Linear Algebra Examples

$x + 2 y + 3 z = 5$ , $3 x - y + 2 z = 8$ , $4 x - 6 y - 4 = - 2$

Step 1

Add $4$ to both sides of the equation.

$x + 2 y + 3 z = 5, 3 x - y + 2 z = 8, 4 x - 6 y = - 2 + 4$

Step 2

Add $- 2$ and $4$ .

$x + 2 y + 3 z = 5, 3 x - y + 2 z = 8, 4 x - 6 y = 2$

Step 3

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 2 & 3 & 5 \\ 3 & - 1 & 2 & 8 \\ 4 & - 6 & 0 & 2 \end{array}]$

Step 4

Find the reduced row echelon form.

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Step 4.1

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

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Step 4.1.1

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

$[\begin{array}{c} 1 & 2 & 3 & 5 \\ 3 - 3 \cdot 1 & - 1 - 3 \cdot 2 & 2 - 3 \cdot 3 & 8 - 3 \cdot 5 \\ 4 & - 6 & 0 & 2 \end{array}]$

Step 4.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 2 & 3 & 5 \\ 0 & - 7 & - 7 & - 7 \\ 4 & - 6 & 0 & 2 \end{array}]$

Step 4.2

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 4.2.1

Perform the row operation $R_{3} = R_{3} - 4 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & 2 & 3 & 5 \\ 0 & - 7 & - 7 & - 7 \\ 4 - 4 \cdot 1 & - 6 - 4 \cdot 2 & 0 - 4 \cdot 3 & 2 - 4 \cdot 5 \end{array}]$

Step 4.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 2 & 3 & 5 \\ 0 & - 7 & - 7 & - 7 \\ 0 & - 14 & - 12 & - 18 \end{array}]$

Step 4.3

Multiply each element of $R_{2}$ by $- \frac{1}{7}$ to make the entry at $2, 2$ a $1$ .

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Step 4.3.1

Multiply each element of $R_{2}$ by $- \frac{1}{7}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 2 & 3 & 5 \\ - \frac{1}{7} \cdot 0 & - \frac{1}{7} \cdot - 7 & - \frac{1}{7} \cdot - 7 & - \frac{1}{7} \cdot - 7 \\ 0 & - 14 & - 12 & - 18 \end{array}]$

Step 4.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 2 & 3 & 5 \\ 0 & 1 & 1 & 1 \\ 0 & - 14 & - 12 & - 18 \end{array}]$

Step 4.4

Perform the row operation $R_{3} = R_{3} + 14 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 4.4.1

Perform the row operation $R_{3} = R_{3} + 14 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & 2 & 3 & 5 \\ 0 & 1 & 1 & 1 \\ 0 + 14 \cdot 0 & - 14 + 14 \cdot 1 & - 12 + 14 \cdot 1 & - 18 + 14 \cdot 1 \end{array}]$

Step 4.4.2

Simplify $R_{3}$ .

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

Step 4.5

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

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Step 4.5.1

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

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

Step 4.5.2

Simplify $R_{3}$ .

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

Step 4.6

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

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Step 4.6.1

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

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

Step 4.6.2

Simplify $R_{2}$ .

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

Step 4.7

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

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Step 4.7.1

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

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

Step 4.7.2

Simplify $R_{1}$ .

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

Step 4.8

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

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Step 4.8.1

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

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

Step 4.8.2

Simplify $R_{1}$ .

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

Step 5

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

$x = 5$

$y = 3$

$z = - 2$

Step 6

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

$(5, 3, - 2)$

Step 7

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.

$X = [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 5 \\ 3 \\ - 2 \end{array}]$