Finite Math Examples

$x - 4 y - 3 z = 0$ , $- 2 x + 9 y + 4 z = - 2$ , $3 x - 13 y - 7 z = 2$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 1 & - 4 & - 3 & 0 \\ - 2 & 9 & 4 & - 2 \\ 3 & - 13 & - 7 & 2 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & - 4 & - 3 & 0 \\ - 2 + 2 \cdot 1 & 9 + 2 \cdot - 4 & 4 + 2 \cdot - 3 & - 2 + 2 \cdot 0 \\ 3 & - 13 & - 7 & 2 \end{array}]$

Step 2.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 4 & - 3 & 0 \\ 0 & 1 & - 2 & - 2 \\ 3 & - 13 & - 7 & 2 \end{array}]$

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} 1 & - 4 & - 3 & 0 \\ 0 & 1 & - 2 & - 2 \\ 3 - 3 \cdot 1 & - 13 - 3 \cdot - 4 & - 7 - 3 \cdot - 3 & 2 - 3 \cdot 0 \end{array}]$

Step 2.2.2

Simplify $R_{3}$ .

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

Simplify $R_{3}$ .

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

Step 2.4

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

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Step 2.4.1

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

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

Step 2.4.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - 11 & - 8 \\ 0 & 1 & - 2 & - 2 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 3

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

$x - 11 z = - 8$

$y - 2 z = - 2$

$0 = 0$

Step 4

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

$(- 8 + 11 z, - 2 + 2 z, z)$