7. Evaluate: 1×3 by 3×1
\( \begin{bmatrix} 4 & -1 & 3 \end{bmatrix} \begin{bmatrix} 2 \\ 5 \\ -2 \end{bmatrix} \)
Dim: \(1 \times 3\) \(\times\) \(3 \times 1\) | Defined? [✓] Yes [ ] No
Result Size: \(1 \times 1\)
Work & Product Answer:
\(4(2) + (-1)(5) + 3(-2) = 8 - 5 - 6\) \([-3]\)
8. Evaluate: 3×2 by 3×2
\( \begin{bmatrix} 5 & -2 \\ 1 & 4 \\ 0 & 3 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -2 & 6 \\ 4 & 0 \end{bmatrix} \)
Dim: \(3 \times 2\) \(\times\) \(3 \times 2\) | Defined? [ ] Yes [✓] No
Result Size / Status: Undefined
Answer / Explanation:
Undefined: Inner dimensions \(2 \neq 3\). Columns of 1st (\(2\)) \(\neq\) rows of 2nd (\(3\)).
9. Given \(P = \begin{bmatrix} 2 & -3 \\ 1 & 4 \end{bmatrix}\) and \(Q = \begin{bmatrix} 5 & 1 \\ 0 & -2 \end{bmatrix}\), find \(P \times Q\):
Product Size: 2 × 2
Row-by-Column Calculations:
R1 \(\times\) C1: \(2(5) + (-3)(0) = 10\)
R1 \(\times\) C2: \(2(1) + (-3)(-2) = 2 + 6 = \mathbf{8}\)
R2 \(\times\) C1: \(1(5) + 4(0) = \mathbf{5}\)
R2 \(\times\) C2: \(1(1) + 4(-2) = 1 - 8 = \mathbf{-7}\)
Answer \(PQ\):
\( \begin{bmatrix} 10 & 8 \\ 5 & -7 \end{bmatrix} \)
Solve each \(2 \times 2\) system using the designated algebraic or graphical method. Show complete steps.
10. Solve by Graphing:
\(\begin{cases} \color{#2563eb}{y = -\frac{1}{2}x + 4} \\ \color{#059669}{y = x + 1} \end{cases}\)
2 4 6 -2 -4 -6 2 4 6 -2 -4 -6
Intersection Point \((x, y)\):
(2, 3)
11. Solve by Elimination:
\(\begin{cases} 4x + 3y = 10 \\ 2x - 5y = -8 \end{cases}\)
Multiply to cancel a variable, add equations, and solve:
Step 1: Multiply eq 2 by \(-2\):
\(-2(2x - 5y = -8) \implies \mathbf{-4x + 10y = 16}\)
Step 2: Add to equation 1:
\( (4x + 3y = 10) + (-4x + 10y = 16) \)
\(13y = 26 \implies \mathbf{y = 2}\)
Step 3: Substitute \(y = 2\) into eq 1:
\(4x + 3(2) = 10 \implies 4x = 4 \implies \mathbf{x = 1}\)
Solution \((x, y)\):
(1, 2)
TEACHER ANSWER KEY • UNIT 3 TEST REVIEW Page 2 of 3
TEACHER ANSWER KEY
Answer Key • Calculator RREF
Write the augmented matrix \([A \,|\, B]\), record the RREF output from your matrix calculator, and state the solution.
12. Solve the \(2 \times 2\) System: 2 × 3 Augmented Matrix
\[ \begin{cases} 2x + 5y = 1 \\ 3x - 2y = 11 \end{cases} \]
1. Augmented Matrix:
\( \left[\begin{array}{cc|c} 2 & 5 & 1 \\ 3 & -2 & 11 \end{array}\right] \)
2. RREF Output:
\( \begin{bmatrix} 1 & 0 & 3 \\ 0 & 1 & -1 \end{bmatrix} \)
3. Solution:
(3, -1) \(x=3, y=-1\)
13. Solve the \(3 \times 3\) System: Missing Variable (0 in row 2!)
3 × 4 Augmented Matrix
\[ \begin{cases} x + 2y + z = 3 \\ 2x - 3z = -5 \\ 3x + y - 2z = -1 \end{cases} \]
1. Augmented Matrix:
\( \left[\begin{array}{ccc|c} 1 & 2 & 1 & 3 \\ 2 & \mathbf{0} & -3 & -5 \\ 3 & 1 & -2 & -1 \end{array}\right] \)
2. RREF Output:
\( \begin{bmatrix} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 3 \end{bmatrix} \)
3. Solution:
(2, -1, 3) \(x=2\) \(y=-1\) \(z=3\)
14. The Athletic Concession Stand 0 in row 3!
3 × 4 Augmented Matrix
The high school booster club tracks food and beverage sales at three sporting events: At Game 1, they sold 40 hot dogs, 30 bags of popcorn, and 50 sodas for $350. At Game 2, they sold 50 hot dogs, 20 bags of popcorn, and 60 sodas for $380. At Game 3, they ran out of supplies for hot dogs, but they still sold 50 bags of popcorn and 70 sodas for $290. Represent this scenario with an augmented matrix and solve using a matrix calculator to determine the individual price of each item.
Augmented Matrix Used:
\( \left[\begin{array}{ccc|c} 40 & 30 & 50 & 350 \\ 50 & 20 & 60 & 380 \\ \mathbf{0} & 50 & 70 & 290 \end{array}\right] \)
RREF: \(\begin{bmatrix} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 2 \end{bmatrix}\)
Final Calculated Item Prices:
Hot Dog:
$4.00
Popcorn:
$3.00
Soda:
$2.00
TEACHER ANSWER KEY • UNIT 3 TEST REVIEW Page 3 of 3
