Linear Algebra Examples

$x + y - 2 z = 9 c$ , $x + 2 y - 3 z = 3$ , $- 1 x + 4 z = - 6$

Step 1

Subtract $9 c$ from both sides of the equation.

$x + y - 2 z - 9 c = 0, x + 2 y - 3 z = 3, - 1 x + 4 z = - 6$

Step 2

Rewrite $- 1 x$ as $- x$ .

$x + y - 2 z - 9 c = 0, x + 2 y - 3 z = 3, - x + 4 z = - 6$

Step 3

Write the system of equations in matrix form.

$[\begin{array}{c} - 9 & 1 & 1 & - 2 & 0 \\ 0 & 1 & 2 & - 3 & 3 \\ 0 & - 1 & 0 & 4 & - 6 \end{array}]$

Step 4

Find the reduced row echelon form.

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Step 4.1

Multiply each element of $R_{1}$ by $- \frac{1}{9}$ to make the entry at $1, 1$ a $1$ .

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Step 4.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{9}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{9} \cdot - 9 & - \frac{1}{9} \cdot 1 & - \frac{1}{9} \cdot 1 & - \frac{1}{9} \cdot - 2 & - \frac{1}{9} \cdot 0 \\ 0 & 1 & 2 & - 3 & 3 \\ 0 & - 1 & 0 & 4 & - 6 \end{array}]$

Step 4.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{9} & - \frac{1}{9} & \frac{2}{9} & 0 \\ 0 & 1 & 2 & - 3 & 3 \\ 0 & - 1 & 0 & 4 & - 6 \end{array}]$

Step 4.2

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

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Step 4.2.1

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

$[\begin{array}{c} 1 & - \frac{1}{9} & - \frac{1}{9} & \frac{2}{9} & 0 \\ 0 & 1 & 2 & - 3 & 3 \\ 0 + 0 & - 1 + 1 \cdot 1 & 0 + 1 \cdot 2 & 4 - 3 & - 6 + 1 \cdot 3 \end{array}]$

Step 4.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{1}{9} & - \frac{1}{9} & \frac{2}{9} & 0 \\ 0 & 1 & 2 & - 3 & 3 \\ 0 & 0 & 2 & 1 & - 3 \end{array}]$

Step 4.3

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

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Step 4.3.1

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

$[\begin{array}{c} 1 & - \frac{1}{9} & - \frac{1}{9} & \frac{2}{9} & 0 \\ 0 & 1 & 2 & - 3 & 3 \\ \frac{0}{2} & \frac{0}{2} & \frac{2}{2} & \frac{1}{2} & \frac{- 3}{2} \end{array}]$

Step 4.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{1}{9} & - \frac{1}{9} & \frac{2}{9} & 0 \\ 0 & 1 & 2 & - 3 & 3 \\ 0 & 0 & 1 & \frac{1}{2} & - \frac{3}{2} \end{array}]$

Step 4.4

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

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Step 4.4.1

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

$[\begin{array}{c} 1 & - \frac{1}{9} & - \frac{1}{9} & \frac{2}{9} & 0 \\ 0 - 2 \cdot 0 & 1 - 2 \cdot 0 & 2 - 2 \cdot 1 & - 3 - 2 (\frac{1}{2}) & 3 - 2 (- \frac{3}{2}) \\ 0 & 0 & 1 & \frac{1}{2} & - \frac{3}{2} \end{array}]$

Step 4.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{9} & - \frac{1}{9} & \frac{2}{9} & 0 \\ 0 & 1 & 0 & - 4 & 6 \\ 0 & 0 & 1 & \frac{1}{2} & - \frac{3}{2} \end{array}]$

Step 4.5

Perform the row operation $R_{1} = R_{1} + \frac{1}{9} R_{3}$ to make the entry at $1, 3$ a $0$ .

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Step 4.5.1

Perform the row operation $R_{1} = R_{1} + \frac{1}{9} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 + \frac{1}{9} \cdot 0 & - \frac{1}{9} + \frac{1}{9} \cdot 0 & - \frac{1}{9} + \frac{1}{9} \cdot 1 & \frac{2}{9} + \frac{1}{9} \cdot \frac{1}{2} & 0 + \frac{1}{9} (- \frac{3}{2}) \\ 0 & 1 & 0 & - 4 & 6 \\ 0 & 0 & 1 & \frac{1}{2} & - \frac{3}{2} \end{array}]$

Step 4.5.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{9} & 0 & \frac{5}{18} & - \frac{1}{6} \\ 0 & 1 & 0 & - 4 & 6 \\ 0 & 0 & 1 & \frac{1}{2} & - \frac{3}{2} \end{array}]$

Step 4.6

Perform the row operation $R_{1} = R_{1} + \frac{1}{9} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 4.6.1

Perform the row operation $R_{1} = R_{1} + \frac{1}{9} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{1}{9} \cdot 0 & - \frac{1}{9} + \frac{1}{9} \cdot 1 & 0 + \frac{1}{9} \cdot 0 & \frac{5}{18} + \frac{1}{9} \cdot - 4 & - \frac{1}{6} + \frac{1}{9} \cdot 6 \\ 0 & 1 & 0 & - 4 & 6 \\ 0 & 0 & 1 & \frac{1}{2} & - \frac{3}{2} \end{array}]$

Step 4.6.2

Simplify $R_{1}$ .

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

Step 5

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

$c - \frac{z}{6} = \frac{1}{2}$

$x - 4 z = 6$

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

Step 6

Add $\frac{z}{6}$ to both sides of the equation.

$c = \frac{1}{2} + \frac{z}{6}$

$x - 4 z = 6$

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

Step 7

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

$x = 6 + 4 z$

$c = \frac{1}{2} + \frac{z}{6}$

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

Step 8

Subtract $\frac{z}{2}$ from both sides of the equation.

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

$c = \frac{1}{2} + \frac{z}{6}$

$x = 6 + 4 z$

Step 9

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

$(\frac{1}{2} + \frac{z}{6}, 6 + 4 z, - \frac{3}{2} - \frac{z}{2}, z)$

Step 10

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.

$X = [\begin{array}{c} c \\ x \\ y \\ z \end{array}] = [\begin{array}{c} \frac{1}{2} + \frac{z}{6} \\ 6 + 4 z \\ - \frac{3}{2} - \frac{z}{2} \\ z \end{array}]$