How do I solve a system of linear equations using Gaussian elimination?

Question

I'm studying linear algebra and I need to solve systems of equations using Gaussian elimination. For example:

\begin{align*} x + 2y + 3z &= 9 \\2x - y + z &= 8 \\3x - z &= 3 \end{align*}

I know I need to form an augmented matrix and use row operations to reach row-echelon form. But what are the allowed row operations? And how do I know when I've reached the correct form? Should I aim for row-echelon form or reduced row-echelon form (RREF)?

Aisha Mohammed · Accepted Answer

Here is the complete solution. Allowed Row Operations: 1. Swap two rows (R i ↔ R j) 2. Scale a row by a non-zero constant (c R i → R i) 3. Add a multiple of one row to another (R i + c R j → R i) Augmented Matrix: ( 1 2 3 ∣ 9 2 − 1 1 ∣ 8 3 0 − 1 ∣ 3 ) Step 1: Eliminate x from rows 2 and 3. R 2 → R 2 − 2 R 1: ( 1 2 3 ∣ 9 0 − 5 − 5 ∣ − 10 3 0 − 1 ∣ 3 ) R 3 → R 3 − 3 R 1: ( 1 2 3 ∣ 9 0 − 5 − 5 ∣ − 10 0 − 6 − 10 ∣ − 24 ) Step 2: Eliminate y from row 3. R 3 → 5 R 3 − 6 R 2: ( 1 2 3 ∣ 9 0 − 5 − 5 ∣ − 10 0 0 − 20 ∣ − 60 ) Step 3: Back-substitution. From R 3: − 20 z = − 60 ⇒ z = 3 From R 2: − 5 y − 5 ( 3 ) = − 10 ⇒ − 5 y = 5 ⇒ y = − 1 From R 1: x + 2 ( − 1 ) + 3 ( 3 ) = 9 ⇒ x − 2 + 9 = 9 ⇒ x = 2 Answer: ( x , y , z ) = ( 2 , − 1 , 3 ) Row-Echelon vs. Reduced Row-Echelon (RREF): | Form | Requirements | When to Use | |---|---|---| | Row-Echelon | Leading entry in each row is 1 (pivot), zeros below each pivot | Back-substitution | | RREF | Row-echelon + zeros above each pivot as well | Directly reading solutions | Gaussian elimination stops at row-echelon form, then uses back-substitution. Gauss-Jordan elimination continues to RREF, which gives the solution directly without back-substitution.

How do I solve a system of linear equations using Gaussian elimination?

1 Answer

How do I solve a system of linear equations using Gaussian elimination?

1 Answer