Solving Systems of Equations Using Matrices — Solutions

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1. Fluency Show Solution

   A =

   3	2
   1	−1

   ,  X =

   x

   y

   ,  B =

   11

   1

2. Fluency Show Solution

   A =

   2	1
   1	1

   ,  B =

   7

   5

   det(A) = 2×1 − 1×1 = 1

   A⁻¹ = (1/1)

   1	−1
   −1	2

   X = A⁻¹B =

   1×7 − 1×5

   −1×7 + 2×5

   =

   2

   3

   x = 2, y = 3. Check: 2(2)+3=7 ✓ and 2+3=5 ✓

3. Fluency Show Solution

   A =

   4	−1
   2	1

   ,  B =

   9

   3

   det(A) = 4×1 − (−1)×2 = 4 + 2 = 6

   A⁻¹ = ⅙

   1	1
   −2	4

   X = ⅙

   9 + 3

   −18 + 12

   = ⅙

   12

   −6

   =

   2

   −1

   x = 2, y = −1. Check: 4(2)−(−1)=9 ✓ and 2(2)+(−1)=3 ✓

4. Fluency Show Solution

   A =

   1	3
   2	−1

   ,  B =

   10

   1

   det(A) = −1 − 6 = −7

   A⁻¹ = −&frac17;

   −1	−3
   −2	1

   = &frac17;

   1	3
   2	−1

   X = &frac17;

   10 + 3

   20 − 1

   = &frac17;

   13

   19

   x = 13/7, y = 19/7. Check: 13/7 + 3(19/7) = 13/7 + 57/7 = 70/7 = 10 ✓

5. Understanding Show Solution

   det(A) = 1×4 − 2×2 = 4 − 4 = 0.

   The determinant is zero, so A is singular (no inverse exists). The matrix method cannot be applied. Geometrically, the two equations represent parallel lines (no intersection → no solution) or the same line (infinitely many solutions). To determine which, check if the equations are consistent.

   Equation 1: x + 2y = e. Equation 2: 2x + 4y = f = 2e (if consistent) or f ≠ 2e (no solution).

6. Understanding Show Solution

   A =

   3	5
   1	2

   ,  B =

   1

   0

   det(A) = 3×2 − 5×1 = 6 − 5 = 1

   A⁻¹ =

   2	−5
   −1	3

   X = A⁻¹B =

   2×1 − 5×0

   −1×1 + 3×0

   =

   2

   −1

   x = 2, y = −1. Verify: 3(2)+5(−1)=6−5=1 ✓ and 2+2(−1)=0 ✓

7. Understanding Show Solution

   Let L = number of lattes, C = number of cappuccinos.

   System: L + C = 40 and 5L + 4C = 182

   A =

   1	1
   5	4

   ,  B =

   40

   182

   det(A) = 4 − 5 = −1

   A⁻¹ = −1 ×

   4	−1
   −5	1

   =

   −4	1
   5	−1

   X =

   −160 + 182

   200 − 182

   =

   22

   18

   22 lattes and 18 cappuccinos. Check: 22+18=40 ✓ and 5(22)+4(18)=110+72=182 ✓

8. Understanding Show Solution

   A =

   2	−3
   1	1

   ,  B =

   4

   3

   det(A) = 2(1) − (−3)(1) = 2 + 3 = 5

   A⁻¹ = ⅕

   1	3
   −1	2

   X = ⅕

   4 + 9

   −4 + 6

   = ⅕

   13

   2

   x = 13/5 = 2.6, y = 2/5 = 0.4. Check: 2(2.6)−3(0.4)=5.2−1.2=4 ✓ and 2.6+0.4=3 ✓

9. Problem Solving Show Solution

   (a) Let a = adult tickets, c = concession tickets.

   a + c = 120   and   15a + 9c = 1470

   A =

   1	1
   15	9

   ,  B =

   120

   1470

   (b) det(A) = 9 − 15 = −6

   A⁻¹ = −⅙

   9	−1
   −15	1

   X = −⅙

   9(120) − 1470

   −15(120) + 1470

   = −⅙

   −390

   −330

   =

   65

   55

   65 adult tickets and 55 concession tickets.

   (c) Check: 65 + 55 = 120 ✓ and 15(65) + 9(55) = 975 + 495 = 1470 ✓

10. Problem Solving Show Solution

    Let x = kg of Alloy X, y = kg of Alloy Y.

    Equation 1 (total mass): x + y = 100

    Equation 2 (copper content): 0.60x + 0.30y = 0.45 × 100 = 45

    A =

    1	1
    0.60	0.30

    ,  B =

    100

    45

    det(A) = 0.30 − 0.60 = −0.30

    A⁻¹ = (1/−0.30)

    0.30	−1
    −0.60	1

    X = (1/−0.30)

    0.30(100) − 45

    −0.60(100) + 45

    = (1/−0.30)

    −15

    −15

    =

    50

    50

    50 kg of Alloy X and 50 kg of Alloy Y.

    Check: 50 + 50 = 100 ✓ and 0.60(50) + 0.30(50) = 30 + 15 = 45 ✓ (45% of 100 = 45 kg copper)