Linear Algebra Examples

3 x + 3 y + 3 z = 6

$3 x + 3 y + 3 z = 6$ ,

x - y = - 3

$x - y = - 3$ ,

- 4 x + y - z = - 1

$- 4 x + y - z = - 1$

Step 1

Write the system of equations in matrix form.

⎡ ⎢ ⎣ \begin{matrix} 3 & 3 & 3 & 6 1 & - 1 & 0 & - 3 - 4 & 1 & - 1 & - 1 \end{matrix} ⎤ ⎥ ⎦

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

Step 2

Find the reduced row echelon form.

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Step 2.1

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{3}

$\frac{1}{3}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

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Step 2.1.1

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{3}

$\frac{1}{3}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} \frac{3}{3} & \frac{3}{3} & \frac{3}{3} & \frac{6}{3} 1 & - 1 & 0 & - 3 - 4 & 1 & - 1 & - 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

Step 2.1.2

Simplify

R_{1}

$R_{1}$ .

⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 & 2 1 & - 1 & 0 & - 3 - 4 & 1 & - 1 & - 1 \end{matrix} ⎤ ⎥ ⎦

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

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

$0$ .

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

Step 2.2.2

Simplify

$R_{2}$ .

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

Step 2.3

Perform the row operation

$R_{3} = R_{3} + 4 R_{1}$ to make the entry at

$3, 1$ a

$0$ .

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Step 2.3.1

Perform the row operation

$R_{3} = R_{3} + 4 R_{1}$ to make the entry at

$3, 1$ a

$0$ .

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

Step 2.3.2

Simplify

$R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 2 \\ 0 & - 2 & - 1 & - 5 \\ 0 & 5 & 3 & 7 \end{array}]$

Step 2.4

Multiply each element of

$R_{2}$ by

$- \frac{1}{2}$ to make the entry at

$2, 2$ a

$1$ .

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Step 2.4.1

Multiply each element of

$R_{2}$ by

$- \frac{1}{2}$ to make the entry at

$2, 2$ a

$1$ .

$[\begin{array}{c} 1 & 1 & 1 & 2 \\ - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot - 2 & - \frac{1}{2} \cdot - 1 & - \frac{1}{2} \cdot - 5 \\ 0 & 5 & 3 & 7 \end{array}]$

Step 2.4.2

Simplify

$R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 2 \\ 0 & 1 & \frac{1}{2} & \frac{5}{2} \\ 0 & 5 & 3 & 7 \end{array}]$

Step 2.5

Perform the row operation

$R_{3} = R_{3} - 5 R_{2}$ to make the entry at

$3, 2$ a

$0$ .

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Step 2.5.1

Perform the row operation

$R_{3} = R_{3} - 5 R_{2}$ to make the entry at

$3, 2$ a

$0$ .

$[\begin{array}{c} 1 & 1 & 1 & 2 \\ 0 & 1 & \frac{1}{2} & \frac{5}{2} \\ 0 - 5 \cdot 0 & 5 - 5 \cdot 1 & 3 - 5 (\frac{1}{2}) & 7 - 5 (\frac{5}{2}) \end{array}]$

Step 2.5.2

Simplify

$R_{3}$ .

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

Step 2.6

Multiply each element of

$R_{3}$ by

$2$ to make the entry at

$3, 3$ a

$1$ .

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Step 2.6.1

Multiply each element of

$R_{3}$ by

$2$ to make the entry at

$3, 3$ a

$1$ .

$[\begin{array}{c} 1 & 1 & 1 & 2 \\ 0 & 1 & \frac{1}{2} & \frac{5}{2} \\ 2 \cdot 0 & 2 \cdot 0 & 2 (\frac{1}{2}) & 2 (- \frac{11}{2}) \end{array}]$

Step 2.6.2

Simplify

$R_{3}$ .

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

Step 2.7

Perform the row operation

$R_{2} = R_{2} - \frac{1}{2} R_{3}$ to make the entry at

$2, 3$ a

$0$ .

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Step 2.7.1

Perform the row operation

$R_{2} = R_{2} - \frac{1}{2} R_{3}$ to make the entry at

$2, 3$ a

$0$ .

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

Step 2.7.2

Simplify

$R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 2 \\ 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & - 11 \end{array}]$

Step 2.8

Perform the row operation

$R_{1} = R_{1} - R_{3}$ to make the entry at

$1, 3$ a

$0$ .

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Step 2.8.1

Perform the row operation

$R_{1} = R_{1} - R_{3}$ to make the entry at

$1, 3$ a

$0$ .

$[\begin{array}{c} 1 - 0 & 1 - 0 & 1 - 1 & 2 + 11 \\ 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & - 11 \end{array}]$

Step 2.8.2

Simplify

$R_{1}$ .

$[\begin{array}{c} 1 & 1 & 0 & 13 \\ 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & - 11 \end{array}]$

Step 2.9

Perform the row operation

$R_{1} = R_{1} - R_{2}$ to make the entry at

$1, 2$ a

$0$ .

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Step 2.9.1

Perform the row operation

$R_{1} = R_{1} - R_{2}$ to make the entry at

$1, 2$ a

$0$ .

$[\begin{array}{c} 1 - 0 & 1 - 1 & 0 - 0 & 13 - 8 \\ 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & - 11 \end{array}]$

Step 2.9.2

Simplify

$R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 5 \\ 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & - 11 \end{array}]$

Step 3

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

$x = 5$

$y = 8$

$z = - 11$

Step 4

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

$(5, 8, - 11)$

Step 5

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.

$X = [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 5 \\ 8 \\ - 11 \end{array}]$

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