有限数学示例

$x - y + 4 z + w = - 1.3$ , $- 2 x - 5 y - 16 z - 8 w = 7.6$ , $4 x - y + 12 z = - 1$ , $3 x - 2 y + 7 z + 6 w = - 6.7$

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 $4 z$ 。

$x - y + w + 4 z = - 1.3$

$- 2 x - 5 y - 16 z - 8 w = 7.6$

$4 x - y + 12 z = - 1$

$3 x - 2 y + 7 z + 6 w = - 6.7$

解题步骤 1.2

移动 $- y$ 。

$x + w - y + 4 z = - 1.3$

$- 2 x - 5 y - 16 z - 8 w = 7.6$

$4 x - y + 12 z = - 1$

$3 x - 2 y + 7 z + 6 w = - 6.7$

解题步骤 1.3

将 $x$ 和 $w$ 重新排序。

$w + x - y + 4 z = - 1.3$

$- 2 x - 5 y - 16 z - 8 w = 7.6$

$4 x - y + 12 z = - 1$

$3 x - 2 y + 7 z + 6 w = - 6.7$

解题步骤 1.4

移动 $- 16 z$ 。

$w + x - y + 4 z = - 1.3$

$- 2 x - 5 y - 8 w - 16 z = 7.6$

$4 x - y + 12 z = - 1$

$3 x - 2 y + 7 z + 6 w = - 6.7$

解题步骤 1.5

移动 $- 5 y$ 。

$w + x - y + 4 z = - 1.3$

$- 2 x - 8 w - 5 y - 16 z = 7.6$

$4 x - y + 12 z = - 1$

$3 x - 2 y + 7 z + 6 w = - 6.7$

解题步骤 1.6

将 $- 2 x$ 和 $- 8 w$ 重新排序。

$w + x - y + 4 z = - 1.3$

$- 8 w - 2 x - 5 y - 16 z = 7.6$

$4 x - y + 12 z = - 1$

$3 x - 2 y + 7 z + 6 w = - 6.7$

解题步骤 1.7

移动 $7 z$ 。

$w + x - y + 4 z = - 1.3$

$- 8 w - 2 x - 5 y - 16 z = 7.6$

$4 x - y + 12 z = - 1$

$3 x - 2 y + 6 w + 7 z = - 6.7$

解题步骤 1.8

移动 $- 2 y$ 。

$w + x - y + 4 z = - 1.3$

$- 8 w - 2 x - 5 y - 16 z = 7.6$

$4 x - y + 12 z = - 1$

$3 x + 6 w - 2 y + 7 z = - 6.7$

解题步骤 1.9

将 $3 x$ 和 $6 w$ 重新排序。

$w + x - y + 4 z = - 1.3$

$- 8 w - 2 x - 5 y - 16 z = 7.6$

$4 x - y + 12 z = - 1$

$6 w + 3 x - 2 y + 7 z = - 6.7$

$w + x - y + 4 z = - 1.3$

$- 8 w - 2 x - 5 y - 16 z = 7.6$

$4 x - y + 12 z = - 1$

$6 w + 3 x - 2 y + 7 z = - 6.7$

解题步骤 2

Write the system as a matrix.

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ - 8 & - 2 & - 5 & - 16 & 7.6 \\ 0 & 4 & - 1 & 12 & - 1 \\ 6 & 3 & - 2 & 7 & - 6.7 \end{array}]$

解题步骤 3

求行简化阶梯形矩阵。

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解题步骤 3.1

Perform the row operation $R_{2} = R_{2} + 8 R_{1}$ to make the entry at $2, 1$ a $0$ .

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解题步骤 3.1.1

Perform the row operation $R_{2} = R_{2} + 8 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ - 8 + 8 \cdot 1 & - 2 + 8 \cdot 1 & - 5 + 8 \cdot - 1 & - 16 + 8 \cdot 4 & 7.6 + 8 \cdot - 1.3 \\ 0 & 4 & - 1 & 12 & - 1 \\ 6 & 3 & - 2 & 7 & - 6.7 \end{array}]$

解题步骤 3.1.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ 0 & 6 & - 13 & 16 & - 2.8 \\ 0 & 4 & - 1 & 12 & - 1 \\ 6 & 3 & - 2 & 7 & - 6.7 \end{array}]$

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ 0 & 6 & - 13 & 16 & - 2.8 \\ 0 & 4 & - 1 & 12 & - 1 \\ 6 & 3 & - 2 & 7 & - 6.7 \end{array}]$

解题步骤 3.2

Perform the row operation $R_{4} = R_{4} - 6 R_{1}$ to make the entry at $4, 1$ a $0$ .

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解题步骤 3.2.1

Perform the row operation $R_{4} = R_{4} - 6 R_{1}$ to make the entry at $4, 1$ a $0$ .

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ 0 & 6 & - 13 & 16 & - 2.8 \\ 0 & 4 & - 1 & 12 & - 1 \\ 6 - 6 \cdot 1 & 3 - 6 \cdot 1 & - 2 - 6 \cdot - 1 & 7 - 6 \cdot 4 & - 6.7 - 6 \cdot - 1.3 \end{array}]$

解题步骤 3.2.2

化简 $R_{4}$ 。

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ 0 & 6 & - 13 & 16 & - 2.8 \\ 0 & 4 & - 1 & 12 & - 1 \\ 0 & - 3 & 4 & - 17 & 1.1 \end{array}]$

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ 0 & 6 & - 13 & 16 & - 2.8 \\ 0 & 4 & - 1 & 12 & - 1 \\ 0 & - 3 & 4 & - 17 & 1.1 \end{array}]$

解题步骤 3.3

Multiply each element of $R_{2}$ by $\frac{1}{6}$ to make the entry at $2, 2$ a $1$ .

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解题步骤 3.3.1

Multiply each element of $R_{2}$ by $\frac{1}{6}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ \frac{0}{6} & \frac{6}{6} & \frac{- 13}{6} & \frac{16}{6} & \frac{- 2.8}{6} \\ 0 & 4 & - 1 & 12 & - 1 \\ 0 & - 3 & 4 & - 17 & 1.1 \end{array}]$

解题步骤 3.3.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ 0 & 1 & - \frac{13}{6} & \frac{8}{3} & - 0.4\overline{6} \\ 0 & 4 & - 1 & 12 & - 1 \\ 0 & - 3 & 4 & - 17 & 1.1 \end{array}]$

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ 0 & 1 & - \frac{13}{6} & \frac{8}{3} & - 0.4\overline{6} \\ 0 & 4 & - 1 & 12 & - 1 \\ 0 & - 3 & 4 & - 17 & 1.1 \end{array}]$

解题步骤 3.4

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 3.4.1

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

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ 0 & 1 & - \frac{13}{6} & \frac{8}{3} & - 0.4\overline{6} \\ 0 - 4 \cdot 0 & 4 - 4 \cdot 1 & - 1 - 4 (- \frac{13}{6}) & 12 - 4 (\frac{8}{3}) & - 1 - 4 \cdot - 0.4\overline{6} \\ 0 & - 3 & 4 & - 17 & 1.1 \end{array}]$

解题步骤 3.4.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ 0 & 1 & - \frac{13}{6} & \frac{8}{3} & - 0.4\overline{6} \\ 0 & 0 & \frac{23}{3} & \frac{4}{3} & 0.8\overline{6} \\ 0 & - 3 & 4 & - 17 & 1.1 \end{array}]$

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ 0 & 1 & - \frac{13}{6} & \frac{8}{3} & - 0.4\overline{6} \\ 0 & 0 & \frac{23}{3} & \frac{4}{3} & 0.8\overline{6} \\ 0 & - 3 & 4 & - 17 & 1.1 \end{array}]$

解题步骤 3.5

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

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解题步骤 3.5.1

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

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ 0 & 1 & - \frac{13}{6} & \frac{8}{3} & - 0.4\overline{6} \\ 0 & 0 & \frac{23}{3} & \frac{4}{3} & 0.8\overline{6} \\ 0 + 3 \cdot 0 & - 3 + 3 \cdot 1 & 4 + 3 (- \frac{13}{6}) & - 17 + 3 (\frac{8}{3}) & 1.1 + 3 \cdot - 0.4\overline{6} \end{array}]$

解题步骤 3.5.2

化简 $R_{4}$ 。

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ 0 & 1 & - \frac{13}{6} & \frac{8}{3} & - 0.4\overline{6} \\ 0 & 0 & \frac{23}{3} & \frac{4}{3} & 0.8\overline{6} \\ 0 & 0 & - \frac{5}{2} & - 9 & - 0.3 \end{array}]$

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ 0 & 1 & - \frac{13}{6} & \frac{8}{3} & - 0.4\overline{6} \\ 0 & 0 & \frac{23}{3} & \frac{4}{3} & 0.8\overline{6} \\ 0 & 0 & - \frac{5}{2} & - 9 & - 0.3 \end{array}]$

解题步骤 3.6

Multiply each element of $R_{3}$ by $\frac{3}{23}$ to make the entry at $3, 3$ a $1$ .

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解题步骤 3.6.1

Multiply each element of $R_{3}$ by $\frac{3}{23}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ 0 & 1 & - \frac{13}{6} & \frac{8}{3} & - 0.4\overline{6} \\ \frac{3}{23} \cdot 0 & \frac{3}{23} \cdot 0 & \frac{3}{23} \cdot \frac{23}{3} & \frac{3}{23} \cdot \frac{4}{3} & \frac{3}{23} \cdot 0.8\overline{6} \\ 0 & 0 & - \frac{5}{2} & - 9 & - 0.3 \end{array}]$

解题步骤 3.6.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ 0 & 1 & - \frac{13}{6} & \frac{8}{3} & - 0.4\overline{6} \\ 0 & 0 & 1 & \frac{4}{23} & 0.11304347 \\ 0 & 0 & - \frac{5}{2} & - 9 & - 0.3 \end{array}]$

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ 0 & 1 & - \frac{13}{6} & \frac{8}{3} & - 0.4\overline{6} \\ 0 & 0 & 1 & \frac{4}{23} & 0.11304347 \\ 0 & 0 & - \frac{5}{2} & - 9 & - 0.3 \end{array}]$

解题步骤 3.7

Perform the row operation $R_{4} = R_{4} + \frac{5}{2} R_{3}$ to make the entry at $4, 3$ a $0$ .

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解题步骤 3.7.1

Perform the row operation $R_{4} = R_{4} + \frac{5}{2} R_{3}$ to make the entry at $4, 3$ a $0$ .

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ 0 & 1 & - \frac{13}{6} & \frac{8}{3} & - 0.4\overline{6} \\ 0 & 0 & 1 & \frac{4}{23} & 0.11304347 \\ 0 + \frac{5}{2} \cdot 0 & 0 + \frac{5}{2} \cdot 0 & - \frac{5}{2} + \frac{5}{2} \cdot 1 & - 9 + \frac{5}{2} \cdot \frac{4}{23} & - 0.3 + \frac{5}{2} \cdot 0.11304347 \end{array}]$

解题步骤 3.7.2

化简 $R_{4}$ 。

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ 0 & 1 & - \frac{13}{6} & \frac{8}{3} & - 0.4\overline{6} \\ 0 & 0 & 1 & \frac{4}{23} & 0.11304347 \\ 0 & 0 & 0 & - \frac{197}{23} & - 0.0173913 \end{array}]$

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ 0 & 1 & - \frac{13}{6} & \frac{8}{3} & - 0.4\overline{6} \\ 0 & 0 & 1 & \frac{4}{23} & 0.11304347 \\ 0 & 0 & 0 & - \frac{197}{23} & - 0.0173913 \end{array}]$

解题步骤 3.8

Multiply each element of $R_{4}$ by $- \frac{23}{197}$ to make the entry at $4, 4$ a $1$ .

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解题步骤 3.8.1

Multiply each element of $R_{4}$ by $- \frac{23}{197}$ to make the entry at $4, 4$ a $1$ .

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ 0 & 1 & - \frac{13}{6} & \frac{8}{3} & - 0.4\overline{6} \\ 0 & 0 & 1 & \frac{4}{23} & 0.11304347 \\ - \frac{23}{197} \cdot 0 & - \frac{23}{197} \cdot 0 & - \frac{23}{197} \cdot 0 & - \frac{23}{197} (- \frac{197}{23}) & - \frac{23}{197} \cdot - 0.0173913 \end{array}]$

解题步骤 3.8.2

化简 $R_{4}$ 。

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ 0 & 1 & - \frac{13}{6} & \frac{8}{3} & - 0.4\overline{6} \\ 0 & 0 & 1 & \frac{4}{23} & 0.11304347 \\ 0 & 0 & 0 & 1 & 0.00203045 \end{array}]$

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ 0 & 1 & - \frac{13}{6} & \frac{8}{3} & - 0.4\overline{6} \\ 0 & 0 & 1 & \frac{4}{23} & 0.11304347 \\ 0 & 0 & 0 & 1 & 0.00203045 \end{array}]$

解题步骤 3.9

Perform the row operation $R_{3} = R_{3} - \frac{4}{23} R_{4}$ to make the entry at $3, 4$ a $0$ .

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解题步骤 3.9.1

Perform the row operation $R_{3} = R_{3} - \frac{4}{23} R_{4}$ to make the entry at $3, 4$ a $0$ .

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ 0 & 1 & - \frac{13}{6} & \frac{8}{3} & - 0.4\overline{6} \\ 0 - \frac{4}{23} \cdot 0 & 0 - \frac{4}{23} \cdot 0 & 1 - \frac{4}{23} \cdot 0 & \frac{4}{23} - \frac{4}{23} \cdot 1 & 0.11304347 - \frac{4}{23} \cdot 0.00203045 \\ 0 & 0 & 0 & 1 & 0.00203045 \end{array}]$

解题步骤 3.9.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ 0 & 1 & - \frac{13}{6} & \frac{8}{3} & - 0.4\overline{6} \\ 0 & 0 & 1 & 0 & 0.11269035 \\ 0 & 0 & 0 & 1 & 0.00203045 \end{array}]$

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ 0 & 1 & - \frac{13}{6} & \frac{8}{3} & - 0.4\overline{6} \\ 0 & 0 & 1 & 0 & 0.11269035 \\ 0 & 0 & 0 & 1 & 0.00203045 \end{array}]$

解题步骤 3.10

Perform the row operation $R_{2} = R_{2} - \frac{8}{3} R_{4}$ to make the entry at $2, 4$ a $0$ .

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解题步骤 3.10.1

Perform the row operation $R_{2} = R_{2} - \frac{8}{3} R_{4}$ to make the entry at $2, 4$ a $0$ .

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ 0 - \frac{8}{3} \cdot 0 & 1 - \frac{8}{3} \cdot 0 & - \frac{13}{6} - \frac{8}{3} \cdot 0 & \frac{8}{3} - \frac{8}{3} \cdot 1 & - 0.4\overline{6} - \frac{8}{3} \cdot 0.00203045 \\ 0 & 0 & 1 & 0 & 0.11269035 \\ 0 & 0 & 0 & 1 & 0.00203045 \end{array}]$

解题步骤 3.10.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ 0 & 1 & - \frac{13}{6} & 0 & - 0.47208121 \\ 0 & 0 & 1 & 0 & 0.11269035 \\ 0 & 0 & 0 & 1 & 0.00203045 \end{array}]$

$[\begin{array}{c} 1 & 1 & - 1 & 4 & - 1.3 \\ 0 & 1 & - \frac{13}{6} & 0 & - 0.47208121 \\ 0 & 0 & 1 & 0 & 0.11269035 \\ 0 & 0 & 0 & 1 & 0.00203045 \end{array}]$

解题步骤 3.11

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

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解题步骤 3.11.1

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

$[\begin{array}{c} 1 - 4 \cdot 0 & 1 - 4 \cdot 0 & - 1 - 4 \cdot 0 & 4 - 4 \cdot 1 & - 1.3 - 4 \cdot 0.00203045 \\ 0 & 1 & - \frac{13}{6} & 0 & - 0.47208121 \\ 0 & 0 & 1 & 0 & 0.11269035 \\ 0 & 0 & 0 & 1 & 0.00203045 \end{array}]$

解题步骤 3.11.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 1 & - 1 & 0 & - 1.30812182 \\ 0 & 1 & - \frac{13}{6} & 0 & - 0.47208121 \\ 0 & 0 & 1 & 0 & 0.11269035 \\ 0 & 0 & 0 & 1 & 0.00203045 \end{array}]$

$[\begin{array}{c} 1 & 1 & - 1 & 0 & - 1.30812182 \\ 0 & 1 & - \frac{13}{6} & 0 & - 0.47208121 \\ 0 & 0 & 1 & 0 & 0.11269035 \\ 0 & 0 & 0 & 1 & 0.00203045 \end{array}]$

解题步骤 3.12

Perform the row operation $R_{2} = R_{2} + \frac{13}{6} R_{3}$ to make the entry at $2, 3$ a $0$ .

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解题步骤 3.12.1

Perform the row operation $R_{2} = R_{2} + \frac{13}{6} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & 1 & - 1 & 0 & - 1.30812182 \\ 0 + \frac{13}{6} \cdot 0 & 1 + \frac{13}{6} \cdot 0 & - \frac{13}{6} + \frac{13}{6} \cdot 1 & 0 + \frac{13}{6} \cdot 0 & - 0.47208121 + \frac{13}{6} \cdot 0.11269035 \\ 0 & 0 & 1 & 0 & 0.11269035 \\ 0 & 0 & 0 & 1 & 0.00203045 \end{array}]$

解题步骤 3.12.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & 1 & - 1 & 0 & - 1.30812182 \\ 0 & 1 & 0 & 0 & - 0.22791878 \\ 0 & 0 & 1 & 0 & 0.11269035 \\ 0 & 0 & 0 & 1 & 0.00203045 \end{array}]$

$[\begin{array}{c} 1 & 1 & - 1 & 0 & - 1.30812182 \\ 0 & 1 & 0 & 0 & - 0.22791878 \\ 0 & 0 & 1 & 0 & 0.11269035 \\ 0 & 0 & 0 & 1 & 0.00203045 \end{array}]$

解题步骤 3.13

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 3.13.1

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

$[\begin{array}{c} 1 + 0 & 1 + 0 & - 1 + 1 \cdot 1 & 0 + 0 & - 1.30812182 + 1 \cdot 0.11269035 \\ 0 & 1 & 0 & 0 & - 0.22791878 \\ 0 & 0 & 1 & 0 & 0.11269035 \\ 0 & 0 & 0 & 1 & 0.00203045 \end{array}]$

解题步骤 3.13.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 1 & 0 & 0 & - 1.19543147 \\ 0 & 1 & 0 & 0 & - 0.22791878 \\ 0 & 0 & 1 & 0 & 0.11269035 \\ 0 & 0 & 0 & 1 & 0.00203045 \end{array}]$

$[\begin{array}{c} 1 & 1 & 0 & 0 & - 1.19543147 \\ 0 & 1 & 0 & 0 & - 0.22791878 \\ 0 & 0 & 1 & 0 & 0.11269035 \\ 0 & 0 & 0 & 1 & 0.00203045 \end{array}]$

解题步骤 3.14

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 3.14.1

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

$[\begin{array}{c} 1 - 0 & 1 - 1 & 0 - 0 & 0 - 0 & - 1.19543147 + 0.22791878 \\ 0 & 1 & 0 & 0 & - 0.22791878 \\ 0 & 0 & 1 & 0 & 0.11269035 \\ 0 & 0 & 0 & 1 & 0.00203045 \end{array}]$

解题步骤 3.14.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 0 & 0 & - 0.96751269 \\ 0 & 1 & 0 & 0 & - 0.22791878 \\ 0 & 0 & 1 & 0 & 0.11269035 \\ 0 & 0 & 0 & 1 & 0.00203045 \end{array}]$

$[\begin{array}{c} 1 & 0 & 0 & 0 & - 0.96751269 \\ 0 & 1 & 0 & 0 & - 0.22791878 \\ 0 & 0 & 1 & 0 & 0.11269035 \\ 0 & 0 & 0 & 1 & 0.00203045 \end{array}]$

$[\begin{array}{c} 1 & 0 & 0 & 0 & - 0.96751269 \\ 0 & 1 & 0 & 0 & - 0.22791878 \\ 0 & 0 & 1 & 0 & 0.11269035 \\ 0 & 0 & 0 & 1 & 0.00203045 \end{array}]$

解题步骤 4

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

$w = - 0.96751269$

$x = - 0.22791878$

$y = 0.11269035$

$z = 0.00203045$

解题步骤 5

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

$(- 0.96751269, - 0.22791878, 0.11269035, 0.00203045)$

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$