Finite Math Examples

$- x + y = - 1$ , $3 x - 2 y = 0$ , $2 x - y = 4$

Step 1

Write the system as a matrix.

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

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

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

Step 2.1.2

Simplify $R_{1}$ .

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

Simplify $R_{2}$ .

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

Simplify $R_{3}$ .

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

Step 2.4

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

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Step 2.4.1

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

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

Step 2.4.2

Simplify $R_{3}$ .

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

Step 2.5

Multiply each element of $R_{3}$ by $\frac{1}{5}$ to make the entry at $3, 3$ a $1$ .

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Step 2.5.1

Multiply each element of $R_{3}$ by $\frac{1}{5}$ to make the entry at $3, 3$ a $1$ .

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

Step 2.5.2

Simplify $R_{3}$ .

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

Step 2.6

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

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Step 2.6.1

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

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

Step 2.6.2

Simplify $R_{2}$ .

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

Step 2.7

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

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Step 2.7.1

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

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

Step 2.7.2

Simplify $R_{1}$ .

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

Step 2.8

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

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Step 2.8.1

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

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

Step 2.8.2

Simplify $R_{1}$ .

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

Step 3

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

$x = 0$

$y = 0$

$0 = 1$

Step 4

The system is inconsistent so there is no solution.

$No solution$