Algebra lineare Esempi

$[\begin{array}{c} 1 & 3 \\ 7 & - 3 \end{array}]$

Passaggio 1

Write as an augmented matrix for $A x = 0$ .

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

Passaggio 2

Trova la forma ridotta a scala per righe di Echelon.

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Passaggio 2.1

Perform the row operation $R_{2} = R_{2} - 7 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Passaggio 2.1.1

Perform the row operation $R_{2} = R_{2} - 7 R_{1}$ to make the entry at $2, 1$ a $0$ .

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

Passaggio 2.1.2

Semplifica $R_{2}$ .

$[\begin{array}{c} 1 & 3 & 0 \\ 0 & - 24 & 0 \end{array}]$

Passaggio 2.2

Multiply each element of $R_{2}$ by $- \frac{1}{24}$ to make the entry at $2, 2$ a $1$ .

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Passaggio 2.2.1

Multiply each element of $R_{2}$ by $- \frac{1}{24}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 3 & 0 \\ - \frac{1}{24} \cdot 0 & - \frac{1}{24} \cdot - 24 & - \frac{1}{24} \cdot 0 \end{array}]$

Passaggio 2.2.2

Semplifica $R_{2}$ .

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

Passaggio 2.3

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

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Passaggio 2.3.1

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

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

Passaggio 2.3.2

Semplifica $R_{1}$ .

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

Passaggio 3

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

$x = 0$

$y = 0$

Passaggio 4

Write a solution vector by solving in terms of the free variables in each row.

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 0 \\ 0 \end{array}]$

Passaggio 5

Write as a solution set.

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