Linear Algebra Examples

$x + 2 y - 3 z = 5$ , $- y + 4 z = 1$ , $3 x - 4 y + 3 x = - 3$

Step 1

Add $3 x$ and $3 x$ .

$x + 2 y - 3 z = 5, - y + 4 z = 1, 6 x - 4 y = - 3$

Step 2

Write the system of equations in matrix form.

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

Step 3

Find the reduced row echelon form.

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Step 3.1

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

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Step 3.1.1

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

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

Step 3.1.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 2 & - 3 & 5 \\ 0 & - 1 & 4 & 1 \\ 0 & - 16 & 18 & - 33 \end{array}]$

Step 3.2

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

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Step 3.2.1

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

$[\begin{array}{c} 1 & 2 & - 3 & 5 \\ - 0 & - - 1 & - 1 \cdot 4 & - 1 \cdot 1 \\ 0 & - 16 & 18 & - 33 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 2 & - 3 & 5 \\ 0 & 1 & - 4 & - 1 \\ 0 & - 16 & 18 & - 33 \end{array}]$

Step 3.3

Perform the row operation $R_{3} = R_{3} + 16 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 3.3.1

Perform the row operation $R_{3} = R_{3} + 16 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & 2 & - 3 & 5 \\ 0 & 1 & - 4 & - 1 \\ 0 + 16 \cdot 0 & - 16 + 16 \cdot 1 & 18 + 16 \cdot - 4 & - 33 + 16 \cdot - 1 \end{array}]$

Step 3.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 2 & - 3 & 5 \\ 0 & 1 & - 4 & - 1 \\ 0 & 0 & - 46 & - 49 \end{array}]$

Step 3.4

Multiply each element of $R_{3}$ by $- \frac{1}{46}$ to make the entry at $3, 3$ a $1$ .

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Step 3.4.1

Multiply each element of $R_{3}$ by $- \frac{1}{46}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 2 & - 3 & 5 \\ 0 & 1 & - 4 & - 1 \\ - \frac{1}{46} \cdot 0 & - \frac{1}{46} \cdot 0 & - \frac{1}{46} \cdot - 46 & - \frac{1}{46} \cdot - 49 \end{array}]$

Step 3.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 2 & - 3 & 5 \\ 0 & 1 & - 4 & - 1 \\ 0 & 0 & 1 & \frac{49}{46} \end{array}]$

Step 3.5

Perform the row operation $R_{2} = R_{2} + 4 R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 3.5.1

Perform the row operation $R_{2} = R_{2} + 4 R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & 2 & - 3 & 5 \\ 0 + 4 \cdot 0 & 1 + 4 \cdot 0 & - 4 + 4 \cdot 1 & - 1 + 4 (\frac{49}{46}) \\ 0 & 0 & 1 & \frac{49}{46} \end{array}]$

Step 3.5.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 2 & - 3 & 5 \\ 0 & 1 & 0 & \frac{75}{23} \\ 0 & 0 & 1 & \frac{49}{46} \end{array}]$

Step 3.6

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 3.6.1

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

$[\begin{array}{c} 1 + 3 \cdot 0 & 2 + 3 \cdot 0 & - 3 + 3 \cdot 1 & 5 + 3 (\frac{49}{46}) \\ 0 & 1 & 0 & \frac{75}{23} \\ 0 & 0 & 1 & \frac{49}{46} \end{array}]$

Step 3.6.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 2 & 0 & \frac{377}{46} \\ 0 & 1 & 0 & \frac{75}{23} \\ 0 & 0 & 1 & \frac{49}{46} \end{array}]$

Step 3.7

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 3.7.1

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

$[\begin{array}{c} 1 - 2 \cdot 0 & 2 - 2 \cdot 1 & 0 - 2 \cdot 0 & \frac{377}{46} - 2 (\frac{75}{23}) \\ 0 & 1 & 0 & \frac{75}{23} \\ 0 & 0 & 1 & \frac{49}{46} \end{array}]$

Step 3.7.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{77}{46} \\ 0 & 1 & 0 & \frac{75}{23} \\ 0 & 0 & 1 & \frac{49}{46} \end{array}]$

Step 4

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

$x = \frac{77}{46}$

$y = \frac{75}{23}$

$z = \frac{49}{46}$

Step 5

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

$(\frac{77}{46}, \frac{75}{23}, \frac{49}{46})$

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.

$X = [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} \frac{77}{46} \\ \frac{75}{23} \\ \frac{49}{46} \end{array}]$