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Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: N-VM.C.6, N-VM.C.7, N-VM.C.8, N-VM.C.9, N-VM.C.10 · Level: Grade 12 Precalculus
Problem 1 Data in a Matrix
The matrix has rows (48, 30, 25) and (36, 41, 52):
| 48 | 30 | 25 |
| 36 | 41 | 52 |
It has 2 rows and 3 columns, so it is 2 × 3. The number in the second row and third column is 52, the bagels sold on Sunday.
Answer: 2 × 3; entry 52 (Sunday bagels)
Problem 2 Adding and Subtracting
A + B:
| −1 | 5 |
| 6 | 6 |
A − B:
| 7 | −9 |
| 4 | −6 |
Answer: A + B has rows (−1, 5), (6, 6); A − B has rows (7, −9), (4, −6)
Problem 3 Scalar Multiple
Multiply every entry by −3:
| −6 | 3 | −12 |
| 0 | −15 | 6 |
Answer: rows (−6, 3, −12) and (0, −15, 6)
Problem 4 Can the Product Be Found?
(a) The width of A (2) matches the height of B (2), so AB exists. It is as tall as A and as wide as B: 4 × 5.
(b) For BA, the width of B (5) would have to match the height of A (4). It does not, so BA cannot be formed.
Answer: AB is 4 × 5; BA does not exist
Problem 5 A Determinant and Its Area
det = 8(6) − (−1)(3) = 48 + 3 = 51. The area of the parallelogram is |det| = 51 square units.
Answer: det = 51; area 51 square units
Problem 6 Identity and Zero
The identity matrix leaves any matrix unchanged, and adding the zero matrix changes nothing, so both results equal A:
| 7 | −3 |
| 2 | 9 |
Answer: IA = A + O = A, with rows (7, −3) and (2, 9)
Problem 7 A 2 × 2 Product
Row 1: 2(5) + (−1)(−2) = 12 and 2(0) + (−1)(1) = −1. Row 2: 3(5) + 4(−2) = 7 and 3(0) + 4(1) = 4.
| 12 | −1 |
| 7 | 4 |
Answer: AB has rows (12, −1) and (7, 4)
Problem 8 Order Matters
AB:
| 8 | 9 |
| 6 | 15 |
BA:
| 4 | 5 |
| 2 | 19 |
The matrices differ, so matrix multiplication is not commutative here.
Answer: AB has rows (8, 9), (6, 15) and BA has rows (4, 5), (2, 19); not equal
Problem 9 A 2 × 3 Times a 3 × 2
AB is 2 × 2. Row 1: 1(2) + 0(0) + 2(4) = 10 and 1(1) + 0(−3) + 2(5) = 11. Row 2: (−1)(2) + 3(0) + 1(4) = 2 and (−1)(1) + 3(−3) + 1(5) = −5.
Answer: AB is 2 × 2 with rows (10, 11) and (2, −5)
Problem 10 Finding an Inverse
det A = 4(3) − 1(6) = 6. Swap and negate: rows (3, −1) and (−6, 4). Divide by 6:
| 12 | −16 |
| −1 | 23 |
Check: row 1 times column 1 is 4(12) + 1(−1) = 1; row 1 times column 2 is 4(−16) + 1(23) = 0; the other entries work the same way, so AA⁻¹ = I.
Answer: A⁻¹ has rows (12, −16) and (−1, 23)
Problem 11 A 3 × 3 Product
Row 1: 1(2) + 0(1) + 2(−1) = 0; 1(1) + 0(0) + 2(2) = 5; 1(0) + 0(3) + 2(1) = 2.
Row 2: 0 + 1 + 1 = 2; 0 + 0 − 2 = −2; 0 + 3 − 1 = 2.
Row 3: 6 + 2 + 0 = 8; 3 + 0 + 0 = 3; 0 + 6 + 0 = 6.
| 0 | 5 | 2 |
| 2 | −2 | 2 |
| 8 | 3 | 6 |
Answer: rows (0, 5, 2), (2, −2, 2), (8, 3, 6)
Problem 12 No Inverse
A matrix has no inverse when its determinant is 0. det = 2(6) − k(3) = 12 − 3k. Setting 12 − 3k = 0 gives k = 4.
Check: with k = 4 the rows (2, 4) and (3, 6) are proportional (the second row is 1.5 times the first), and det = 12 − 12 = 0.
Answer: k = 4
Problem 13 A Matrix Equation
Subtract A: 2X = B − A, which has rows (8, −4) and (8, 4). Divide by 2:
| 4 | −2 |
| 4 | 2 |
Check: 2(4) + 1 = 9, 2(−2) + 3 = −1, 2(4) − 5 = 3, 2(2) + 7 = 11.
Answer: X has rows (4, −2) and (4, 2)
Problem 14 Powers of a Matrix
A² = A·A:
| −1 | 0 |
| 0 | −1 |
Then A⁴ = (A²)² has rows (1, 0) and (0, 1), the identity matrix I. Since A⁴ = I, A10 = (A⁴)²·A² = A² = −I.
Answer: A² has rows (−1, 0), (0, −1); A⁴ = I; A10 = A²
Problem 15 Fundraiser Stands
Q has rows (40, 25, 15) and (30, 60, 10). P is the column (1.50, 0.75, 2.00).
Row 1: 40(1.50) + 25(0.75) + 15(2.00) = 60 + 18.75 + 30 = 108.75. Row 2: 30(1.50) + 60(0.75) + 10(2.00) = 45 + 45 + 20 = 110.
| 108.75 |
| 110 |
Each entry is the total revenue of one stand.
Answer: Stand 1 earned $108.75 and Stand 2 earned $110.00
Problem 16 Camp Enrollment
(a) Week 1 + Week 2:
| 20 | 32 |
| 36 | 24 |
| 40 | 60 |
(b) Multiply by 1.25:
| 25 | 40 |
| 45 | 30 |
| 50 | 75 |
Answer: (a) rows (20, 32), (36, 24), (40, 60); (b) rows (25, 40), (45, 30), (50, 75)
Problem 17 Secret Number Code
(a) K times the column (7, 4): 2(7) + 3(4) = 26 and 1(7) + 4(4) = 23. The code is (26, 23).
(b) det K = 2(4) − 3(1) = 5, so K⁻¹ has rows (45, −35) and (−15, 25).
(c) 45(27) − 35(26) = 108 − 785 = 6 and −15(27) + 25(26) = −27 + 525 = 5.
Answer: (26, 23); K⁻¹ as above; the original pair is (6, 5)
Problem 18 City and Suburb Moves
Population column after one year: P times (50,000; 30,000). City: 0.8(50,000) + 0.3(30,000) = 40,000 + 9,000 = 49,000. Suburb: 0.2(50,000) + 0.7(30,000) = 10,000 + 21,000 = 31,000.
After two years: City: 0.8(49,000) + 0.3(31,000) = 39,200 + 9,300 = 48,500. Suburb: 0.2(49,000) + 0.7(31,000) = 9,800 + 21,700 = 31,500.
The total stays 80,000.
Answer: after one year 49,000 and 31,000; after two years 48,500 and 31,500
Problem 19 Solving AX = B
det A = 3(5) − 2(4) = 7, so A⁻¹ = 17 times rows (5, −2) and (−4, 3). Multiply both sides by A⁻¹ on the left: X = A⁻¹B.
Row 1: 17(5·9 − 2·19, 5·(−6) − 2·(−8)) = 17(7, −14) = (1, −2). Row 2: 17(−4·9 + 3·19, −4·(−6) + 3·(−8)) = 17(21, 0) = (3, 0).
| 1 | −2 |
| 3 | 0 |
Check: AX has first row 3(1) + 2(3) = 9 and 3(−2) + 2(0) = −6, matching B.
Answer: X has rows (1, −2) and (3, 0)
Problem 20 Inverse of a Product
AB has rows (5, 4) and (1, 2), with determinant 5(2) − 4(1) = 6. So (AB)⁻¹ = 16 times rows (2, −4) and (−1, 5).
A⁻¹ has rows (1, −2) and (0, 1). det B = 6, so B⁻¹ has rows (13, 0) and (−16, 12).
B⁻¹A⁻¹: row 1 is (13, −23); row 2 is (−16, 13 + 12) = (−16, 56).
| 13 | −23 |
| −16 | 56 |
Both sides give the same matrix (16 times rows (2, −4) and (−1, 5)).
Answer: both equal the matrix with rows (13, −23) and (−16, 56)
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