Algebra lineare Esempi

$- 2 a + 3 b = 15$ , $\frac{4}{3} a + 2 = - 2 b$

Passaggio 1

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

Tocca per altri passaggi...

Passaggio 1.1

Somma $2 b$ a entrambi i lati dell'equazione.

$- 2 a + 3 b = 15$

$\frac{4}{3} a + 2 + 2 b = 0$

Passaggio 1.2

$\frac{4}{3}$ e $a$ .

$- 2 a + 3 b = 15$

$\frac{4 a}{3} + 2 + 2 b = 0$

Passaggio 1.3

Sottrai $2$ da entrambi i lati dell'equazione.

$- 2 a + 3 b = 15$

$\frac{4 a}{3} + 2 b = - 2$

Passaggio 1.4

Riordina i termini.

$- 2 a + 3 b = 15$

$\frac{4}{3} a + 2 b = - 2$

$- 2 a + 3 b = 15$

$\frac{4}{3} a + 2 b = - 2$

Passaggio 2

Write the system as a matrix.

$[\begin{array}{c} - 2 & 3 & 15 \\ \frac{4}{3} & 2 & - 2 \end{array}]$

Passaggio 3

Trova la forma ridotta a scala per righe di Echelon.

Tocca per altri passaggi...

Passaggio 3.1

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

Tocca per altri passaggi...

Passaggio 3.1.1

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

$[\begin{array}{c} - \frac{1}{2} \cdot - 2 & - \frac{1}{2} \cdot 3 & - \frac{1}{2} \cdot 15 \\ \frac{4}{3} & 2 & - 2 \end{array}]$

Passaggio 3.1.2

Semplifica $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{3}{2} & - \frac{15}{2} \\ \frac{4}{3} & 2 & - 2 \end{array}]$

Passaggio 3.2

Perform the row operation $R_{2} = R_{2} - \frac{4}{3} R_{1}$ to make the entry at $2, 1$ a $0$ .

Tocca per altri passaggi...

Passaggio 3.2.1

Perform the row operation $R_{2} = R_{2} - \frac{4}{3} R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{3}{2} & - \frac{15}{2} \\ \frac{4}{3} - \frac{4}{3} \cdot 1 & 2 - \frac{4}{3} (- \frac{3}{2}) & - 2 - \frac{4}{3} (- \frac{15}{2}) \end{array}]$

Passaggio 3.2.2

Semplifica $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{3}{2} & - \frac{15}{2} \\ 0 & 4 & 8 \end{array}]$

Passaggio 3.3

Multiply each element of $R_{2}$ by $\frac{1}{4}$ to make the entry at $2, 2$ a $1$ .

Tocca per altri passaggi...

Passaggio 3.3.1

Multiply each element of $R_{2}$ by $\frac{1}{4}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{3}{2} & - \frac{15}{2} \\ \frac{0}{4} & \frac{4}{4} & \frac{8}{4} \end{array}]$

Passaggio 3.3.2

Semplifica $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{3}{2} & - \frac{15}{2} \\ 0 & 1 & 2 \end{array}]$

Passaggio 3.4

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

Tocca per altri passaggi...

Passaggio 3.4.1

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

$[\begin{array}{c} 1 + \frac{3}{2} \cdot 0 & - \frac{3}{2} + \frac{3}{2} \cdot 1 & - \frac{15}{2} + \frac{3}{2} \cdot 2 \\ 0 & 1 & 2 \end{array}]$

Passaggio 3.4.2

Semplifica $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - \frac{9}{2} \\ 0 & 1 & 2 \end{array}]$

Passaggio 4

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

$a = - \frac{9}{2}$

$b = 2$

Passaggio 5

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

$(- \frac{9}{2}, 2)$