Linear Algebra Examples

$(x + y + 6 z) = 4$ , $(x + y + 3 z) = 4$ , $(x + 2 y + 4 z) = 9$

Step 1

Remove parentheses.

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

Step 2

Write the system of equations in matrix form.

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

Step 3

Find the reduced row echelon form.

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Step 3.1

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

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Step 3.1.1

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

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

Step 3.1.2

Simplify $R_{2}$ .

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

Simplify $R_{3}$ .

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

Step 3.3

Swap $R_{3}$ with $R_{2}$ to put a nonzero entry at $2, 2$ .

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

Step 3.4

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

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Step 3.4.1

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

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

Step 3.4.2

Simplify $R_{3}$ .

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

Step 3.5

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 3.5.1

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

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

Step 3.5.2

Simplify $R_{2}$ .

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

Step 3.6

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

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Step 3.6.1

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

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

Step 3.6.2

Simplify $R_{1}$ .

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

Step 3.7

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

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Step 3.7.1

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

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

Step 3.7.2

Simplify $R_{1}$ .

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

Step 4

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

$x = - 1$

$y = 5$

$z = 0$

Step 5

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

$(- 1, 5, 0)$

Step 6

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.

$X = [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} - 1 \\ 5 \\ 0 \end{array}]$