MAT1511 Assignment 5 2025 - Due July 2025

MAT1511
ASSIGNMENT 5 2025
DUE: JULY 2025 (MEMO)

, MAR1511 Assignment 5 2025

Question 1

1.1 Solve by matrices and row operations



⎧5x − 3y + 2z = 13,
⎨ 2x − y − 3z = 1,
​ ​




⎩
4x − 2y + 4z = 12



5 −3 2 13
​ 2 −1 −3 ​ ​ ​ 1 ​ ​
.
4 −2 4 12

elementary row operations to obtain an upper-triangular form:

1.
Pivot on row 1. Use the first row (pivot 5) to eliminate below:

Multiply Row 1 by 2/5 and subtract from Row 2:

R2 ← R2 − 52 R1 :
​ ​ ​ ​
[2, −1, −3 ∣ 1] − 25 [5, −3, 2 ∣ 13] = [0, −0.5, 9.5 ∣ 10.5].
​




Multiply Row 1 by 4/5 and subtract from Row 3:

R3 ← R3 − 45 R1 :
​ ​ ​ ​ [4, −2, 4 ∣ 12] − 45 [5, −3, 2 ∣ 13] = [0, 0, 10 ∣ 10].
​




(Alternatively,

The matrix is now

5 −3 2 13
​ 0 −0.5 9.5
​ ​ ​ 10.5 ​ ​ .
0 0 10 10
2.
Simplify Row 2 and Row 3. Multiply Row 2 by 2 to clear the fraction:

R2 ← 2R2 :
​ ​ [0, −1, 19, 21].

Row 3 is already simple: [0, 0, 10 ∣ 10].
3. Back-substitute for z . From Row 3: 10z = 10 gives z = 1.
4. Back-substitute for y . Substitute z = 1 into Row 2: −1 ⋅ y + 19(1) = 21. Hence −y + 19 = 21, so −y =
2, i.e. y = −2.
5. Back-substitute for x. Substitute y = −2, z = 1 into Row 1 (original equation): 5x − 3(−2) + 2(1) = 13,
which is 5x + 6 + 2 = 13. So 5x + 8 = 13, hence 5x = 5, giving x = 1.

Thus the unique solution is x = 1, y = −2, z = 1.


1.2 Determinant of A