Linear Algebra Examples

x + y + z = 2

$x + y + z = 2$ ,

$4 x + 5 y + z = 12$ ,

$2 x = - 4$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 1 & 1 & 2 \\ 4 & 5 & 1 & 12 \\ 2 & 0 & 0 & - 4 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

Perform the row operation

$R_{2} = R_{2} - 4 R_{1}$ to make the entry at

$2, 1$ a

$0$ .

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Step 2.1.1

Perform the row operation

$R_{2} = R_{2} - 4 R_{1}$ to make the entry at

$2, 1$ a

$0$ .

$[\begin{array}{c} 1 & 1 & 1 & 2 \\ 4 - 4 \cdot 1 & 5 - 4 \cdot 1 & 1 - 4 \cdot 1 & 12 - 4 \cdot 2 \\ 2 & 0 & 0 & - 4 \end{array}]$

Step 2.1.2

Simplify

$R_{2}$ .

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

Step 2.2

Perform the row operation

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

$3, 1$ a

$0$ .

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Step 2.2.1

Perform the row operation

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

$3, 1$ a

$0$ .

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

Step 2.2.2

Simplify

$R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 2 \\ 0 & 1 & - 3 & 4 \\ 0 & - 2 & - 2 & - 8 \end{array}]$

Step 2.3

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 2.3.1

Perform the row operation

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

$3, 2$ a

$0$ .

$[\begin{array}{c} 1 & 1 & 1 & 2 \\ 0 & 1 & - 3 & 4 \\ 0 + 2 \cdot 0 & - 2 + 2 \cdot 1 & - 2 + 2 \cdot - 3 & - 8 + 2 \cdot 4 \end{array}]$

Step 2.3.2

Simplify

$R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 2 \\ 0 & 1 & - 3 & 4 \\ 0 & 0 & - 8 & 0 \end{array}]$

Step 2.4

Multiply each element of

$R_{3}$ by

$- \frac{1}{8}$ to make the entry at

$3, 3$ a

$1$ .

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Step 2.4.1

Multiply each element of

$R_{3}$ by

$- \frac{1}{8}$ to make the entry at

$3, 3$ a

$1$ .

$[\begin{array}{c} 1 & 1 & 1 & 2 \\ 0 & 1 & - 3 & 4 \\ - \frac{1}{8} \cdot 0 & - \frac{1}{8} \cdot 0 & - \frac{1}{8} \cdot - 8 & - \frac{1}{8} \cdot 0 \end{array}]$

Step 2.4.2

Simplify

$R_{3}$ .

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

Step 2.5

Perform the row operation

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

$2, 3$ a

$0$ .

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Step 2.5.1

Perform the row operation

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

$2, 3$ a

$0$ .

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

Step 2.5.2

Simplify

$R_{2}$ .

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

Step 2.6

Perform the row operation

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

$1, 3$ a

$0$ .

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Step 2.6.1

Perform the row operation

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

$1, 3$ a

$0$ .

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

Step 2.6.2

Simplify

$R_{1}$ .

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

Step 2.7

Perform the row operation

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

$1, 2$ a

$0$ .

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Step 2.7.1

Perform the row operation

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

$1, 2$ a

$0$ .

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

Step 2.7.2

Simplify

$R_{1}$ .

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

Step 3

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

$x = - 2$

$y = 4$

$z = 0$

Step 4

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

$(- 2, 4, 0)$

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.

$X = [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} - 2 \\ 4 \\ 0 \end{array}]$