Question

linear equations using gauss elimination method x+y+z=2 2x-3y=5 -3x+2z=1

Answer 1

Answer on Question #63499 – Math – Linear Algebra

Question

Solve the system of linear equations using Gauss elimination method:

$$\begin{array}{l} x + y + z = 2 \\ 2x - 3y = 5 \\ -3x + 2z = 1. \end{array}$$

Solution

Write down the augmented matrix:

$$\begin{array}{c c c c c} 1 & 1 & 1 & 2 \\ 2 & -3 & 0 & 5 \\ -3 & 0 & 2 & 1. \end{array}$$

Reduce this to row echelon form by using elementary row operations:

a) $-2 * R1 + R2 \rightarrow R2; 3 * R1 + R3 \rightarrow R3$:

$$\begin{array}{c c c c c} 1 & 1 & 1 & 2 \\ 0 & -5 & -2 & 1 \\ 0 & 3 & 5 & 7 \end{array}$$

b) $3 * R2 + 5 * R3 \rightarrow R3$:

$$\begin{array}{c c c c c} 1 & 1 & 1 & 2 \\ 0 & -5 & -2 & 1 \\ 0 & 0 & 19 & 38 \end{array}$$

c) $1/19 * R3 \rightarrow R3$:

$$\begin{array}{c c c c c} 1 & 1 & 1 & 2 \\ 0 & -5 & -2 & 1 \\ 0 & 0 & 1 & 2. \end{array}$$

Changing back to system of equations, we have:

$$\begin{array}{l} x + y + z = 2 \\ -5y - 2z = 1 \\ z = 2. \end{array}$$

Now, the solution can be obtained by back substitution:

$$\begin{array}{l} z = 2, \\ y = \frac{(1 + 2z)}{-5} = (1 + 2 * 2)/(-5) = -1, \\ x = 2 - y - z = 2 - (-1) - 2 = 1. \end{array}$$

The solution is $(x, y, z) = (1, -1, 2)$.

**Answer:** $(x, y, z) = (1, -1, 2)$.

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