Álgebra lineal Ejemplos

$9 c - 3 d = - 33$ , $6 c + 5 d = - 8$

Paso 1

Write the system as a matrix.

$[\begin{array}{c} 9 & - 3 & - 33 \\ 6 & 5 & - 8 \end{array}]$

Paso 2

Obtén la forma escalonada reducida por filas.

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Paso 2.1

Multiply each element of $R_{1}$ by $\frac{1}{9}$ to make the entry at $1, 1$ a $1$ .

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Paso 2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{9}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{9}{9} & \frac{- 3}{9} & \frac{- 33}{9} \\ 6 & 5 & - 8 \end{array}]$

Paso 2.1.2

Simplifica $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{3} & - \frac{11}{3} \\ 6 & 5 & - 8 \end{array}]$

Paso 2.2

Perform the row operation $R_{2} = R_{2} - 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Paso 2.2.1

Perform the row operation $R_{2} = R_{2} - 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{3} & - \frac{11}{3} \\ 6 - 6 \cdot 1 & 5 - 6 (- \frac{1}{3}) & - 8 - 6 (- \frac{11}{3}) \end{array}]$

Paso 2.2.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{3} & - \frac{11}{3} \\ 0 & 7 & 14 \end{array}]$

Paso 2.3

Multiply each element of $R_{2}$ by $\frac{1}{7}$ to make the entry at $2, 2$ a $1$ .

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Paso 2.3.1

Multiply each element of $R_{2}$ by $\frac{1}{7}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{3} & - \frac{11}{3} \\ \frac{0}{7} & \frac{7}{7} & \frac{14}{7} \end{array}]$

Paso 2.3.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{3} & - \frac{11}{3} \\ 0 & 1 & 2 \end{array}]$

Paso 2.4

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

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Paso 2.4.1

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

$[\begin{array}{c} 1 + \frac{1}{3} \cdot 0 & - \frac{1}{3} + \frac{1}{3} \cdot 1 & - \frac{11}{3} + \frac{1}{3} \cdot 2 \\ 0 & 1 & 2 \end{array}]$

Paso 2.4.2

Simplifica $R_{1}$ .

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

Paso 3

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

$c = - 3$

$d = 2$

Paso 4

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

$(- 3, 2)$