Linear Algebra Examples

$3 a + 5 b + 2 d = 0$ , $4 a + 7 b + 5 d = 0$ , $a + b - 4 d = 0$ , $2 a + 9 b + 6 d = 0$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 3 & 5 & 2 & 0 \\ 4 & 7 & 5 & 0 \\ 1 & 1 & - 4 & 0 \\ 2 & 9 & 6 & 0 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

$[\begin{array}{c} \frac{3}{3} & \frac{5}{3} & \frac{2}{3} & \frac{0}{3} \\ 4 & 7 & 5 & 0 \\ 1 & 1 & - 4 & 0 \\ 2 & 9 & 6 & 0 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{5}{3} & \frac{2}{3} & 0 \\ 4 & 7 & 5 & 0 \\ 1 & 1 & - 4 & 0 \\ 2 & 9 & 6 & 0 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & \frac{5}{3} & \frac{2}{3} & 0 \\ 4 - 4 \cdot 1 & 7 - 4 (\frac{5}{3}) & 5 - 4 (\frac{2}{3}) & 0 - 4 \cdot 0 \\ 1 & 1 & - 4 & 0 \\ 2 & 9 & 6 & 0 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{5}{3} & \frac{2}{3} & 0 \\ 0 & \frac{1}{3} & \frac{7}{3} & 0 \\ 1 & 1 & - 4 & 0 \\ 2 & 9 & 6 & 0 \end{array}]$

Step 2.3

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

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Step 2.3.1

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

$[\begin{array}{c} 1 & \frac{5}{3} & \frac{2}{3} & 0 \\ 0 & \frac{1}{3} & \frac{7}{3} & 0 \\ 1 - 1 & 1 - \frac{5}{3} & - 4 - \frac{2}{3} & 0 - 0 \\ 2 & 9 & 6 & 0 \end{array}]$

Step 2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{5}{3} & \frac{2}{3} & 0 \\ 0 & \frac{1}{3} & \frac{7}{3} & 0 \\ 0 & - \frac{2}{3} & - \frac{14}{3} & 0 \\ 2 & 9 & 6 & 0 \end{array}]$

Step 2.4

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 2.4.1

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

$[\begin{array}{c} 1 & \frac{5}{3} & \frac{2}{3} & 0 \\ 0 & \frac{1}{3} & \frac{7}{3} & 0 \\ 0 & - \frac{2}{3} & - \frac{14}{3} & 0 \\ 2 - 2 \cdot 1 & 9 - 2 (\frac{5}{3}) & 6 - 2 (\frac{2}{3}) & 0 - 2 \cdot 0 \end{array}]$

Step 2.4.2

Simplify $R_{4}$ .

$[\begin{array}{c} 1 & \frac{5}{3} & \frac{2}{3} & 0 \\ 0 & \frac{1}{3} & \frac{7}{3} & 0 \\ 0 & - \frac{2}{3} & - \frac{14}{3} & 0 \\ 0 & \frac{17}{3} & \frac{14}{3} & 0 \end{array}]$

Step 2.5

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

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Step 2.5.1

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

$[\begin{array}{c} 1 & \frac{5}{3} & \frac{2}{3} & 0 \\ 3 \cdot 0 & 3 (\frac{1}{3}) & 3 (\frac{7}{3}) & 3 \cdot 0 \\ 0 & - \frac{2}{3} & - \frac{14}{3} & 0 \\ 0 & \frac{17}{3} & \frac{14}{3} & 0 \end{array}]$

Step 2.5.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{5}{3} & \frac{2}{3} & 0 \\ 0 & 1 & 7 & 0 \\ 0 & - \frac{2}{3} & - \frac{14}{3} & 0 \\ 0 & \frac{17}{3} & \frac{14}{3} & 0 \end{array}]$

Step 2.6

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

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Step 2.6.1

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

$[\begin{array}{c} 1 & \frac{5}{3} & \frac{2}{3} & 0 \\ 0 & 1 & 7 & 0 \\ 0 + \frac{2}{3} \cdot 0 & - \frac{2}{3} + \frac{2}{3} \cdot 1 & - \frac{14}{3} + \frac{2}{3} \cdot 7 & 0 + \frac{2}{3} \cdot 0 \\ 0 & \frac{17}{3} & \frac{14}{3} & 0 \end{array}]$

Step 2.6.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{5}{3} & \frac{2}{3} & 0 \\ 0 & 1 & 7 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & \frac{17}{3} & \frac{14}{3} & 0 \end{array}]$

Step 2.7

Perform the row operation $R_{4} = R_{4} - \frac{17}{3} R_{2}$ to make the entry at $4, 2$ a $0$ .

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Step 2.7.1

Perform the row operation $R_{4} = R_{4} - \frac{17}{3} R_{2}$ to make the entry at $4, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{5}{3} & \frac{2}{3} & 0 \\ 0 & 1 & 7 & 0 \\ 0 & 0 & 0 & 0 \\ 0 - \frac{17}{3} \cdot 0 & \frac{17}{3} - \frac{17}{3} \cdot 1 & \frac{14}{3} - \frac{17}{3} \cdot 7 & 0 - \frac{17}{3} \cdot 0 \end{array}]$

Step 2.7.2

Simplify $R_{4}$ .

$[\begin{array}{c} 1 & \frac{5}{3} & \frac{2}{3} & 0 \\ 0 & 1 & 7 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & - 35 & 0 \end{array}]$

Step 2.8

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

$[\begin{array}{c} 1 & \frac{5}{3} & \frac{2}{3} & 0 \\ 0 & 1 & 7 & 0 \\ 0 & 0 & - 35 & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 2.9

Multiply each element of $R_{3}$ by $- \frac{1}{35}$ to make the entry at $3, 3$ a $1$ .

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Step 2.9.1

Multiply each element of $R_{3}$ by $- \frac{1}{35}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & \frac{5}{3} & \frac{2}{3} & 0 \\ 0 & 1 & 7 & 0 \\ - \frac{1}{35} \cdot 0 & - \frac{1}{35} \cdot 0 & - \frac{1}{35} \cdot - 35 & - \frac{1}{35} \cdot 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 2.9.2

Simplify $R_{3}$ .

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

Step 2.10

Perform the row operation $R_{2} = R_{2} - 7 R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 2.10.1

Perform the row operation $R_{2} = R_{2} - 7 R_{3}$ to make the entry at $2, 3$ a $0$ .

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

Step 2.10.2

Simplify $R_{2}$ .

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

Step 2.11

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

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Step 2.11.1

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

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

Step 2.11.2

Simplify $R_{1}$ .

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

Step 2.12

Perform the row operation $R_{1} = R_{1} - \frac{5}{3} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 2.12.1

Perform the row operation $R_{1} = R_{1} - \frac{5}{3} R_{2}$ to make the entry at $1, 2$ a $0$ .

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

Step 2.12.2

Simplify $R_{1}$ .

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

Step 3

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

$a = 0$

$b = 0$

$d = 0$

$0 = 0$

Step 4

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

$(0, 0, 0)$