Q107 | PDF | System Of Linear Equations | Analysis - Free Printable

You've uploaded a worksheet titled “Intermediate Algebra Skill: Solving 3×3 Linear Systems by Gaussian Elimination”. It contains 14 systems of three linear equations with three variables (x, y, z). Since you asked to “solve the problem and explain the solution,” I’ll solve Problem #1 step-by-step using Gaussian Elimination, as requested in the title. If you’d like solutions for other problems, let me know!

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## ✔ Problem 1:

Solve the system:
\[
\begin{cases}
4x + 2y - 6z = 34 \quad \text{(Equation 1)} \\
2x + y + 3z = 3 \quad \text{(Equation 2)} \\
6x + 3y - 3z = 37 \quad \text{(Equation 3)}
\end{cases}
\]

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## 🧭 Step 1: Write the Augmented Matrix

We represent the system as an augmented matrix:

\[
\left[\begin{array}{ccc|c}
4 & 2 & -6 & 34 \\
2 & 1 & 3 & 3 \\
6 & 3 & -3 & 37
\end{array}\right]
\]

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## 🧮 Step 2: Use Row Operations to Get Upper Triangular Form (Row-Echelon Form)

Our goal is to get zeros below the main diagonal.

➤ First Pivot: Make the element at (1,1) = 1 or keep it as pivot.

It’s easier to swap rows to get smaller numbers, but we can proceed.

Let’s eliminate the x-term from Rows 2 and 3 using Row 1.

#### 🔁 Row Operation 1: R2 ← R2 - (1/2)R1

Why? Because 2 (in R2) is half of 4 (in R1).

Compute:

- New R2:
- x: \(2 - \frac{1}{2}(4) = 0\)
- y: \(1 - \frac{1}{2}(2) = 0\)
- z: \(3 - \frac{1}{2}(-6) = 3 + 3 = 6\)
- Constant: \(3 - \frac{1}{2}(34) = 3 - 17 = -14\)

So new R2: \([0\quad 0\quad 6\quad |\quad -14]\)

Wait — this gives us a row with no y-term! That’s unusual. Let’s check our arithmetic again.

Actually, let's try a different approach to avoid fractions and errors.

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## 🔄 Alternate Strategy: Eliminate x from R2 and R3 using R1

We can use multiples to avoid fractions.

➤ Eliminate x from R2: R2 ← 2*R2 - R1

Multiply R2 by 2:
→ [4, 2, 6 | 6]
Subtract R1: [4, 2, -6 | 34]
→ New R2: [0, 0, 12 | -28]

Simplify by dividing by 4: → [0, 0, 3 | -7] ✔

➤ Eliminate x from R3: R3 ← 2*R3 - 3*R1

Compute:

2*R3: [12, 6, -6 | 74]
3*R1: [12, 6, -18 | 102]
Subtract: [0, 0, 12 | -28]

Same as above! So new R3: [0, 0, 12 | -28] → divide by 4 → [0, 0, 3 | -7]

Now our matrix becomes:

\[
\left[\begin{array}{ccc|c}
4 & 2 & -6 & 34 \\
0 & 0 & 3 & -7 \\
0 & 0 & 3 & -7
\end{array}\right]
\]

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## ⚠️ Step 3: Analyze the Result

Look at Rows 2 and 3 — they are identical after simplification.

This means we have dependent equations — one is redundant.

So effectively, we only have two independent equations for three variables → infinitely many solutions? Or maybe inconsistent?

Wait — let’s write the simplified system:

From Row 1: \(4x + 2y - 6z = 34\)
From Row 2: \(3z = -7\) → \(z = -\frac{7}{3}\)

But Row 3 also says \(3z = -7\) → same thing.

So we have:

- \(z = -\frac{7}{3}\)
- Plug into Row 1: \(4x + 2y - 6(-\frac{7}{3}) = 34\)
→ \(4x + 2y + 14 = 34\)
→ \(4x + 2y = 20\)
→ Divide by 2: \(2x + y = 10\) → Equation A

Now, what about Row 2 originally? We used R2 to eliminate x, but we lost y entirely. Let’s go back to original Equation 2:

Original Eq2: \(2x + y + 3z = 3\)

Plug in \(z = -\frac{7}{3}\):

→ \(2x + y + 3(-\frac{7}{3}) = 3\)
→ \(2x + y - 7 = 3\)
→ \(2x + y = 10\) → Same as Equation A!

So all three equations reduce to:

- \(2x + y = 10\)
- \(z = -\frac{7}{3}\)

This means we have one free variable — say, let \(x = t\), then \(y = 10 - 2t\), and \(z = -\frac{7}{3}\)

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## ✔ Final Answer (Parametric Form):

The system has infinitely many solutions given by:

\[
\boxed{
\begin{cases}
x = t \\
y = 10 - 2t \\
z = -\dfrac{7}{3}
\end{cases}
}
\quad \text{for any real number } t
\]

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## 📌 Summary of Gaussian Elimination Steps Used:

1. Wrote augmented matrix.
2. Used row operations to eliminate x from Rows 2 and 3.
3. Found that Rows 2 and 3 became identical → dependent system.
4. Solved for z directly.
5. Substituted back to find relation between x and y.
6. Expressed solution parametrically since one variable is free.

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## ❗ Note:

This system is dependent — not inconsistent (no contradiction), and not unique (infinite solutions). This is common when one equation is a linear combination of others.

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If you'd like me to solve any other problem from this list (e.g., #2, #8, #14, etc.), just say which one! I can solve them all if needed.

Would you like to continue with Problem #2?