Linear Algebra Examples

$52 f + 48 b + 36 m = 67936$ , $f - b - m = 398$ , $b - 3 m = 453$

Step 1

Move variables to the left and constant terms to the right.

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Step 1.1

Reorder $52 f$ and $48 b$ .

$48 b + 52 f + 36 m = 67936$

$f - b - m = 398$

$b - 3 m = 453$

Step 1.2

Reorder $f$ and $- b$ .

$48 b + 52 f + 36 m = 67936$

$- b + f - m = 398$

$b - 3 m = 453$

$48 b + 52 f + 36 m = 67936$

$- b + f - m = 398$

$b - 3 m = 453$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 48 & 52 & 36 & 67936 \\ - 1 & 1 & - 1 & 398 \\ 1 & 0 & - 3 & 453 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Multiply each element of $R_{1}$ by $\frac{1}{48}$ to make the entry at $1, 1$ a $1$ .

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Step 3.1.1

Multiply each element of $R_{1}$ by $\frac{1}{48}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{48}{48} & \frac{52}{48} & \frac{36}{48} & \frac{67936}{48} \\ - 1 & 1 & - 1 & 398 \\ 1 & 0 & - 3 & 453 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{13}{12} & \frac{3}{4} & \frac{4246}{3} \\ - 1 & 1 & - 1 & 398 \\ 1 & 0 & - 3 & 453 \end{array}]$

Step 3.2

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

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Step 3.2.1

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

$[\begin{array}{c} 1 & \frac{13}{12} & \frac{3}{4} & \frac{4246}{3} \\ - 1 + 1 \cdot 1 & 1 + \frac{13}{12} & - 1 + \frac{3}{4} & 398 + \frac{4246}{3} \\ 1 & 0 & - 3 & 453 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{13}{12} & \frac{3}{4} & \frac{4246}{3} \\ 0 & \frac{25}{12} & - \frac{1}{4} & \frac{5440}{3} \\ 1 & 0 & - 3 & 453 \end{array}]$

Step 3.3

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

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Step 3.3.1

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

$[\begin{array}{c} 1 & \frac{13}{12} & \frac{3}{4} & \frac{4246}{3} \\ 0 & \frac{25}{12} & - \frac{1}{4} & \frac{5440}{3} \\ 1 - 1 & 0 - \frac{13}{12} & - 3 - \frac{3}{4} & 453 - \frac{4246}{3} \end{array}]$

Step 3.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{13}{12} & \frac{3}{4} & \frac{4246}{3} \\ 0 & \frac{25}{12} & - \frac{1}{4} & \frac{5440}{3} \\ 0 & - \frac{13}{12} & - \frac{15}{4} & - \frac{2887}{3} \end{array}]$

Step 3.4

Multiply each element of $R_{2}$ by $\frac{12}{25}$ to make the entry at $2, 2$ a $1$ .

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Step 3.4.1

Multiply each element of $R_{2}$ by $\frac{12}{25}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{13}{12} & \frac{3}{4} & \frac{4246}{3} \\ \frac{12}{25} \cdot 0 & \frac{12}{25} \cdot \frac{25}{12} & \frac{12}{25} (- \frac{1}{4}) & \frac{12}{25} \cdot \frac{5440}{3} \\ 0 & - \frac{13}{12} & - \frac{15}{4} & - \frac{2887}{3} \end{array}]$

Step 3.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{13}{12} & \frac{3}{4} & \frac{4246}{3} \\ 0 & 1 & - \frac{3}{25} & \frac{4352}{5} \\ 0 & - \frac{13}{12} & - \frac{15}{4} & - \frac{2887}{3} \end{array}]$

Step 3.5

Perform the row operation $R_{3} = R_{3} + \frac{13}{12} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 3.5.1

Perform the row operation $R_{3} = R_{3} + \frac{13}{12} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{13}{12} & \frac{3}{4} & \frac{4246}{3} \\ 0 & 1 & - \frac{3}{25} & \frac{4352}{5} \\ 0 + \frac{13}{12} \cdot 0 & - \frac{13}{12} + \frac{13}{12} \cdot 1 & - \frac{15}{4} + \frac{13}{12} (- \frac{3}{25}) & - \frac{2887}{3} + \frac{13}{12} \cdot \frac{4352}{5} \end{array}]$

Step 3.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{13}{12} & \frac{3}{4} & \frac{4246}{3} \\ 0 & 1 & - \frac{3}{25} & \frac{4352}{5} \\ 0 & 0 & - \frac{97}{25} & - \frac{97}{5} \end{array}]$

Step 3.6

Multiply each element of $R_{3}$ by $- \frac{25}{97}$ to make the entry at $3, 3$ a $1$ .

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Step 3.6.1

Multiply each element of $R_{3}$ by $- \frac{25}{97}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & \frac{13}{12} & \frac{3}{4} & \frac{4246}{3} \\ 0 & 1 & - \frac{3}{25} & \frac{4352}{5} \\ - \frac{25}{97} \cdot 0 & - \frac{25}{97} \cdot 0 & - \frac{25}{97} (- \frac{97}{25}) & - \frac{25}{97} (- \frac{97}{5}) \end{array}]$

Step 3.6.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{13}{12} & \frac{3}{4} & \frac{4246}{3} \\ 0 & 1 & - \frac{3}{25} & \frac{4352}{5} \\ 0 & 0 & 1 & 5 \end{array}]$

Step 3.7

Perform the row operation $R_{2} = R_{2} + \frac{3}{25} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 3.7.1

Perform the row operation $R_{2} = R_{2} + \frac{3}{25} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & \frac{13}{12} & \frac{3}{4} & \frac{4246}{3} \\ 0 + \frac{3}{25} \cdot 0 & 1 + \frac{3}{25} \cdot 0 & - \frac{3}{25} + \frac{3}{25} \cdot 1 & \frac{4352}{5} + \frac{3}{25} \cdot 5 \\ 0 & 0 & 1 & 5 \end{array}]$

Step 3.7.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{13}{12} & \frac{3}{4} & \frac{4246}{3} \\ 0 & 1 & 0 & 871 \\ 0 & 0 & 1 & 5 \end{array}]$

Step 3.8

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

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Step 3.8.1

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

$[\begin{array}{c} 1 - \frac{3}{4} \cdot 0 & \frac{13}{12} - \frac{3}{4} \cdot 0 & \frac{3}{4} - \frac{3}{4} \cdot 1 & \frac{4246}{3} - \frac{3}{4} \cdot 5 \\ 0 & 1 & 0 & 871 \\ 0 & 0 & 1 & 5 \end{array}]$

Step 3.8.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{13}{12} & 0 & \frac{16939}{12} \\ 0 & 1 & 0 & 871 \\ 0 & 0 & 1 & 5 \end{array}]$

Step 3.9

Perform the row operation $R_{1} = R_{1} - \frac{13}{12} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 3.9.1

Perform the row operation $R_{1} = R_{1} - \frac{13}{12} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{13}{12} \cdot 0 & \frac{13}{12} - \frac{13}{12} \cdot 1 & 0 - \frac{13}{12} \cdot 0 & \frac{16939}{12} - \frac{13}{12} \cdot 871 \\ 0 & 1 & 0 & 871 \\ 0 & 0 & 1 & 5 \end{array}]$

Step 3.9.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 468 \\ 0 & 1 & 0 & 871 \\ 0 & 0 & 1 & 5 \end{array}]$

Step 4

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

$b = 468$

$f = 871$

$m = 5$

Step 5

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

$(468, 871, 5)$