Linear Algebra Examples

$6 x + 3 y - 3 z = 12$ , $4 x + 2 y - 2 z = 8$ , $3 x + 2 y - z = 8$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 6 & 3 & - 3 & 12 \\ 4 & 2 & - 2 & 8 \\ 3 & 2 & - 1 & 8 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

Multiply each element of $R_{1}$ by $\frac{1}{6}$ to make the entry at $1, 1$ a $1$ .

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Step 2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{6}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{6}{6} & \frac{3}{6} & \frac{- 3}{6} & \frac{12}{6} \\ 4 & 2 & - 2 & 8 \\ 3 & 2 & - 1 & 8 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

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

Step 2.2

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 2.2.1

Perform the row operation $R_{2} = R_{2} - 4 R_{1}$ to make the entry at $2, 1$ a $0$ .

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

Step 2.2.2

Simplify $R_{2}$ .

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

Simplify $R_{3}$ .

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

Step 2.4

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

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

Step 2.5

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

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Step 2.5.1

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

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

Step 2.5.2

Simplify $R_{2}$ .

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

Step 2.6

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

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Step 2.6.1

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

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

Step 2.6.2

Simplify $R_{1}$ .

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

Step 3

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

$x - z = 0$

$y + z = 4$

Step 4

Add $z$ to both sides of the equation.

$x = z$

$y + z = 4$

Step 5

Subtract $z$ from both sides of the equation.

$y = 4 - z$

$x = z$

Step 6

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

$(z, 4 - z, z)$

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.

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