有限数学示例

$4 a + 4 y - 4 Z = 24$ , $2 a - y + Z = - 9$ , $a - 2 y + 3 Z = 1$

解题步骤 1

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

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解题步骤 1.1

移动 $4 y$ 。

$4 a - 4 Z + 4 y = 24$

$2 a - y + Z = - 9$

$a - 2 y + 3 Z = 1$

解题步骤 1.2

将 $4 a$ 和 $- 4 Z$ 重新排序。

$- 4 Z + 4 a + 4 y = 24$

$2 a - y + Z = - 9$

$a - 2 y + 3 Z = 1$

解题步骤 1.3

移动 $- y$ 。

$- 4 Z + 4 a + 4 y = 24$

$2 a + Z - y = - 9$

$a - 2 y + 3 Z = 1$

解题步骤 1.4

将 $2 a$ 和 $Z$ 重新排序。

$- 4 Z + 4 a + 4 y = 24$

$Z + 2 a - y = - 9$

$a - 2 y + 3 Z = 1$

解题步骤 1.5

移动 $- 2 y$ 。

$- 4 Z + 4 a + 4 y = 24$

$Z + 2 a - y = - 9$

$a + 3 Z - 2 y = 1$

解题步骤 1.6

将 $a$ 和 $3 Z$ 重新排序。

$- 4 Z + 4 a + 4 y = 24$

$Z + 2 a - y = - 9$

$3 Z + a - 2 y = 1$

$- 4 Z + 4 a + 4 y = 24$

$Z + 2 a - y = - 9$

$3 Z + a - 2 y = 1$

解题步骤 2

Write the system as a matrix.

$[\begin{array}{c} - 4 & 4 & 4 & 24 \\ 1 & 2 & - 1 & - 9 \\ 3 & 1 & - 2 & 1 \end{array}]$

解题步骤 3

求行简化阶梯形矩阵。

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解题步骤 3.1

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

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解题步骤 3.1.1

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

$[\begin{array}{c} - \frac{1}{4} \cdot - 4 & - \frac{1}{4} \cdot 4 & - \frac{1}{4} \cdot 4 & - \frac{1}{4} \cdot 24 \\ 1 & 2 & - 1 & - 9 \\ 3 & 1 & - 2 & 1 \end{array}]$

解题步骤 3.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - 1 & - 1 & - 6 \\ 1 & 2 & - 1 & - 9 \\ 3 & 1 & - 2 & 1 \end{array}]$

解题步骤 3.2

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

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解题步骤 3.2.1

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

$[\begin{array}{c} 1 & - 1 & - 1 & - 6 \\ 1 - 1 & 2 + 1 & - 1 + 1 & - 9 + 6 \\ 3 & 1 & - 2 & 1 \end{array}]$

解题步骤 3.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - 1 & - 1 & - 6 \\ 0 & 3 & 0 & - 3 \\ 3 & 1 & - 2 & 1 \end{array}]$

解题步骤 3.3

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

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解题步骤 3.3.1

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

$[\begin{array}{c} 1 & - 1 & - 1 & - 6 \\ 0 & 3 & 0 & - 3 \\ 3 - 3 \cdot 1 & 1 - 3 \cdot - 1 & - 2 - 3 \cdot - 1 & 1 - 3 \cdot - 6 \end{array}]$

解题步骤 3.3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - 1 & - 1 & - 6 \\ 0 & 3 & 0 & - 3 \\ 0 & 4 & 1 & 19 \end{array}]$

解题步骤 3.4

Multiply each element of $R_{2}$ by $\frac{1}{3}$ to make the entry at $2, 2$ a $1$ .

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解题步骤 3.4.1

Multiply each element of $R_{2}$ by $\frac{1}{3}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 1 & - 1 & - 6 \\ \frac{0}{3} & \frac{3}{3} & \frac{0}{3} & \frac{- 3}{3} \\ 0 & 4 & 1 & 19 \end{array}]$

解题步骤 3.4.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - 1 & - 1 & - 6 \\ 0 & 1 & 0 & - 1 \\ 0 & 4 & 1 & 19 \end{array}]$

解题步骤 3.5

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

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解题步骤 3.5.1

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

$[\begin{array}{c} 1 & - 1 & - 1 & - 6 \\ 0 & 1 & 0 & - 1 \\ 0 - 4 \cdot 0 & 4 - 4 \cdot 1 & 1 - 4 \cdot 0 & 19 - 4 \cdot - 1 \end{array}]$

解题步骤 3.5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - 1 & - 1 & - 6 \\ 0 & 1 & 0 & - 1 \\ 0 & 0 & 1 & 23 \end{array}]$

解题步骤 3.6

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

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解题步骤 3.6.1

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

$[\begin{array}{c} 1 + 0 & - 1 + 0 & - 1 + 1 \cdot 1 & - 6 + 1 \cdot 23 \\ 0 & 1 & 0 & - 1 \\ 0 & 0 & 1 & 23 \end{array}]$

解题步骤 3.6.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - 1 & 0 & 17 \\ 0 & 1 & 0 & - 1 \\ 0 & 0 & 1 & 23 \end{array}]$

解题步骤 3.7

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

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解题步骤 3.7.1

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

$[\begin{array}{c} 1 + 0 & - 1 + 1 \cdot 1 & 0 + 0 & 17 - 1 \\ 0 & 1 & 0 & - 1 \\ 0 & 0 & 1 & 23 \end{array}]$

解题步骤 3.7.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 0 & 16 \\ 0 & 1 & 0 & - 1 \\ 0 & 0 & 1 & 23 \end{array}]$

解题步骤 4

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

$Z = 16$

$a = - 1$

$y = 23$

解题步骤 5

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

$(16, - 1, 23)$

有限数学 示例

有限数学示例