Matrices & Determinant. Grade 12 Mathematics: Matrices - Part A (Introduction & Operations) Subtopics Navigator - Pt. A Introduction to Matrices Types of Matrices Matrix Operations Matrix Multiplication Transpose of a Matrix Cumulative Exercises Conclusion Lesson Objectives Understand matrix notation and basic terminology Identify different types of matrices Perform matrix addition, subtraction, and scalar multiplication Master matrix multiplication with correct dimensions Find the transpose of a matrix Apply matrix operations to solve problems Lesson Introduction A matrix is a rectangular array of numbers, symbols, or expressions arranged in rows and columns. Matrices are powerful mathematical tools used to solve systems of equations, transform geometric figures, and represent data in various fields including computer graphics, economics, and engineering. Matrix Notation: A matrix is typically denoted by a capital letter and its elements by corresponding lowercase letters with subscripts. For example, matrix A with m rows and n columns is written as: [latex]A = begin{bmatrix} a_{11} & a_{12} & cdots & a_{1n} \ a_{21} & a_{22} & cdots & a_{2n} \ vdots & vdots & ddots & vdots \ a_{m1} & a_{m2} & cdots & a_{mn} end{bmatrix}[/latex] The dimension or order of the matrix is m × n (read as "m by n"). Types of Matrices Matrices can be classified based on their dimensions and special properties: Type Definition Example Row Matrix Matrix with only one row [latex]begin{bmatrix} 1 & 2 & 3 end{bmatrix}[/latex] Column Matrix Matrix with only one column [latex]begin{bmatrix} 4 \ 5 \ 6 end{bmatrix}[/latex] Square Matrix Matrix with equal number of rows and columns [latex]begin{bmatrix} 1 & 2 \ 3 & 4 end{bmatrix}[/latex] Zero Matrix Matrix with all elements equal to zero [latex]begin{bmatrix} 0 & 0 \ 0 & 0 end{bmatrix}[/latex] Identity Matrix Square matrix with 1's on main diagonal and 0's elsewhere [latex]begin{bmatrix} 1 & 0 \ 0 & 1 end{bmatrix}[/latex] Diagonal Matrix Square matrix with non-zero elements only on main diagonal [latex]begin{bmatrix} 2 & 0 \ 0 & 5 end{bmatrix}[/latex] Exercises (Types of Matrices) Identify the type of matrix: [latex]begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 end{bmatrix}[/latex] Create a 3×3 identity matrix Which of these is a row matrix: [latex]begin{bmatrix} 1 \ 2 end{bmatrix}[/latex] or [latex]begin{bmatrix} 3 & 4 end{bmatrix}[/latex]? Matrix Operations Basic matrix operations include addition, subtraction, and scalar multiplication: Matrix Addition and Subtraction: Two matrices can be added or subtracted only if they have the same dimensions. If A = [aij] and B = [bij] are both m × n matrices, then: A + B = [aij + bij] and A - B = [aij - bij] Example 1: Let [latex]A = begin{bmatrix} 2 & 3 \ 1 & 4 end{bmatrix}[/latex] and [latex]B = begin{bmatrix} 5 & 1 \ 2 & 3 end{bmatrix}[/latex] Then [latex]A + B = begin{bmatrix} 2+5 & 3+1 \ 1+2 & 4+3 end{bmatrix} = begin{bmatrix} 7 & 4 \ 3 & 7 end{bmatrix}[/latex] And [latex]A - B = begin{bmatrix} 2-5 & 3-1 \ 1-2 & 4-3 end{bmatrix} = begin{bmatrix} -3 & 2 \ -1 & 1 end{bmatrix}[/latex] Scalar Multiplication: When a matrix is multiplied by a scalar (a single number), each element is multiplied by that scalar. If A = [aij] is an m × n matrix and k is a scalar, then kA = [kaij] Example 2: Let [latex]A = begin{bmatrix} 1 & 4 \ 2 & 5 \ 3 & 6 end{bmatrix}[/latex] and k = 3 Then [latex]3A = begin{bmatrix} 3×1 & 3×4 \ 3×2 & 3×5 \ 3×3 & 3×6 end{bmatrix} = begin{bmatrix} 3 & 12 \ 6 & 15 \ 9 & 18 end{bmatrix}[/latex] Matrix Operations If [latex]A = begin{bmatrix} 2 & 5 \ 1 & 3 end{bmatrix}[/latex] and [latex]B = begin{bmatrix} 4 & 1 \ 2 & 6 end{bmatrix}[/latex], find A + B and A - B Calculate [latex]2begin{bmatrix} 3 & 1 & 4 \ 2 & 5 & 7 end{bmatrix}[/latex] If [latex]A = begin{bmatrix} 1 & 0 \ 2 & 4 end{bmatrix}[/latex] and [latex]B = begin{bmatrix} 3 & 1 \ 1 & 2 end{bmatrix}[/latex], find 2A - 3B Matrix Multiplication Matrix multiplication is more complex than addition and subtraction. Two matrices can be multiplied only if the number of columns in the first matrix equals the number of rows in the second matrix. If A is an m × n matrix and B is an n × p matrix, then their product AB is an m × p matrix. The element in the ith row and jth column of AB is the dot product of the ith row of A and the jth column of B. Example 3: Matrix Multiplication Let [latex]A = begin{bmatrix} 1 & 2 & 3 \ 4 & 5 & 6 end{bmatrix}[/latex] (2×3 matrix) and [latex]B = begin{bmatrix} 7 & 8 \ 9 & 10 \ 11 & 12 end{bmatrix}[/latex] (3×2 matrix) Since A has 3 columns and B has 3 rows, they can be multiplied. The product AB will be a 2×2 matrix. To find AB[1,1] (first row, first column): Multiply elements of first row of A with corresponding elements of first column of B and sum: (1×7) + (2×9) + (3×11) = 7 + 18 + 33 = 58 Similarly: AB[1,2] = (1×8) + (2×10) + (3×12) = 8 + 20 + 36 = 64 AB[2,1] = (4×7) + (5×9) + (6×11) = 28 + 45 + 66 = 139 AB[2,2] = (4×8) + (5×10) + (6×12) = 32 + 50 + 72 = 154 So, [latex]AB = begin{bmatrix} 58 & 64 \ 139 & 154 end{bmatrix}[/latex] Important Note: Matrix multiplication is not commutative. In general, AB ≠ BA. Example 4: Let [latex]A = begin{bmatrix} 1 & 2 \ 3 & 4 end{bmatrix}[/latex] and [latex]B = begin{bmatrix} 5 & 6 \ 7 & 8 end{bmatrix}[/latex] [latex]AB = begin{bmatrix} 1×5+2×7 & 1×6+2×8 \ 3×5+4×7 & 3×6+4×8 end{bmatrix} = begin{bmatrix} 19 & 22 \ 43 & 50 end{bmatrix}[/latex] [latex]BA = begin{bmatrix} 5×1+6×3 & 5×2+6×4 \ 7×1+8×3 & 7×2+8×4 end{bmatrix} = begin{bmatrix} 23 & 34 \ 31 & 46 end{bmatrix}[/latex] Clearly, AB ≠ BA Matrix Multiplication If [latex]A = begin{bmatrix} 2 & 3 \ 1 & 4 end{bmatrix}[/latex] and [latex]B = begin{bmatrix} 5 & 1 \ 2 & 3 end{bmatrix}[/latex], find AB and BA Multiply: [latex]begin{bmatrix} 1 & 0 & 2 \ 3 & 1 & 4 end{bmatrix} begin{bmatrix} 2 & 1 \ 0 & 3 \ 1 & 2 end{bmatrix}[/latex] Explain why matrix multiplication is not commutative with an example Transpose of a Matrix The transpose of a matrix is obtained by interchanging its rows and columns. The transpose of matrix A is denoted as AT or A'. If A is an m × n matrix, then AT is an n × m matrix where (AT)ij = Aji. Example 5: Finding Transpose Let [latex]A = begin{bmatrix} 1 & 2 & 3 \ 4 & 5 & 6 end{bmatrix}[/latex] (2×3 matrix) Then [latex]A^T = begin{bmatrix} 1 & 4 \ 2 & 5 \ 3 & 6 end{bmatrix}[/latex] (3×2 matrix) Let [latex]B = begin{bmatrix} 7 & 8 & 9 end{bmatrix}[/latex] (1×3 row matrix) Then [latex]B^T = begin{bmatrix} 7 \ 8 \ 9 end{bmatrix}[/latex] (3×1 column matrix) Properties of Transpose: (AT)T = A (A + B)T = AT + BT (kA)T = kAT (where k is a scalar) (AB)T = BTAT Transpose of a Matrix Find the transpose of [latex]begin{bmatrix} 2 & 5 & 1 \ 3 & 0 & 4 end{bmatrix}[/latex] If [latex]A = begin{bmatrix} 1 & 2 \ 3 & 4 end{bmatrix}[/latex], verify that (AT)T = A For matrices A and B from question 1 of the multiplication section, verify that (AB)T = BTAT Cumulative Exercises If [latex]A = begin{bmatrix} 2 & 1 \ 3 & 4 end{bmatrix}[/latex] and [latex]B = begin{bmatrix} 5 & 2 \ 1 & 3 end{bmatrix}[/latex], calculate 2A + 3B Multiply: [latex]begin{bmatrix} 1 & 2 & 3 end{bmatrix} begin{bmatrix} 4 \ 5 \ 6 end{bmatrix}[/latex] Find the transpose of [latex]begin{bmatrix} 1 & 0 & 2 \ 3 & 1 & 4 \ 2 & 5 & 0 end{bmatrix}[/latex] If [latex]A = begin{bmatrix} 2 & 1 \ 0 & 3 end{bmatrix}[/latex] and [latex]B = begin{bmatrix} 4 & 2 \ 1 & 5 end{bmatrix}[/latex], show that (A + B)T = AT + BT Calculate: [latex]begin{bmatrix} 2 & 3 \ 1 & 4 end{bmatrix} begin{bmatrix} 5 \ 6 end{bmatrix}[/latex] If [latex]A = begin{bmatrix} 1 & 2 \ 3 & 4 end{bmatrix}[/latex], [latex]B = begin{bmatrix} 2 & 0 \ 1 & 3 end{bmatrix}[/latex], and [latex]C = begin{bmatrix} 1 & 1 \ 0 & 2 end{bmatrix}[/latex], verify that A(B + C) = AB + AC Find the product: [latex]begin{bmatrix} 2 & 1 \ 3 & 0 \ 1 & 4 end{bmatrix} begin{bmatrix} 1 & 2 & 0 \ 3 & 1 & 2 end{bmatrix}[/latex] If [latex]A = begin{bmatrix} 1 & 3 \ 2 & 4 end{bmatrix}[/latex], find ATA Determine whether matrix multiplication is commutative using specific examples Create two matrices that cannot be added together and explain why Show/Hide Answers Problem: If [latex]A = begin{bmatrix} 2 & 1 \ 3 & 4 end{bmatrix}[/latex] and [latex]B = begin{bmatrix} 5 & 2 \ 1 & 3 end{bmatrix}[/latex], calculate 2A + 3B Step 1: [latex]2A = begin{bmatrix} 4 & 2 \ 6 & 8 end{bmatrix}[/latex] Step 2: [latex]3B = begin{bmatrix} 15 & 6 \ 3 & 9 end{bmatrix}[/latex] Step 3: [latex]2A + 3B = begin{bmatrix} 4+15 & 2+6 \ 6+3 & 8+9 end{bmatrix} = begin{bmatrix} 19 & 8 \ 9 & 17 end{bmatrix}[/latex] Answer: [latex]begin{bmatrix} 19 & 8 \ 9 & 17 end{bmatrix}[/latex] Problem: Multiply: [latex]begin{bmatrix} 1 & 2 & 3 end{bmatrix} begin{bmatrix} 4 \ 5 \ 6 end{bmatrix}[/latex] Step 1: This is a 1×3 matrix multiplied by a 3×1 matrix, resulting in a 1×1 matrix Step 2: (1×4) + (2×5) + (3×6) = 4 + 10 + 18 = 32 Answer: [32] Problem: Find the transpose of [latex]begin{bmatrix} 1 & 0 & 2 \ 3 & 1 & 4 \ 2 & 5 & 0 end{bmatrix}[/latex] Step 1: Interchange rows and columns Step 2: [latex]begin{bmatrix} 1 & 3 & 2 \ 0 & 1 & 5 \ 2 & 4 & 0 end{bmatrix}[/latex] Answer: [latex]begin{bmatrix} 1 & 3 & 2 \ 0 & 1 & 5 \ 2 & 4 & 0 end{bmatrix}[/latex] Problem: If [latex]A = begin{bmatrix} 2 & 1 \ 0 & 3 end{bmatrix}[/latex] and [latex]B = begin{bmatrix} 4 & 2 \ 1 & 5 end{bmatrix}[/latex], show that (A + B)T = AT + BT Step 1: [latex]A + B = begin{bmatrix} 2+4 & 1+2 \ 0+1 & 3+5 end{bmatrix} = begin{bmatrix} 6 & 3 \ 1 & 8 end{bmatrix}[/latex] Step 2: [latex](A + B)^T = begin{bmatrix} 6 & 1 \ 3 & 8 end{bmatrix}[/latex] Step 3: [latex]A^T = begin{bmatrix} 2 & 0 \ 1 & 3 end{bmatrix}[/latex], [latex]B^T = begin{bmatrix} 4 & 1 \ 2 & 5 end{bmatrix}[/latex] Step 4: [latex]A^T + B^T = begin{bmatrix} 2+4 & 0+1 \ 1+2 & 3+5 end{bmatrix} = begin{bmatrix} 6 & 1 \ 3 & 8 end{bmatrix}[/latex] Step 5: Both sides equal [latex]begin{bmatrix} 6 & 1 \ 3 & 8 end{bmatrix}[/latex], so the property is verified. Answer: Verified Problem: Calculate: [latex]begin{bmatrix} 2 & 3 \ 1 & 4 end{bmatrix} begin{bmatrix} 5 \ 6 end{bmatrix}[/latex] Step 1: (2×5) + (3×6) = 10 + 18 = 28 Step 2: (1×5) + (4×6) = 5 + 24 = 29 Answer: [latex]begin{bmatrix} 28 \ 29 end{bmatrix}[/latex] Problem: Verify that A(B + C) = AB + AC for given matrices Step 1: [latex]B + C = begin{bmatrix} 2+1 & 0+1 \ 1+0 & 3+2 end{bmatrix} = begin{bmatrix} 3 & 1 \ 1 & 5 end{bmatrix}[/latex] Step 2: [latex]A(B + C) = begin{bmatrix} 1 & 2 \ 3 & 4 end{bmatrix} begin{bmatrix} 3 & 1 \ 1 & 5 end{bmatrix} = begin{bmatrix} 5 & 11 \ 13 & 23 end{bmatrix}[/latex] Step 3: [latex]AB = begin{bmatrix} 4 & 6 \ 10 & 12 end{bmatrix}[/latex], [latex]AC = begin{bmatrix} 1 & 5 \ 3 & 11 end{bmatrix}[/latex] Step 4: [latex]AB + AC = begin{bmatrix} 4+1 & 6+5 \ 10+3 & 12+11 end{bmatrix} = begin{bmatrix} 5 & 11 \ 13 & 23 end{bmatrix}[/latex] Step 5: Both sides equal, so distributive property is verified. Answer: Verified Problem: Find the product: [latex]begin{bmatrix} 2 & 1 \ 3 & 0 \ 1 & 4 end{bmatrix} begin{bmatrix} 1 & 2 & 0 \ 3 & 1 & 2 end{bmatrix}[/latex] Step 1: Row 1: (2×1)+(1×3)=5, (2×2)+(1×1)=5, (2×0)+(1×2)=2 Step 2: Row 2: (3×1)+(0×3)=3, (3×2)+(0×1)=6, (3×0)+(0×2)=0 Step 3: Row 3: (1×1)+(4×3)=13, (1×2)+(4×1)=6, (1×0)+(4×2)=8 Answer: [latex]begin{bmatrix} 5 & 5 & 2 \ 3 & 6 & 0 \ 13 & 6 & 8 end{bmatrix}[/latex] Problem: If [latex]A = begin{bmatrix} 1 & 3 \ 2 & 4 end{bmatrix}[/latex], find ATA Step 1: [latex]A^T = begin{bmatrix} 1 & 2 \ 3 & 4 end{bmatrix}[/latex] Step 2: [latex]A^TA = begin{bmatrix} 1 & 2 \ 3 & 4 end{bmatrix} begin{bmatrix} 1 & 3 \ 2 & 4 end{bmatrix} = begin{bmatrix} 5 & 11 \ 11 & 25 end{bmatrix}[/latex] Answer: [latex]begin{bmatrix} 5 & 11 \ 11 & 25 end{bmatrix}[/latex] Problem: Determine whether matrix multiplication is commutative using specific examples Step 1: Use matrices from example 4 in the lesson Step 2: Show that AB ≠ BA Step 3: Conclude that matrix multiplication is not commutative in general Answer: Matrix multiplication is not commutative as shown by specific examples where AB ≠ BA Problem: Create two matrices that cannot be added together and explain why Step 1: Example: [latex]A = begin{bmatrix} 1 & 2 end{bmatrix}[/latex] (1×2) and [latex]B = begin{bmatrix} 3 \ 4 end{bmatrix}[/latex] (2×1) Step 2: They have different dimensions (1×2 vs 2×1) Step 3: Matrix addition requires matrices to have the same dimensions Answer: Matrices with different dimensions cannot be added Conclusion/Recap In this lesson, we've covered the fundamentals of matrices including their types, basic operations (addition, subtraction, scalar multiplication), matrix multiplication, and the transpose operation. We've learned that matrix multiplication is not commutative and requires specific dimension compatibility. These concepts form the foundation for more advanced matrix topics we'll cover in Part B. Clip It! Share your ANSWER in the Chat. Indicate TITLE e.g Linear Equation 1. .....2. e.t.c
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