Question:

A company makes two products. Product A requires 2 hours of labor and 3 kg of material. Product B requires 1 hour of labor and 2 kg of material. The company plans to make 50 units of A and 30 units of B. Use matrix multiplication to find total labor hours and material needed.

💡 Show Solution

Step 1: Set up requirement matrix R (resources × products): R = [2 1] (hours) [3 2] (kg)

Step 2: Set up production matrix P (products × 1): P = [50] (product A) [30] (product B)

Step 3: Calculate RP (total resources): [2 1][50] = [2(50) + 1(30)] [3 2][30] [3(50) + 2(30)]

Step 4: Compute: Row 1: 2(50) + 1(30) = 100 + 30 = 130 hours Row 2: 3(50) + 2(30) = 150 + 60 = 210 kg

Step 5: Result: [130] (total labor hours) [210] (total material in kg)

Answer: 130 hours of labor, 210 kg of material

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