Introduction to Matrices — Solutions

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1. Matrix A has order 2 × 3.

   A =

   4	−2	7
   1	0	−3

   Show Solution

   (a) A has 2 rows and 3 columns, so its order is 2 × 3.

   (b) a₁₃ = element in row 1, column 3 = 7

   (c) a₂₁ = element in row 2, column 1 = 1

   (d) Number of elements = 2 × 3 = 6 elements

2. Fluency Show Solution

   (a) 2 × 2 identity matrix:

   1	0
   0	1

   (b) 2 × 3 zero matrix:

   0	0	0
   0	0	0

   (c) 1 × 4 row matrix:

   3

   −1

   5

   2

3. Fluency Show Solution

   (a) 3 rows, 2 columns → order 3 × 2

   (b) 4 rows, 1 column → order 4 × 1 (column matrix)

   (c) 3 rows, 3 columns → order 3 × 3 (this is a 3×3 identity matrix)

   (d) 1 row, 4 columns → order 1 × 4 (row matrix)

4. Fluency Show Solution

   Both matrices are 2 × 1 (column matrices), so equality is valid. Equate each corresponding element:

   Row 1: 2x = 6 → x = 3

   Row 2: y + 1 = 4 → y = 3

   Verification: 2(3) = 6 ✓   3 + 1 = 4 ✓

5. Understanding Show Solution

   From the first matrix equation, both matrices are 2 × 1:

   Equation 1: a + 2b = 10

   Equation 2: 3c = 12 → c = 4

   From the second matrix equation:

   Equation 3: 2a = 8 → a = 4

   Equation 4: b − 1 = 1 → b = 2

   Check Equation 1: a + 2b = 4 + 2(2) = 4 + 4 = 8 ≠ 10

   Wait — let us recheck. From Equation 3: a = 4. From Equation 1: 4 + 2b = 10 → 2b = 6 → b = 3. But Equation 4 gives b − 1 = 1 → b = 2. This is a contradiction, meaning the two matrix equations are inconsistent (no single value of b satisfies both). However, as presented in this problem, we use each equation independently:

   Taking the values most directly determined: from Equation 3, a = 4; from Equation 4, b = 2; from Equation 2, c = 4.

   Note: a + 2b = 4 + 4 = 8 ≠ 10 shows an inconsistency, highlighting that overdetermined systems may have no solution. In this exercise, the intent is to use the second system to find a and b: a = 4, b = 2, c = 4.

6. Understanding Show Solution

   (a) Matrix S has 3 rows and 2 columns → order is 3 × 2.

   (b) s₃₂ = element in row 3, column 2 = 90.

   In context: there are 90 units of Product C stored in Warehouse 2.

   (c) Total stock of Product B = row 2 of S = s₂₁ + s₂₂ = 80 + 30 = 110 units

7. Understanding Show Solution

   Using m_ij = i + j for each position:

   1+1=2	1+2=3	1+3=4
   2+1=3	2+2=4	2+3=5
   3+1=4	3+2=5	3+3=6

   =

   2	3	4
   3	4	5
   4	5	6

   Special property: The matrix is symmetric — it is equal to its own transpose (m_ij = m_ji for all i, j). Each row is an arithmetic sequence with common difference 1, and each column is also an arithmetic sequence with common difference 1.

8. Understanding Show Solution

   (a) Number of elements = 4 × 3 = 12 elements

   (b) If the product of all elements in row 2 is 0, then at least one element in row 2 must equal zero. (The product of any set of numbers is 0 if and only if at least one factor is 0.)

   (c) No, M cannot be a square matrix. A square matrix requires the same number of rows and columns. M has order 4 × 3 (4 rows, 3 columns), so m ≠ n. A square matrix would need order n × n.

9. Problem Solving Show Solution

   (a) Order: 3 × 4 (3 students, 4 subjects). t₂₃ = element in row 2, column 3 = 83 (Student 2’s score in Subject 3).

   (b) Row totals:

   • Student 1: 72 + 85 + 68 + 91 = 316
   • Student 2: 90 + 78 + 83 + 75 = 326
   • Student 3: 64 + 70 + 88 + 79 = 301

   Student 2 has the highest total score (326).

   (c) Subject column averages (out of 100):

   • Subject 1: (72 + 90 + 64) ÷ 3 = 226 ÷ 3 ≈ 75.3
   • Subject 2: (85 + 78 + 70) ÷ 3 = 233 ÷ 3 ≈ 77.7
   • Subject 3: (68 + 83 + 88) ÷ 3 = 239 ÷ 3 ≈ 79.7
   • Subject 4: (91 + 75 + 79) ÷ 3 = 245 ÷ 3 ≈ 81.7

   Subject 4 has the highest class average (≈81.7).

   (d) Improvement from Subject 1 to Subject 4 (change = Subject 4 − Subject 1):

   • Student 1: 91 − 72 = +19
   • Student 2: 75 − 90 = −15
   • Student 3: 79 − 64 = +15

   Student 1 showed the greatest improvement (+19 marks).

10. Problem Solving Show Solution

    (a) Both matrices are 2 × 1. Equating corresponding elements gives two equations:

    Equation 1: 2x + y = 7

    Equation 2: x − y = 1

    (b) Solving simultaneously — add the equations:

    (2x + y) + (x − y) = 7 + 1

    3x = 8 → x = 8/3

    Wait — let’s try again: 3x = 8, so x = 8/3. Then from Equation 2: y = x − 1 = 8/3 − 1 = 5/3.

    Alternatively: adding gives 3x = 8, x = 8/3 ≈ 2.67; y = 8/3 − 1 = 5/3 ≈ 1.67.

    So x = 8/3 and y = 5/3.

    (c) Verification:

    2x + y = 2(8/3) + 5/3 = 16/3 + 5/3 = 21/3 = 7 ✓

    x − y = 8/3 − 5/3 = 3/3 = 1 ✓

    Both equations are satisfied. The solution is confirmed.