线性代数示例

$[\begin{array}{c} 0 & 10 & 20 & 30 \\ 25 & 45 & 65 & 85 \end{array}]$

解题步骤 1

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

$[\begin{array}{c} 0 & 10 & 20 & 30 & 0 \\ 25 & 45 & 65 & 85 & 0 \end{array}]$

解题步骤 2

求行简化阶梯形矩阵。

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解题步骤 2.1

Swap $R_{2}$ with $R_{1}$ to put a nonzero entry at $1, 1$ .

$[\begin{array}{c} 25 & 45 & 65 & 85 & 0 \\ 0 & 10 & 20 & 30 & 0 \end{array}]$

解题步骤 2.2

Multiply each element of $R_{1}$ by $\frac{1}{25}$ to make the entry at $1, 1$ a $1$ .

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解题步骤 2.2.1

Multiply each element of $R_{1}$ by $\frac{1}{25}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{25}{25} & \frac{45}{25} & \frac{65}{25} & \frac{85}{25} & \frac{0}{25} \\ 0 & 10 & 20 & 30 & 0 \end{array}]$

解题步骤 2.2.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{9}{5} & \frac{13}{5} & \frac{17}{5} & 0 \\ 0 & 10 & 20 & 30 & 0 \end{array}]$

解题步骤 2.3

Multiply each element of $R_{2}$ by $\frac{1}{10}$ to make the entry at $2, 2$ a $1$ .

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解题步骤 2.3.1

Multiply each element of $R_{2}$ by $\frac{1}{10}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{9}{5} & \frac{13}{5} & \frac{17}{5} & 0 \\ \frac{0}{10} & \frac{10}{10} & \frac{20}{10} & \frac{30}{10} & \frac{0}{10} \end{array}]$

解题步骤 2.3.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{9}{5} & \frac{13}{5} & \frac{17}{5} & 0 \\ 0 & 1 & 2 & 3 & 0 \end{array}]$

解题步骤 2.4

Perform the row operation $R_{1} = R_{1} - \frac{9}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

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解题步骤 2.4.1

Perform the row operation $R_{1} = R_{1} - \frac{9}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{9}{5} \cdot 0 & \frac{9}{5} - \frac{9}{5} \cdot 1 & \frac{13}{5} - \frac{9}{5} \cdot 2 & \frac{17}{5} - \frac{9}{5} \cdot 3 & 0 - \frac{9}{5} \cdot 0 \\ 0 & 1 & 2 & 3 & 0 \end{array}]$

解题步骤 2.4.2

化简 $R_{1}$ 。

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

解题步骤 3

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

$x_{1} - x_{3} - 2 x_{4} = 0$

$x_{2} + 2 x_{3} + 3 x_{4} = 0$

解题步骤 4

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

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}] = [\begin{array}{c} x_{3} + 2 x_{4} \\ - 2 x_{3} - 3 x_{4} \\ x_{3} \\ x_{4} \end{array}]$

解题步骤 5

Write the solution as a linear combination of vectors.

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

解题步骤 6

Write as a solution set.

${x_{3} [\begin{array}{c} 1 \\ - 2 \\ 1 \\ 0 \end{array}] + x_{4} [\begin{array}{c} 2 \\ - 3 \\ 0 \\ 1 \end{array}] | x_{3}, x_{4} \in R}$

线性代数 示例

线性代数示例