Gauss elimination method | PDF - Free Printable

Looking at the problem in the image:

We are given a system of three equations:

1. x + (2/3)y = 5/3
2. y + z = 2
3. -(3/2)y - (3/2)z = -3

And we’re told to multiply the 3rd equation by (-2/3) for A₃₁ = -(-2/3). Wait — actually, looking more carefully, this seems to be part of a larger example where they’re solving a system using elimination.

But then there’s a note: “Multiply the 3rd equation by (-2/3)” — and after that, they show the augmented matrix becomes:

[1 0 1 | 3]
[0 1 1 | 2]
[0 0 1 | 1]

Then they say: “The last equation gives z = 1. The second equation now gives y = 1. Finally the first equation gives x = 1.”

So the solution is (x, y, z) = (1, 1, 1)

Let me verify this step-by-step.

Original system (from the text):

Equation 1: x + (2/3)y = 5/3
Equation 2: y + z = 2
Equation 3: -(3/2)y - (3/2)z = -3

Step 1: Multiply Equation 3 by (-2/3)

Compute:
(-2/3) * [-(3/2)y - (3/2)z] = (-2/3)*(-3)
→ (2/3)*(3/2)y + (2/3)*(3/2)z = 2
→ y + z = 2

Wait — that’s the same as Equation 2!

That can’t be right for elimination unless we’re doing something else.

Actually, looking again — perhaps the original system was different? Because in the example below (EXAMPLE 2.2.12), they have:

x + y + z = 3
y + 2z = 5
3x + 4y + 4z = 11

And they solve it via Gaussian elimination.

But the top part with the fractions might be a separate problem.

Wait — let's re-read the user’s image description.

It says:

“5. Multiply the 3rd equation by (-2/3) for A₃₁ = -(-2/3).”

Then shows:

x + (2/3)y = 5/3
y + z = 2
-(3/2)y - (3/2)z = -3

Then after multiplying 3rd eq by (-2/3), they get:

[1 0 1 | 3]
[0 1 1 | 2]
[0 0 1 | 1]

Which implies that after operations, they’ve eliminated variables.

But if I take the third equation: -(3/2)y - (3/2)z = -3

Multiply by (-2/3):

Left side: (-2/3) * [-(3/2)y - (3/2)z] = (2/3)(3/2)y + (2/3)(3/2)z = y + z
Right side: (-2/3)*(-3) = 2

So new equation: y + z = 2 — which is identical to the second equation.

That suggests dependency — but then how do they get z=1?

Perhaps the original system was meant to be different? Or maybe there’s a typo in the problem statement?

Alternatively, maybe the “multiply the 3rd equation by (-2/3)” is meant to eliminate the y-term from the third equation relative to the second? But since both become y+z=2, that doesn’t help.

Wait — perhaps the initial system is not what’s written? Let me look at the augmented matrix shown after operation:

[1 0 1 | 3]
[0 1 1 | 2]
[0 0 1 | 1]

This corresponds to:

x + z = 3
y + z = 2
z = 1

So back-substitute:

z = 1
y + 1 = 2 → y = 1
x + 1 = 3 → x = 1

So solution is (1,1,1)

Now, let’s check if this satisfies the original equations given:

Original Eq1: x + (2/3)y = 5/3 → 1 + (2/3)(1) = 1 + 2/3 = 5/3 ✓
Original Eq2: y + z = 2 → 1 + 1 = 2 ✓
Original Eq3: -(3/2)y - (3/2)z = -3 → -(3/2)(1) - (3/2)(1) = -3/2 - 3/2 = -3 ✓

Perfect! So even though multiplying the third equation by (-2/3) gives y+z=2 (same as eq2), that must have been an intermediate step in a larger elimination process — perhaps they had already used row operations to zero out other entries.

In fact, looking at the augmented matrix before the operation, it was probably:

[1 2/3 0 | 5/3]
[0 1 1 | 2 ]
[0 -3/2 -3/2| -3 ]

Then, to eliminate the y-term in row 3, they might have added (3/2) times row 2 to row 3? But the instruction says “multiply the 3rd equation by (-2/3)”.

Actually, if you multiply row 3 by (-2/3), you get:

Row3_new = (-2/3) * [0, -3/2, -3/2 | -3] = [0, 1, 1 | 2]

Which is exactly row 2. So that would make rows 2 and 3 identical — which is fine, but then to get to the final matrix, they must have done additional steps.

Perhaps the “multiply by (-2/3)” is misstated, or it’s part of a sequence.

But regardless — the final result shown is correct: after all operations, they have:

z = 1
y = 1
x = 1

And it satisfies all original equations.

So the solution is (x, y, z) = (1, 1, 1)

Final Answer:
(1, 1, 1)