Linear Algebra Examples

$- x + y + z = - 2$ , $- x + 5 y - 15 z = - 26$ , $6 x - 4 y - 14 z = 0$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} - 1 & 1 & 1 & - 2 \\ - 1 & 5 & - 15 & - 26 \\ 6 & - 4 & - 14 & 0 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

$[\begin{array}{c} - - 1 & - 1 \cdot 1 & - 1 \cdot 1 & - - 2 \\ - 1 & 5 & - 15 & - 26 \\ 6 & - 4 & - 14 & 0 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 2 \\ - 1 & 5 & - 15 & - 26 \\ 6 & - 4 & - 14 & 0 \end{array}]$

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} 1 & - 1 & - 1 & 2 \\ - 1 + 1 \cdot 1 & 5 - 1 & - 15 - 1 & - 26 + 1 \cdot 2 \\ 6 & - 4 & - 14 & 0 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 2 \\ 0 & 4 & - 16 & - 24 \\ 6 & - 4 & - 14 & 0 \end{array}]$

Step 2.3

Perform the row operation $R_{3} = R_{3} - 6 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 2.3.1

Perform the row operation $R_{3} = R_{3} - 6 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 2 \\ 0 & 4 & - 16 & - 24 \\ 6 - 6 \cdot 1 & - 4 - 6 \cdot - 1 & - 14 - 6 \cdot - 1 & 0 - 6 \cdot 2 \end{array}]$

Step 2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 2 \\ 0 & 4 & - 16 & - 24 \\ 0 & 2 & - 8 & - 12 \end{array}]$

Step 2.4

Multiply each element of $R_{2}$ by $\frac{1}{4}$ to make the entry at $2, 2$ a $1$ .

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Step 2.4.1

Multiply each element of $R_{2}$ by $\frac{1}{4}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 2 \\ \frac{0}{4} & \frac{4}{4} & \frac{- 16}{4} & \frac{- 24}{4} \\ 0 & 2 & - 8 & - 12 \end{array}]$

Step 2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 2 \\ 0 & 1 & - 4 & - 6 \\ 0 & 2 & - 8 & - 12 \end{array}]$

Step 2.5

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

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Step 2.5.1

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

$[\begin{array}{c} 1 & - 1 & - 1 & 2 \\ 0 & 1 & - 4 & - 6 \\ 0 - 2 \cdot 0 & 2 - 2 \cdot 1 & - 8 - 2 \cdot - 4 & - 12 - 2 \cdot - 6 \end{array}]$

Step 2.5.2

Simplify $R_{3}$ .

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

Step 2.6

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

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Step 2.6.1

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

$[\begin{array}{c} 1 + 0 & - 1 + 1 \cdot 1 & - 1 - 4 & 2 - 6 \\ 0 & 1 & - 4 & - 6 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 2.6.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - 5 & - 4 \\ 0 & 1 & - 4 & - 6 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 3

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

$x - 5 z = - 4$

$y - 4 z = - 6$

Step 4

Add $5 z$ to both sides of the equation.

$x = - 4 + 5 z$

$y - 4 z = - 6$

Step 5

Add $4 z$ to both sides of the equation.

$y = - 6 + 4 z$

$x = - 4 + 5 z$

Step 6

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

$(- 4 + 5 z, - 6 + 4 z, z)$

Step 7

Decompose a solution vector by re-arranging each equation represented in the row-reduced form of the augmented matrix by solving for the dependent variable in each row yields the vector equality.

$X = [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} - 4 + 5 z \\ - 6 + 4 z \\ z \end{array}]$