Linear Algebra Examples

$y = - 3 x - 3 z + 11$ , $x = - 2 y - z + 7$ , $z = x - y$

Step 1

Move all terms containing variables to the left side of the equation.

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Step 1.1

Add $3 x$ to both sides of the equation.

$y + 3 x = - 3 z + 11, x = - 2 y - z + 7, z = x - y$

Step 1.2

Add $3 z$ to both sides of the equation.

$y + 3 x + 3 z = 11, x = - 2 y - z + 7, z = x - y$

Step 2

Move all terms containing variables to the left side of the equation.

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Step 2.1

Add $2 y$ to both sides of the equation.

$y + 3 x + 3 z = 11, x + 2 y = - z + 7, z = x - y$

Step 2.2

Add $z$ to both sides of the equation.

$y + 3 x + 3 z = 11, x + 2 y + z = 7, z = x - y$

Step 3

Move all terms containing variables to the left side of the equation.

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Step 3.1

Subtract $x$ from both sides of the equation.

$y + 3 x + 3 z = 11, x + 2 y + z = 7, z - x = - y$

Step 3.2

Add $y$ to both sides of the equation.

$y + 3 x + 3 z = 11, x + 2 y + z = 7, z - x + y = 0$

Step 4

Write the system of equations in matrix form.

$[\begin{array}{c} 3 & 1 & 3 & 11 \\ 1 & 2 & 1 & 7 \\ - 1 & 1 & 1 & 0 \end{array}]$

Step 5

Find the reduced row echelon form.

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Step 5.1

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

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Step 5.1.1

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

$[\begin{array}{c} \frac{3}{3} & \frac{1}{3} & \frac{3}{3} & \frac{11}{3} \\ 1 & 2 & 1 & 7 \\ - 1 & 1 & 1 & 0 \end{array}]$

Step 5.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{1}{3} & 1 & \frac{11}{3} \\ 1 & 2 & 1 & 7 \\ - 1 & 1 & 1 & 0 \end{array}]$

Step 5.2

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

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Step 5.2.1

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

$[\begin{array}{c} 1 & \frac{1}{3} & 1 & \frac{11}{3} \\ 1 - 1 & 2 - \frac{1}{3} & 1 - 1 & 7 - \frac{11}{3} \\ - 1 & 1 & 1 & 0 \end{array}]$

Step 5.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{3} & 1 & \frac{11}{3} \\ 0 & \frac{5}{3} & 0 & \frac{10}{3} \\ - 1 & 1 & 1 & 0 \end{array}]$

Step 5.3

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

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Step 5.3.1

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

$[\begin{array}{c} 1 & \frac{1}{3} & 1 & \frac{11}{3} \\ 0 & \frac{5}{3} & 0 & \frac{10}{3} \\ - 1 + 1 \cdot 1 & 1 + \frac{1}{3} & 1 + 1 \cdot 1 & 0 + \frac{11}{3} \end{array}]$

Step 5.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{1}{3} & 1 & \frac{11}{3} \\ 0 & \frac{5}{3} & 0 & \frac{10}{3} \\ 0 & \frac{4}{3} & 2 & \frac{11}{3} \end{array}]$

Step 5.4

Multiply each element of $R_{2}$ by $\frac{3}{5}$ to make the entry at $2, 2$ a $1$ .

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Step 5.4.1

Multiply each element of $R_{2}$ by $\frac{3}{5}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{1}{3} & 1 & \frac{11}{3} \\ \frac{3}{5} \cdot 0 & \frac{3}{5} \cdot \frac{5}{3} & \frac{3}{5} \cdot 0 & \frac{3}{5} \cdot \frac{10}{3} \\ 0 & \frac{4}{3} & 2 & \frac{11}{3} \end{array}]$

Step 5.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{3} & 1 & \frac{11}{3} \\ 0 & 1 & 0 & 2 \\ 0 & \frac{4}{3} & 2 & \frac{11}{3} \end{array}]$

Step 5.5

Perform the row operation $R_{3} = R_{3} - \frac{4}{3} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 5.5.1

Perform the row operation $R_{3} = R_{3} - \frac{4}{3} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{3} & 1 & \frac{11}{3} \\ 0 & 1 & 0 & 2 \\ 0 - \frac{4}{3} \cdot 0 & \frac{4}{3} - \frac{4}{3} \cdot 1 & 2 - \frac{4}{3} \cdot 0 & \frac{11}{3} - \frac{4}{3} \cdot 2 \end{array}]$

Step 5.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{1}{3} & 1 & \frac{11}{3} \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 2 & 1 \end{array}]$

Step 5.6

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

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Step 5.6.1

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

$[\begin{array}{c} 1 & \frac{1}{3} & 1 & \frac{11}{3} \\ 0 & 1 & 0 & 2 \\ \frac{0}{2} & \frac{0}{2} & \frac{2}{2} & \frac{1}{2} \end{array}]$

Step 5.6.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{1}{3} & 1 & \frac{11}{3} \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & \frac{1}{2} \end{array}]$

Step 5.7

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

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Step 5.7.1

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

$[\begin{array}{c} 1 - 0 & \frac{1}{3} - 0 & 1 - 1 & \frac{11}{3} - \frac{1}{2} \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & \frac{1}{2} \end{array}]$

Step 5.7.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{1}{3} & 0 & \frac{19}{6} \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & \frac{1}{2} \end{array}]$

Step 5.8

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

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Step 5.8.1

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

$[\begin{array}{c} 1 - \frac{1}{3} \cdot 0 & \frac{1}{3} - \frac{1}{3} \cdot 1 & 0 - \frac{1}{3} \cdot 0 & \frac{19}{6} - \frac{1}{3} \cdot 2 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & \frac{1}{2} \end{array}]$

Step 5.8.2

Simplify $R_{1}$ .

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

Step 6

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

$x = \frac{5}{2}$

$y = 2$

$z = \frac{1}{2}$

Step 7

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

$(\frac{5}{2}, 2, \frac{1}{2})$

Step 8

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.

$X = [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} \frac{5}{2} \\ 2 \\ \frac{1}{2} \end{array}]$