ACT Math study guide · Chapter 5
Vectors and Matrices
A vector has both magnitude and direction, and is written as an ordered pair (x, y). Vectors add and subtract by combining their corresponding components separately. The magnitude of a vector is found with the same formula as the distance formula: √(x² + y²), since a vector's magnitude is its length from the origin.
A matrix is added, subtracted, or scaled entry by entry. Adding two matrices combines corresponding entries; multiplying a matrix by a scalar multiplies every entry by that scalar. The determinant of a 2×2 matrix [[a, b], [c, d]] is ad − bc, a single number used in later work such as finding an inverse or classifying a system of equations.
Matrix multiplication is not entry-by-entry. Each entry of the product comes from a row of the first matrix dotted with a column of the second: multiply corresponding entries and add the results. Multiplication is only defined when the number of columns in the first matrix equals the number of rows in the second, and the result takes the number of rows of the first matrix and the number of columns of the second.
Where this appears as a model: two velocities combined. A boat moving at (3,4) miles per hour relative to the water, on a current moving at (1,0) miles per hour, has a resultant ground velocity of (4,4), a speed of √32≈5.66 miles per hour.
▸ Matrix multiplication
Worked Example Let A = [[2, 1], [0, 3]] and B = [[4, −1], [2, 5]]. Find AB.
Each entry of the product is a row of A dotted with a column of B.
Row 1, Col 1: (2)(4) + (1)(2) = 10
Row 1, Col 2: (2)(−1) + (1)(5) = 3
Row 2, Col 1: (0)(4) + (3)(2) = 6
Row 2, Col 2: (0)(−1) + (3)(5) = 15
AB = [[10, 3], [6, 15]]
▸ Systems as matrix equations
A system of two linear equations can be written as a single matrix equation. Each row of the coefficient matrix holds one equation's coefficients, in the same order as the variable column, and multiplies out to equal the constant on that equation's right side.
Worked Example Write the system 3x + 2y = 8 and x − y = 1 as a matrix equation.
Each row of the coefficient matrix matches one equation's coefficients against the constant.
[[3, 2], [1, −1]] · [x, y] = [8, 1]
Worked Example Solve the matrix equation 2X + [[1, 2], [3, 4]] = [[7, 8], [9, 10]] for matrix X.
Recognize. This is matrix algebra: isolate X the same way a variable would be isolated, using matrix subtraction and then scalar division, entry by entry.
Set up.
2X = [[7, 8], [9, 10]] − [[1, 2], [3, 4]] = [[6, 6], [6, 6]]
Solve.
X = [[6, 6], [6, 6]] ÷ 2 = [[3, 3], [3, 3]]
Check. 2[[3,3],[3,3]] + [[1,2],[3,4]] = [[6,6],[6,6]] + [[1,2],[3,4]] = [[7,8],[9,10]], which matches the original equation.
Best route. Hand arithmetic on each entry. A familiar approved handheld calculator with matrix functions can also perform this directly; the ACT's embedded Desmos tool has no dedicated matrix mode to rely on here.
Common error. Subtracting the matrices in the wrong order, or forgetting to divide every entry by the scalar 2.
▸ Practice
1. Find the determinant of [[5, 2], [3, 4]].
A) 14 B) 8 C) −2 D) 26
2. Vector u = (5, 12). Find its magnitude.
F) 17 G) 13 H) 7 J) 169
3. M is 2×3 and N is 2×2. Which product is defined, and what are its dimensions?
A) MN, resulting in 2×2 B) NM, resulting in 2×3 C) MN, resulting in 3×3 D) Neither product is defined
4. Vectors a = (2, −3) and b = (−5, 1). Find a + b.
F) (−3, −2) G) (7, −4) H) (−3, 4) J) (3, 2)
5. Multiply [[1, −2], [3, 0]] by the scalar −3.
A) [[3, −6], [9, 0]] B) [[−3, 6], [−9, 0]] C) [[−3, −6], [−9, 0]] D) [[−4, 1], [0, −3]]
6. Which matrix equation represents the system 2x + 3y = 7 and x − y = 1?
F) [[2,1],[3,−1]]·[x,y] = [7,1] G) [[2,3],[1,−1]]·[x,y] = [1,7] H) [[3,2],[−1,1]]·[x,y] = [7,1] J) [[2,3],[1,−1]]·[x,y] = [7,1]
7. Let A = [[2, 5], [−1, 3]] and B = [[4, −2], [3, 1]]. Find A + B.
A) [[6, 3], [2, 4]] B) [[−2, 7], [−4, 2]] C) [[6, −3], [2, 4]] D) [[2, 4], [6, 3]]
8. Let C = [[1, 2], [3, 4]] and D = [[2, 0], [1, 3]]. Find CD.
F) [[4, 6], [10, 12]] G) [[2, 0], [3, 12]] H) [[4, 6], [8, 12]] J) [[2, 6], [10, 4]]
9. Find the determinant of [[7, 3], [2, 5]].
A) 41 B) 29 C) 1 D) 35
10. Let u = (4, −3) and v = (−2, 6). Find u + v.
F) (6, −9) G) (2, 3) H) (−6, 9) J) (2, −3)
11. Find the magnitude of the vector (4, −3).
A) 5 B) 7 C) 25 D) 1
12. Let M = [[3, −2], [5, 4]]. Find −2M.
F) [[6, −4], [10, 8]] G) [[−6, −4], [−10, 8]] H) [[6, 4], [10, −8]] J) [[−6, 4], [−10, −8]]
13. Which matrix equation represents the system 4x − y = 10 and 2x + 3y = 8?
A) [[4,2],[−1,3]]·[x,y] = [10,8] B) [[4,−1],[2,3]]·[x,y] = [8,10] C) [[4,−1],[2,3]]·[x,y] = [10,8] D) [[2,3],[4,−1]]·[x,y] = [10,8]
14. Let u = (7, 1) and v = (2, 5). Find the magnitude of u − v.
F) 6.40 G) 9 H) 3 J) 1
15. Let A = [[4, −1], [2, 3]] and B = [[1, 5], [−3, 2]]. Find A + B.
A) [[3, −6], [5, 1]] B) [[5, −6], [−1, 1]] C) [[5, 4], [−1, 5]] D) [[5, 4], [5, 5]]
16. Find the magnitude of the vector (7, −24).
F) 17 G) 31 H) 25 J) 7
17. Find the determinant of [[9, 4], [2, 6]].
A) 62 B) 38 C) 44 D) 46
18. Let C = [[2, 0], [1, 3]] and D = [[4, 1], [2, 5]]. Find CD.
F) [[6, 1], [6, 8]] G) [[8, 2], [6, 16]] H) [[8, 2], [10, 16]] J) [[8, 1], [10, 15]]
19. Let u = (8, 3) and v = (−2, 6). Find u − v.
A) (6, −3) B) (6, 9) C) (10, 9) D) (10, −3)
20. Solve the system 2x + y = 11 and x − 2y = −7. Find x + y.
F) 6 G) 8 H) 9 J) 7
▸ Answers and Explanations
1: A. (5)(4) − (2)(3) = 20 − 6 = 14. Choice D adds the two products instead of subtracting, computing (5)(4) + (2)(3) = 26.
2: G. √(5²+12²) = √169 = 13. Choice F adds the two components directly (5+12=17) instead of squaring them first.
3: B. Multiplication requires the columns of the first matrix to equal the rows of the second. Cols(M)=3 ≠ rows(N)=2, so MN is undefined; cols(N)=2 = rows(M)=2, so NM is defined and produces a 2×3 result. Choice A assumes MN is defined without checking that the columns of M (3) must match the rows of N (2).
4: F. Add corresponding components: (2+(−5), −3+1) = (−3, −2). Choice G subtracts the components instead of adding them, computing (2−(−5), −3−1) = (7, −4).
5: B. Multiply every entry by −3: 1·(−3)=−3, −2·(−3)=6, 3·(−3)=−9, 0·(−3)=0. Choice A drops the negative sign on the scalar, multiplying every entry by 3 instead of −3.
6: J. Row 1 of the coefficient matrix must match the first equation's coefficients (2, 3) against (x, y), equal to 7; row 2 matches (1, −1) equal to 1. Choice F swaps the coefficients within the first row, reversing which variable each column represents.
7: A. Add corresponding entries: (2+4, 5+(−2); −1+3, 3+1) = [[6, 3], [2, 4]]. Choice C reverses the sign of one entry, computing 5+(−2) as −3 instead of 3.
8: F. Row 1, Col 1: (1)(2)+(2)(1)=4. Row 1, Col 2: (1)(0)+(2)(3)=6. Row 2, Col 1: (3)(2)+(4)(1)=10. Row 2, Col 2: (3)(0)+(4)(3)=12. CD=[[4,6],[10,12]]. Choice H contains an arithmetic slip in computing the entry in row 2, column 1.
9: B. (7)(5)−(3)(2)=35−6=29. Choice A adds the two products instead of subtracting, computing (7)(5) + (3)(2) = 41.
10: G. Add corresponding components: (4+(−2), −3+6)=(2, 3). Choice F subtracts the components instead of adding them.
11: A. √(4²+(−3)²)=√25=5. Choice B adds the absolute values of the components (4+3=7) instead of applying the magnitude formula.
12: J. Multiply every entry by −2: [[−6, 4], [−10, −8]]. Choice F drops the negative sign on the scalar, multiplying every entry by 2 instead of −2.
13: C. Row 1 of the coefficient matrix matches the first equation's coefficients (4, −1) against (x, y), equal to 10; row 2 matches (2, 3) equal to 8. Choice B swaps the order of the constants on the right side, matching them to the wrong equations.
14: F. u−v=(7−2, 1−5)=(5, −4); magnitude=√(25+16)=√41≈6.40. Choice G adds the absolute differences of the components (5+4=9) instead of applying the magnitude formula to u−v.
15: C. Add corresponding entries: (4+1, −1+5, 2+(−3), 3+2) gives [[5, 4], [−1, 5]]. Choice B contains arithmetic slips in two of the four matrix entries.
16: H. √(7²+(−24)²)=√(49+576)=√625=25. Choice G adds the absolute values of the components (7+24=31) instead of applying the magnitude formula.
17: D. (9)(6)−(4)(2)=54−8=46. Choice A adds the two products instead of subtracting, computing (9)(6) + (4)(2) = 62.
18: H. Row 1, Col 1: (2)(4)+(0)(2)=8. Row 1, Col 2: (2)(1)+(0)(5)=2. Row 2, Col 1: (1)(4)+(3)(2)=10. Row 2, Col 2: (1)(1)+(3)(5)=16. CD=[[8,2],[10,16]]. Choice G drops the first term when computing one entry, using only (3)(2) instead of (1)(4) + (3)(2).
19: D. Subtract corresponding components: (8−(−2), 3−6)=(10, −3). Choice A drops the negative sign on the second vector's first component, computing 8−2 instead of 8−(−2).
20: G. From the second equation, x=2y−7; substituting gives 2(2y−7)+y=11, so 5y=25, y=5, x=3; x+y=8. Choice F reports 2x instead of the requested sum x + y.