Linear Algebra Examples

$3 y + 2 z = 2$ , $2 x - y - 3 z = 2$ , $2 x + 2 y - z = 4$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 0 & 3 & 2 & 2 \\ 2 & - 1 & - 3 & 2 \\ 2 & 2 & - 1 & 4 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

Swap $R_{2}$ with $R_{1}$ to put a nonzero entry at $1, 1$ .

$[\begin{array}{c} 2 & - 1 & - 3 & 2 \\ 0 & 3 & 2 & 2 \\ 2 & 2 & - 1 & 4 \end{array}]$

Step 2.2

Multiply each element of $R_{1}$ by $\frac{1}{2}$ to make the entry at $1, 1$ a $1$ .

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Step 2.2.1

Multiply each element of $R_{1}$ by $\frac{1}{2}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{2}{2} & \frac{- 1}{2} & \frac{- 3}{2} & \frac{2}{2} \\ 0 & 3 & 2 & 2 \\ 2 & 2 & - 1 & 4 \end{array}]$

Step 2.2.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{3}{2} & 1 \\ 0 & 3 & 2 & 2 \\ 2 & 2 & - 1 & 4 \end{array}]$

Step 2.3

Perform the row operation $R_{3} = R_{3} - 2 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 2.3.1

Perform the row operation $R_{3} = R_{3} - 2 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{3}{2} & 1 \\ 0 & 3 & 2 & 2 \\ 2 - 2 \cdot 1 & 2 - 2 (- \frac{1}{2}) & - 1 - 2 (- \frac{3}{2}) & 4 - 2 \cdot 1 \end{array}]$

Step 2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{3}{2} & 1 \\ 0 & 3 & 2 & 2 \\ 0 & 3 & 2 & 2 \end{array}]$

Step 2.4

Multiply each element of $R_{2}$ by $\frac{1}{3}$ to make the entry at $2, 2$ a $1$ .

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Step 2.4.1

Multiply each element of $R_{2}$ by $\frac{1}{3}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{3}{2} & 1 \\ \frac{0}{3} & \frac{3}{3} & \frac{2}{3} & \frac{2}{3} \\ 0 & 3 & 2 & 2 \end{array}]$

Step 2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{3}{2} & 1 \\ 0 & 1 & \frac{2}{3} & \frac{2}{3} \\ 0 & 3 & 2 & 2 \end{array}]$

Step 2.5

Perform the row operation $R_{3} = R_{3} - 3 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 2.5.1

Perform the row operation $R_{3} = R_{3} - 3 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{3}{2} & 1 \\ 0 & 1 & \frac{2}{3} & \frac{2}{3} \\ 0 - 3 \cdot 0 & 3 - 3 \cdot 1 & 2 - 3 (\frac{2}{3}) & 2 - 3 (\frac{2}{3}) \end{array}]$

Step 2.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{3}{2} & 1 \\ 0 & 1 & \frac{2}{3} & \frac{2}{3} \\ 0 & 0 & 0 & 0 \end{array}]$

Step 2.6

Perform the row operation $R_{1} = R_{1} + \frac{1}{2} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 2.6.1

Perform the row operation $R_{1} = R_{1} + \frac{1}{2} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{1}{2} \cdot 0 & - \frac{1}{2} + \frac{1}{2} \cdot 1 & - \frac{3}{2} + \frac{1}{2} \cdot \frac{2}{3} & 1 + \frac{1}{2} \cdot \frac{2}{3} \\ 0 & 1 & \frac{2}{3} & \frac{2}{3} \\ 0 & 0 & 0 & 0 \end{array}]$

Step 2.6.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - \frac{7}{6} & \frac{4}{3} \\ 0 & 1 & \frac{2}{3} & \frac{2}{3} \\ 0 & 0 & 0 & 0 \end{array}]$

Step 3

Use the result matrix to declare the final solution to the system of equations.

$x - \frac{7}{6} z = \frac{4}{3}$

$y + \frac{2}{3} z = \frac{2}{3}$

$0 = 0$

Step 4

The solution is the set of ordered pairs that make the system true.

$(\frac{4}{3} + \frac{7 z}{6}, \frac{2}{3} - \frac{2 z}{3}, z)$