10 min readConcept sheet1-mark MCQs and Part II (2 marks) on order, types and aᵢⱼ; Part III (5 marks) on multiplication and transpose

Matrices

A matrix is a rectangular array of numbers in rows and columns. Add only same-order matrices; multiply A by B only when columns of A = rows of B.

Orderm × n means m rows and n columns

Element aᵢⱼThe entry in row i and column j

Product ruleA (m × n) times B (n × p) gives AB of order m × p

How we write it hereRows separated by a semicolon: [1, 2; 3, 4] has rows 1, 2 and 3, 4

Types of matrices

Type	What it is	Example
Row matrix	Only one row (order 1 × n)	[2, −1, 5]
Column matrix	Only one column (order m × 1)	[3; 0; 7]
Square matrix	Number of rows = number of columns	[1, 2; 3, 4]
Diagonal matrix	Square; every element off the leading diagonal (top left to bottom right) is 0	[4, 0; 0, −1]
Scalar matrix	Diagonal matrix with all diagonal elements equal	[5, 0; 0, 5]
Identity (unit) matrix I	Scalar matrix with every diagonal element 1	[1, 0; 0, 1]
Zero (null) matrix O	Every element is 0	[0, 0; 0, 0]
Upper or lower triangular	Square; all elements below (upper) or above (lower) the leading diagonal are 0	[1, 2; 0, 3] is upper triangular

Worked example 1: build a matrix, then find 3A − 2B

Problem

Construct the 2 × 2 matrix A = [aᵢⱼ] with aᵢⱼ = i + 2j. If B = [1, 4; −2, 5], find 3A − 2B.

Working

a₁₁ = 1 + 2 = 3, a₁₂ = 1 + 4 = 5, a₂₁ = 2 + 2 = 4, a₂₂ = 2 + 4 = 6. So A = [3, 5; 4, 6].
Scalar multiplication: multiply every element. 3A = [9, 15; 12, 18] and 2B = [2, 8; −4, 10].
Both are 2 × 2, so we can subtract matching elements.
3A − 2B = [9 − 2, 15 − 8; 12 − (−4), 18 − 10] = [7, 7; 16, 8].

Answer

A = [3, 5; 4, 6] and 3A − 2B = [7, 7; 16, 8]

Worked example 2: AB, BA and (AB)ᵀ

Problem

If A = [1, 2; 3, 4] and B = [2, 0; 1, 3], find AB and BA, and verify that (AB)ᵀ = BᵀAᵀ.

Working

Rule: row i of A times column j of B gives the element in row i, column j of AB.
AB: row 1 × column 1 = 1 × 2 + 2 × 1 = 4; row 1 × column 2 = 1 × 0 + 2 × 3 = 6.
Row 2 × column 1 = 3 × 2 + 4 × 1 = 10; row 2 × column 2 = 3 × 0 + 4 × 3 = 12. So AB = [4, 6; 10, 12].
BA: row 1 × column 1 = 2 × 1 + 0 × 3 = 2; row 1 × column 2 = 2 × 2 + 0 × 4 = 4.
Row 2 × column 1 = 1 × 1 + 3 × 3 = 10; row 2 × column 2 = 1 × 2 + 3 × 4 = 14. So BA = [2, 4; 10, 14].
AB ≠ BA, so matrix multiplication is not commutative.
The rows of AB become columns: (AB)ᵀ = [4, 10; 6, 12]. Also Bᵀ = [2, 1; 0, 3] and Aᵀ = [1, 3; 2, 4].
BᵀAᵀ: row 1 × column 1 = 2 × 1 + 1 × 2 = 4; row 1 × column 2 = 2 × 3 + 1 × 4 = 10.
Row 2 × column 1 = 0 × 1 + 3 × 2 = 6; row 2 × column 2 = 0 × 3 + 3 × 4 = 12. So BᵀAᵀ = [4, 10; 6, 12].
Both sides are equal, so (AB)ᵀ = BᵀAᵀ is verified.

Answer

AB = [4, 6; 10, 12], BA = [2, 4; 10, 14], and (AB)ᵀ = BᵀAᵀ = [4, 10; 6, 12]

Worked example 3: a sweet shop word problem

Problem

Two sweet shops in Tirunelveli sell halwa and mysore pak. In a day, shop P sells 10 kg of halwa and 5 kg of mysore pak; shop Q sells 8 kg and 12 kg. Example prices are ₹ 400 per kg for halwa and ₹ 300 per kg for mysore pak. Use matrices to find how much each shop collects.

Working

Sales matrix S (rows: shops P and Q; columns: halwa and mysore pak) = [10, 5; 8, 12], of order 2 × 2.
Price matrix C (a column) = [400; 300], of order 2 × 1.
Columns of S (2) = rows of C (2), so SC exists, and its order is 2 × 1.
Row P: 10 × 400 + 5 × 300 = 4000 + 1500 = 5500.
Row Q: 8 × 400 + 12 × 300 = 3200 + 3600 = 6800.
SC = [5500; 6800].

Answer

Shop P collects ₹ 5500 and shop Q collects ₹ 6800

Rules to remember

Operation	Rule
Equal matrices	Same order, and every pair of matching elements is equal
Addition A + B	Only for the same order; add matching elements. A + B = B + A, (A + B) + C = A + (B + C), A + O = A, A + (−A) = O
Scalar multiple kA	Multiply every element by k
Multiplication AB	Needs columns of A = rows of B. In general AB ≠ BA. (AB)C = A(BC), A(B + C) = AB + AC, AI = IA = A
Transpose Aᵀ	Rows become columns. (Aᵀ)ᵀ = A, (kA)ᵀ = kAᵀ, (A + B)ᵀ = Aᵀ + Bᵀ, (AB)ᵀ = BᵀAᵀ

In real life

Bills, school timetables and cricket scorecards all store numbers in rows and columns. Multiplying a quantity matrix by a price matrix gives every customer's bill in one step.

Exam angle

1 marks: “A is of order 3 × 4 and B is of order 4 × 2. What is the order of AB?” 3 × 2. The inner numbers (4 and 4) match, and the outer numbers give the order.

2 marks: “If A = [2, −1; 3, 0] and B = [1, 4; −2, 5], find 3A − 2B.” 3A = [6, −3; 9, 0] and 2B = [2, 8; −4, 10]. Subtract matching elements: 3A − 2B = [4, −11; 13, −10].

5 marks: “If A = [2, −1; 1, 3] and B = [1, 0; 4, 2], verify that (AB)ᵀ = BᵀAᵀ.” AB = [−2, −2; 13, 6], so (AB)ᵀ = [−2, 13; −2, 6]. Bᵀ = [1, 4; 0, 2] and Aᵀ = [2, 1; −1, 3], so BᵀAᵀ = [−2, 13; −2, 6]. Both sides are equal.

Remember it

Row into column: run along a row of the first matrix and down a column of the second. The transpose of a product reverses the order: (AB)ᵀ = BᵀAᵀ.