Linear Algebra Examples

$2 x + 3 y + z + w = 15$ , $x + y + z + w = 2$ , $2 x + 3 y + 2 z + w = 3$ , $3 x + y + z + 2 w = 1$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 2 & 3 & 1 & 15 \\ 1 & 1 & 1 & 1 & 2 \\ 1 & 2 & 3 & 2 & 3 \\ 2 & 3 & 1 & 1 & 1 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

$[\begin{array}{c} 1 & 2 & 3 & 1 & 15 \\ 1 - 1 & 1 - 2 & 1 - 3 & 1 - 1 & 2 - 15 \\ 1 & 2 & 3 & 2 & 3 \\ 2 & 3 & 1 & 1 & 1 \end{array}]$

Step 2.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 2 & 3 & 1 & 15 \\ 0 & - 1 & - 2 & 0 & - 13 \\ 1 & 2 & 3 & 2 & 3 \\ 2 & 3 & 1 & 1 & 1 \end{array}]$

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} 1 & 2 & 3 & 1 & 15 \\ 0 & - 1 & - 2 & 0 & - 13 \\ 1 - 1 & 2 - 2 & 3 - 3 & 2 - 1 & 3 - 15 \\ 2 & 3 & 1 & 1 & 1 \end{array}]$

Step 2.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 2 & 3 & 1 & 15 \\ 0 & - 1 & - 2 & 0 & - 13 \\ 0 & 0 & 0 & 1 & - 12 \\ 2 & 3 & 1 & 1 & 1 \end{array}]$

Step 2.3

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

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Step 2.3.1

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

$[\begin{array}{c} 1 & 2 & 3 & 1 & 15 \\ 0 & - 1 & - 2 & 0 & - 13 \\ 0 & 0 & 0 & 1 & - 12 \\ 2 - 2 \cdot 1 & 3 - 2 \cdot 2 & 1 - 2 \cdot 3 & 1 - 2 \cdot 1 & 1 - 2 \cdot 15 \end{array}]$

Step 2.3.2

Simplify $R_{4}$ .

$[\begin{array}{c} 1 & 2 & 3 & 1 & 15 \\ 0 & - 1 & - 2 & 0 & - 13 \\ 0 & 0 & 0 & 1 & - 12 \\ 0 & - 1 & - 5 & - 1 & - 29 \end{array}]$

Step 2.4

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

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Step 2.4.1

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

$[\begin{array}{c} 1 & 2 & 3 & 1 & 15 \\ - 0 & - - 1 & - - 2 & - 0 & - - 13 \\ 0 & 0 & 0 & 1 & - 12 \\ 0 & - 1 & - 5 & - 1 & - 29 \end{array}]$

Step 2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 2 & 3 & 1 & 15 \\ 0 & 1 & 2 & 0 & 13 \\ 0 & 0 & 0 & 1 & - 12 \\ 0 & - 1 & - 5 & - 1 & - 29 \end{array}]$

Step 2.5

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

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Step 2.5.1

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

$[\begin{array}{c} 1 & 2 & 3 & 1 & 15 \\ 0 & 1 & 2 & 0 & 13 \\ 0 & 0 & 0 & 1 & - 12 \\ 0 + 0 & - 1 + 1 \cdot 1 & - 5 + 1 \cdot 2 & - 1 + 0 & - 29 + 1 \cdot 13 \end{array}]$

Step 2.5.2

Simplify $R_{4}$ .

$[\begin{array}{c} 1 & 2 & 3 & 1 & 15 \\ 0 & 1 & 2 & 0 & 13 \\ 0 & 0 & 0 & 1 & - 12 \\ 0 & 0 & - 3 & - 1 & - 16 \end{array}]$

Step 2.6

Swap $R_{4}$ with $R_{3}$ to put a nonzero entry at $3, 3$ .

$[\begin{array}{c} 1 & 2 & 3 & 1 & 15 \\ 0 & 1 & 2 & 0 & 13 \\ 0 & 0 & - 3 & - 1 & - 16 \\ 0 & 0 & 0 & 1 & - 12 \end{array}]$

Step 2.7

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

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Step 2.7.1

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

$[\begin{array}{c} 1 & 2 & 3 & 1 & 15 \\ 0 & 1 & 2 & 0 & 13 \\ - \frac{1}{3} \cdot 0 & - \frac{1}{3} \cdot 0 & - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot - 1 & - \frac{1}{3} \cdot - 16 \\ 0 & 0 & 0 & 1 & - 12 \end{array}]$

Step 2.7.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 2 & 3 & 1 & 15 \\ 0 & 1 & 2 & 0 & 13 \\ 0 & 0 & 1 & \frac{1}{3} & \frac{16}{3} \\ 0 & 0 & 0 & 1 & - 12 \end{array}]$

Step 2.8

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

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Step 2.8.1

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

$[\begin{array}{c} 1 & 2 & 3 & 1 & 15 \\ 0 & 1 & 2 & 0 & 13 \\ 0 - \frac{1}{3} \cdot 0 & 0 - \frac{1}{3} \cdot 0 & 1 - \frac{1}{3} \cdot 0 & \frac{1}{3} - \frac{1}{3} \cdot 1 & \frac{16}{3} - \frac{1}{3} \cdot - 12 \\ 0 & 0 & 0 & 1 & - 12 \end{array}]$

Step 2.8.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 2 & 3 & 1 & 15 \\ 0 & 1 & 2 & 0 & 13 \\ 0 & 0 & 1 & 0 & \frac{28}{3} \\ 0 & 0 & 0 & 1 & - 12 \end{array}]$

Step 2.9

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

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Step 2.9.1

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

$[\begin{array}{c} 1 - 0 & 2 - 0 & 3 - 0 & 1 - 1 & 15 + 12 \\ 0 & 1 & 2 & 0 & 13 \\ 0 & 0 & 1 & 0 & \frac{28}{3} \\ 0 & 0 & 0 & 1 & - 12 \end{array}]$

Step 2.9.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 2 & 3 & 0 & 27 \\ 0 & 1 & 2 & 0 & 13 \\ 0 & 0 & 1 & 0 & \frac{28}{3} \\ 0 & 0 & 0 & 1 & - 12 \end{array}]$

Step 2.10

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 2.10.1

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

$[\begin{array}{c} 1 & 2 & 3 & 0 & 27 \\ 0 - 2 \cdot 0 & 1 - 2 \cdot 0 & 2 - 2 \cdot 1 & 0 - 2 \cdot 0 & 13 - 2 (\frac{28}{3}) \\ 0 & 0 & 1 & 0 & \frac{28}{3} \\ 0 & 0 & 0 & 1 & - 12 \end{array}]$

Step 2.10.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 2 & 3 & 0 & 27 \\ 0 & 1 & 0 & 0 & - \frac{17}{3} \\ 0 & 0 & 1 & 0 & \frac{28}{3} \\ 0 & 0 & 0 & 1 & - 12 \end{array}]$

Step 2.11

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 2.11.1

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

$[\begin{array}{c} 1 - 3 \cdot 0 & 2 - 3 \cdot 0 & 3 - 3 \cdot 1 & 0 - 3 \cdot 0 & 27 - 3 (\frac{28}{3}) \\ 0 & 1 & 0 & 0 & - \frac{17}{3} \\ 0 & 0 & 1 & 0 & \frac{28}{3} \\ 0 & 0 & 0 & 1 & - 12 \end{array}]$

Step 2.11.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 2 & 0 & 0 & - 1 \\ 0 & 1 & 0 & 0 & - \frac{17}{3} \\ 0 & 0 & 1 & 0 & \frac{28}{3} \\ 0 & 0 & 0 & 1 & - 12 \end{array}]$

Step 2.12

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 2.12.1

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

$[\begin{array}{c} 1 - 2 \cdot 0 & 2 - 2 \cdot 1 & 0 - 2 \cdot 0 & 0 - 2 \cdot 0 & - 1 - 2 (- \frac{17}{3}) \\ 0 & 1 & 0 & 0 & - \frac{17}{3} \\ 0 & 0 & 1 & 0 & \frac{28}{3} \\ 0 & 0 & 0 & 1 & - 12 \end{array}]$

Step 2.12.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 & \frac{31}{3} \\ 0 & 1 & 0 & 0 & - \frac{17}{3} \\ 0 & 0 & 1 & 0 & \frac{28}{3} \\ 0 & 0 & 0 & 1 & - 12 \end{array}]$

Step 3

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

$w = \frac{31}{3}$

$x = - \frac{17}{3}$

$y = \frac{28}{3}$

$z = - 12$

Step 4

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

$(\frac{31}{3}, - \frac{17}{3}, \frac{28}{3}, - 12)$

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.

$X = [\begin{array}{c} w \\ x \\ y \\ z \end{array}] = [\begin{array}{c} \frac{31}{3} \\ - \frac{17}{3} \\ \frac{28}{3} \\ - 12 \end{array}]$