Matrix Operations — Solutions

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1. Fluency Show Solution

   Both matrices are 2 × 2 → addition is defined.

   A + B =

   5+4	−3+1
   2+(−6)	7+3

   =

   9	−2
   −4	10

2. Fluency Show Solution

   Both matrices are 2 × 3 → subtraction is defined.

   A − B =

   8−2	3−(−4)	−1−3
   5−(−1)	−2−5	6−2

   =

   6	7	−4
   6	−7	4

3. Fluency Show Solution

   (a) 3A = 3 × every element of A:

   3A =

   9	−6
   3	12

   (b) −2B = −2 × every element of B:

   −2B =

   2	−6
   −4	2

   (c) 3A − 2B = 3A + (−2B):

   3A − 2B =

   9+2	−6+(−6)
   3+(−4)	12+2

   =

   11	−12
   −1	14

4. Fluency Show Solution

   Rearrange: X = B − A

   X =

   7−2	2−5
   4−(−1)	−1−3

   =

   5	−3
   5	−4

   Verify: X + A =

   5+2	−3+5
   5+(−1)	−4+3

   =

   7	2
   4	−1

   = B ✓

5. Understanding Show Solution

   (a) A + B =

   6	1
   3	6

   (b) 2B =

   10	−2
   0	4

   . 2B − C =

   8	−5
   1	3

   (c) A + B + C =

   8	4
   2	7

   (d) B + A =

   6	1
   3	6

   = A + B ✓ (commutative property verified)

6. Understanding Show Solution

   (a) M + D =

   28	27
   35	25

   (b) (M + D) − S =

   28−14	27−11
   35−18	25−6

   =

   14	16
   17	19

   (c) All closing stock levels are positive (14, 16, 17, 19). No stock has dropped below zero, so no stockout occurred.

7. Understanding Show Solution

   First expand: 2

   a	3
   1	b

   =

   2a	6
   2	2b

   Then subtract:

   2a−4	6−(−1)
   2−3	2b−5

   =

   0	7
   −1	3

   Equate corresponding elements:

   Position (1,1): 2a − 4 = 0 → 2a = 4 → a = 2

   Position (1,2): 6 − (−1) = 7 = 7 ✓ (consistent, no unknown)

   Position (2,1): 2 − 3 = −1 = −1 ✓ (consistent)

   Position (2,2): 2b − 5 = 3 → 2b = 8 → b = 4

8. Understanding Show Solution

   Each reading is 2°C too high, so we subtract 2 from every element. This is the same as subtracting 2 × (3×2 matrix of ones), or simply T − 2J where J is the 3×2 matrix of all 1s.

   Corrected matrix T_c = T − 2 =

   28−2	32−2
   25−2	30−2
   31−2	27−2

   =

   26	30
   23	28
   29	25

   This is scalar subtraction: subtracting a constant from every element is equivalent to subtracting the scalar multiplied by the same-order matrix of all 1s.

9. Problem Solving Show Solution

   (a) J + F =

   $1550	$950
   $580	$420

   (b) Monthly average = ½(J + F) =

   $775	$475
   $290	$210

   (c) F − J =

   750−800	500−450
   280−300	220−200

   =

   −$50	+$50
   −$20	+$20

   Column 1 (Actual spending): Food decreased by $50, Rent decreased by $20. No category increased in actual spending. The budgeted column (column 2) both increased by $50 and $20 respectively.

10. Problem Solving Show Solution

    (a) For kA = B, we need k × every element of A to equal the corresponding element of B.

    Checking position (1,1): k × 2 = 6 → k = 3

    Checking position (1,2): k × 4 = 12 → k = 3 ✓

    Checking position (2,1): k × 1 = 3 → k = 3 ✓

    Checking position (2,2): k × 3 = 9 → k = 3 ✓

    The only value is k = 3.

    (b) kA = B + A means kA = 3A + A = 4A (since B = 3A). So k = 4.

    Verify: 4A =

    8	16
    4	12

    and B + A =

    8	16
    4	12

    ✓

    (c) B = 3A. Matrix B is exactly 3 times matrix A — every element of B is 3 times the corresponding element of A. We say B is a scalar multiple of A.