MAT1511
ASSIGNMENT 4 2025
DUE: JULY 2025 (MEMO)
, MAT1511 Assessment 4
Question 1
⎧−x + 2y + 3z = 11,
⎨ 2x − 3y = −6,
⎩
3x − 3y + 3z = 3.
Form the augmented matrix and use row operations. For clarity, multiply the first equation
by –1 to start with a leading 1:
1 −2 −3 ∣ −11
2 −3 0 ∣
−6 .
3 −3 3 ∣ 3
Eliminate x from the other rows:
R2 ← R2 − 2R1 :
R2 : [2 − 2 ⋅ 1 = 0, −3 − 2(−2) = 1, 0 − 2(−3) = 6, −6 − 2(−11) = 16].
R3 ← R3 − 3R1 :
R3 : [3 − 3 = 0, −3 − 3(−2) = 3, 3 − 3(−3) = 12, 3 − 3(−11) = 36].
Now the matrix is
1 −2 −3 ∣ −11
0 1
6 ∣ 16 .
0 3 12 ∣ 36
Next eliminate y from row 3: R3← R3 − 3R2 :
R3 : [0, 3 − 3(1) = 0, 12 − 3(6) = −6, 36 − 3(16) = −12]. Divide R3 by −6:
R3 : [0, 0, 1, 2].
The matrix is now in row-echelon form:
1 −2 −3 ∣ −11
0 1
6 ∣ 16 .
0 0 1 ∣ 2
Back-substitute:
From R3 : z
= 2.
ASSIGNMENT 4 2025
DUE: JULY 2025 (MEMO)
, MAT1511 Assessment 4
Question 1
⎧−x + 2y + 3z = 11,
⎨ 2x − 3y = −6,
⎩
3x − 3y + 3z = 3.
Form the augmented matrix and use row operations. For clarity, multiply the first equation
by –1 to start with a leading 1:
1 −2 −3 ∣ −11
2 −3 0 ∣
−6 .
3 −3 3 ∣ 3
Eliminate x from the other rows:
R2 ← R2 − 2R1 :
R2 : [2 − 2 ⋅ 1 = 0, −3 − 2(−2) = 1, 0 − 2(−3) = 6, −6 − 2(−11) = 16].
R3 ← R3 − 3R1 :
R3 : [3 − 3 = 0, −3 − 3(−2) = 3, 3 − 3(−3) = 12, 3 − 3(−11) = 36].
Now the matrix is
1 −2 −3 ∣ −11
0 1
6 ∣ 16 .
0 3 12 ∣ 36
Next eliminate y from row 3: R3← R3 − 3R2 :
R3 : [0, 3 − 3(1) = 0, 12 − 3(6) = −6, 36 − 3(16) = −12]. Divide R3 by −6:
R3 : [0, 0, 1, 2].
The matrix is now in row-echelon form:
1 −2 −3 ∣ −11
0 1
6 ∣ 16 .
0 0 1 ∣ 2
Back-substitute:
From R3 : z
= 2.