Finite Math Examples

$x + y + z = 102$ , $2 x + 3 y + 4 z = 303$ , $11 x + 17 y + 23 z = 1517$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 1 & 1 & 1 & 102 \\ 2 & 3 & 4 & 303 \\ 11 & 17 & 23 & 1517 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

$[\begin{array}{c} 1 & 1 & 1 & 102 \\ 2 - 2 \cdot 1 & 3 - 2 \cdot 1 & 4 - 2 \cdot 1 & 303 - 2 \cdot 102 \\ 11 & 17 & 23 & 1517 \end{array}]$

Step 2.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 102 \\ 0 & 1 & 2 & 99 \\ 11 & 17 & 23 & 1517 \end{array}]$

Step 2.2

Perform the row operation $R_{3} = R_{3} - 11 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 2.2.1

Perform the row operation $R_{3} = R_{3} - 11 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & 1 & 1 & 102 \\ 0 & 1 & 2 & 99 \\ 11 - 11 \cdot 1 & 17 - 11 \cdot 1 & 23 - 11 \cdot 1 & 1517 - 11 \cdot 102 \end{array}]$

Step 2.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 102 \\ 0 & 1 & 2 & 99 \\ 0 & 6 & 12 & 395 \end{array}]$

Step 2.3

Perform the row operation $R_{3} = R_{3} - 6 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 2.3.1

Perform the row operation $R_{3} = R_{3} - 6 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & 1 & 1 & 102 \\ 0 & 1 & 2 & 99 \\ 0 - 6 \cdot 0 & 6 - 6 \cdot 1 & 12 - 6 \cdot 2 & 395 - 6 \cdot 99 \end{array}]$

Step 2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 102 \\ 0 & 1 & 2 & 99 \\ 0 & 0 & 0 & - 199 \end{array}]$

Step 2.4

Multiply each element of $R_{3}$ by $- \frac{1}{199}$ to make the entry at $3, 4$ a $1$ .

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Step 2.4.1

Multiply each element of $R_{3}$ by $- \frac{1}{199}$ to make the entry at $3, 4$ a $1$ .

$[\begin{array}{c} 1 & 1 & 1 & 102 \\ 0 & 1 & 2 & 99 \\ - \frac{1}{199} \cdot 0 & - \frac{1}{199} \cdot 0 & - \frac{1}{199} \cdot 0 & - \frac{1}{199} \cdot - 199 \end{array}]$

Step 2.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 102 \\ 0 & 1 & 2 & 99 \\ 0 & 0 & 0 & 1 \end{array}]$

Step 2.5

Perform the row operation $R_{2} = R_{2} - 99 R_{3}$ to make the entry at $2, 4$ a $0$ .

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Step 2.5.1

Perform the row operation $R_{2} = R_{2} - 99 R_{3}$ to make the entry at $2, 4$ a $0$ .

$[\begin{array}{c} 1 & 1 & 1 & 102 \\ 0 - 99 \cdot 0 & 1 - 99 \cdot 0 & 2 - 99 \cdot 0 & 99 - 99 \cdot 1 \\ 0 & 0 & 0 & 1 \end{array}]$

Step 2.5.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 102 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

Step 2.6

Perform the row operation $R_{1} = R_{1} - 102 R_{3}$ to make the entry at $1, 4$ a $0$ .

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Step 2.6.1

Perform the row operation $R_{1} = R_{1} - 102 R_{3}$ to make the entry at $1, 4$ a $0$ .

$[\begin{array}{c} 1 - 102 \cdot 0 & 1 - 102 \cdot 0 & 1 - 102 \cdot 0 & 102 - 102 \cdot 1 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

Step 2.6.2

Simplify $R_{1}$ .

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

Step 2.7

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

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Step 2.7.1

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

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

Step 2.7.2

Simplify $R_{1}$ .

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

Step 3

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

$x - z = 0$

$y + 2 z = 0$

$0 = 1$

Step 4

The system is inconsistent so there is no solution.

$No solution$