Linear Algebra Examples

$2 x + 5 y - z = 2$ $3 x + 8 y + z = 4$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 2 & 5 & - 1 & 2 \\ 3 & 8 & 1 & 4 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

$[\begin{array}{c} \frac{2}{2} & \frac{5}{2} & \frac{- 1}{2} & \frac{2}{2} \\ 3 & 8 & 1 & 4 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{5}{2} & - \frac{1}{2} & 1 \\ 3 & 8 & 1 & 4 \end{array}]$

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} 1 & \frac{5}{2} & - \frac{1}{2} & 1 \\ 3 - 3 \cdot 1 & 8 - 3 (\frac{5}{2}) & 1 - 3 (- \frac{1}{2}) & 4 - 3 \cdot 1 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{5}{2} & - \frac{1}{2} & 1 \\ 0 & \frac{1}{2} & \frac{5}{2} & 1 \end{array}]$

Step 2.3

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

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Step 2.3.1

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

$[\begin{array}{c} 1 & \frac{5}{2} & - \frac{1}{2} & 1 \\ 2 \cdot 0 & 2 (\frac{1}{2}) & 2 (\frac{5}{2}) & 2 \cdot 1 \end{array}]$

Step 2.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{5}{2} & - \frac{1}{2} & 1 \\ 0 & 1 & 5 & 2 \end{array}]$

Step 2.4

Perform the row operation $R_{1} = R_{1} - \frac{5}{2} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 2.4.1

Perform the row operation $R_{1} = R_{1} - \frac{5}{2} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{5}{2} \cdot 0 & \frac{5}{2} - \frac{5}{2} \cdot 1 & - \frac{1}{2} - \frac{5}{2} \cdot 5 & 1 - \frac{5}{2} \cdot 2 \\ 0 & 1 & 5 & 2 \end{array}]$

Step 2.4.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - 13 & - 4 \\ 0 & 1 & 5 & 2 \end{array}]$

Step 3

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

$x - 13 z = - 4$

$y + 5 z = 2$

Step 4

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

$(- 4 + 13 z, 2 - 5 z, z)$