线性代数示例

y = 3 x + z - 2

$y = 3 x + z - 2$ ,

z = 3 x + 4

$z = 3 x + 4$ ,

y = 5 z

$y = 5 z$

解题步骤 1

Move all of the variables to the left side of each equation.

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解题步骤 1.1

将所有包含变量的项移到等式左边。

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解题步骤 1.1.1

从等式两边同时减去

3 x

$3 x$ 。

y - 3 x = z - 2

$y - 3 x = z - 2$

z = 3 x + 4

$z = 3 x + 4$

y = 5 z

$y = 5 z$

解题步骤 1.1.2

从等式两边同时减去

z

$z$ 。

y - 3 x - z = - 2

$y - 3 x - z = - 2$

z = 3 x + 4

$z = 3 x + 4$

y = 5 z

$y = 5 z$

y - 3 x - z = - 2

$y - 3 x - z = - 2$

z = 3 x + 4

$z = 3 x + 4$

y = 5 z

$y = 5 z$

解题步骤 1.2

将

y

$y$ 和

- 3 x

$- 3 x$ 重新排序。

- 3 x + y - z = - 2

$- 3 x + y - z = - 2$

z = 3 x + 4

$z = 3 x + 4$

y = 5 z

$y = 5 z$

解题步骤 1.3

从等式两边同时减去

3 x

$3 x$ 。

- 3 x + y - z = - 2

$- 3 x + y - z = - 2$

z - 3 x = 4

$z - 3 x = 4$

y = 5 z

$y = 5 z$

解题步骤 1.4

将

z

$z$ 和

- 3 x

$- 3 x$ 重新排序。

- 3 x + y - z = - 2

$- 3 x + y - z = - 2$

- 3 x + z = 4

$- 3 x + z = 4$

y = 5 z

$y = 5 z$

解题步骤 1.5

从等式两边同时减去

5 z

$5 z$ 。

- 3 x + y - z = - 2

$- 3 x + y - z = - 2$

- 3 x + z = 4

$- 3 x + z = 4$

y - 5 z = 0

$y - 5 z = 0$

- 3 x + y - z = - 2

$- 3 x + y - z = - 2$

- 3 x + z = 4

$- 3 x + z = 4$

y - 5 z = 0

$y - 5 z = 0$

解题步骤 2

以矩阵形式表示方程组。

⎡ ⎢ ⎣ \begin{matrix} - 3 & 1 & - 1 - 3 & 0 & 1 0 & 1 & - 5 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} - 2 40 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} - 3 & 1 & - 1 \\ - 3 & 0 & 1 \\ 0 & 1 & - 5 \end{array}] [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} - 2 \\ 4 \\ 0 \end{array}]$

解题步骤 3

Find the determinant of the coefficient matrix

⎡ ⎢ ⎣ \begin{matrix} - 3 & 1 & - 1 - 3 & 0 & 1 0 & 1 & - 5 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} - 3 & 1 & - 1 \\ - 3 & 0 & 1 \\ 0 & 1 & - 5 \end{array}]$ .

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解题步骤 3.1

Write

⎡ ⎢ ⎣ \begin{matrix} - 3 & 1 & - 1 - 3 & 0 & 1 0 & 1 & - 5 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} - 3 & 1 & - 1 \\ - 3 & 0 & 1 \\ 0 & 1 & - 5 \end{array}]$ in determinant notation.

∣ ∣ ∣ ∣ \begin{matrix} - 3 & 1 & - 1 - 3 & 0 & 1 0 & 1 & - 5 \end{matrix} ∣ ∣ ∣ ∣

$| \begin{array}{c} - 3 & 1 & - 1 \\ - 3 & 0 & 1 \\ 0 & 1 & - 5 \end{array} |$

解题步骤 3.2

Choose the row or column with the most

0

$0$ elements. If there are no

0

$0$ elements choose any row or column. Multiply every element in column

1

$1$ by its cofactor and add.

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解题步骤 3.2.1

Consider the corresponding sign chart.

∣ ∣ ∣ ∣ \begin{matrix} + & - & + - & + & - + & - & + \end{matrix} ∣ ∣ ∣ ∣

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

解题步骤 3.2.2

The cofactor is the minor with the sign changed if the indices match a

-

$-$ position on the sign chart.

解题步骤 3.2.3

The minor for

a_{11}

$a_{11}$ is the determinant with row

1

$1$ and column

1

$1$ deleted.

∣ ∣ ∣ \begin{matrix} 0 & 1 1 & - 5 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 0 & 1 \\ 1 & - 5 \end{array} |$

解题步骤 3.2.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

- 3 ∣ ∣ ∣ \begin{matrix} 0 & 1 1 & - 5 \end{matrix} ∣ ∣ ∣

$- 3 | \begin{array}{c} 0 & 1 \\ 1 & - 5 \end{array} |$

解题步骤 3.2.5

The minor for

a_{21}

$a_{21}$ is the determinant with row

2

$2$ and column

1

$1$ deleted.

∣ ∣ ∣ \begin{matrix} 1 & - 1 1 & - 5 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} |$

解题步骤 3.2.6

Multiply element

a_{21}

$a_{21}$ by its cofactor.

3 ∣ ∣ ∣ \begin{matrix} 1 & - 1 1 & - 5 \end{matrix} ∣ ∣ ∣

$3 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} |$

解题步骤 3.2.7

The minor for

a_{31}

$a_{31}$ is the determinant with row

3

$3$ and column

1

$1$ deleted.

∣ ∣ ∣ \begin{matrix} 1 & - 1 0 & 1 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

解题步骤 3.2.8

Multiply element

a_{31}

$a_{31}$ by its cofactor.

0 ∣ ∣ ∣ \begin{matrix} 1 & - 1 0 & 1 \end{matrix} ∣ ∣ ∣

$0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

解题步骤 3.2.9

Add the terms together.

- 3 ∣ ∣ ∣ \begin{matrix} 0 & 1 1 & - 5 \end{matrix} ∣ ∣ ∣ + 3 ∣ ∣ ∣ \begin{matrix} 1 & - 1 1 & - 5 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 1 & - 1 0 & 1 \end{matrix} ∣ ∣ ∣

$- 3 | \begin{array}{c} 0 & 1 \\ 1 & - 5 \end{array} | + 3 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

- 3 ∣ ∣ ∣ \begin{matrix} 0 & 1 1 & - 5 \end{matrix} ∣ ∣ ∣ + 3 ∣ ∣ ∣ \begin{matrix} 1 & - 1 1 & - 5 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 1 & - 1 0 & 1 \end{matrix} ∣ ∣ ∣

$- 3 | \begin{array}{c} 0 & 1 \\ 1 & - 5 \end{array} | + 3 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

解题步骤 3.3

将

0

$0$ 乘以

∣ ∣ ∣ \begin{matrix} 1 & - 1 0 & 1 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$ 。

- 3 ∣ ∣ ∣ \begin{matrix} 0 & 1 1 & - 5 \end{matrix} ∣ ∣ ∣ + 3 ∣ ∣ ∣ \begin{matrix} 1 & - 1 1 & - 5 \end{matrix} ∣ ∣ ∣ + 0

$- 3 | \begin{array}{c} 0 & 1 \\ 1 & - 5 \end{array} | + 3 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0$

解题步骤 3.4

计算

∣ ∣ ∣ \begin{matrix} 0 & 1 1 & - 5 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 0 & 1 \\ 1 & - 5 \end{array} |$ 。

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解题步骤 3.4.1

可以使用公式

∣ ∣ ∣ \begin{matrix} a & b c & d \end{matrix} ∣ ∣ ∣ = a d - c b

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ 求

2 \times 2

$2 \times 2$ 矩阵的行列式。

- 3 (0 \cdot - 5 - 1 \cdot 1) + 3 ∣ ∣ ∣ \begin{matrix} 1 & - 1 1 & - 5 \end{matrix} ∣ ∣ ∣ + 0

$- 3 (0 \cdot - 5 - 1 \cdot 1) + 3 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0$

解题步骤 3.4.2

化简行列式。

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解题步骤 3.4.2.1

化简每一项。

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解题步骤 3.4.2.1.1

将

0

$0$ 乘以

- 5

$- 5$ 。

- 3 (0 - 1 \cdot 1) + 3 ∣ ∣ ∣ \begin{matrix} 1 & - 1 1 & - 5 \end{matrix} ∣ ∣ ∣ + 0

$- 3 (0 - 1 \cdot 1) + 3 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0$

解题步骤 3.4.2.1.2

将

- 1

$- 1$ 乘以

1

$1$ 。

- 3 (0 - 1) + 3 ∣ ∣ ∣ \begin{matrix} 1 & - 1 1 & - 5 \end{matrix} ∣ ∣ ∣ + 0

$- 3 (0 - 1) + 3 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0$

- 3 (0 - 1) + 3 ∣ ∣ ∣ \begin{matrix} 1 & - 1 1 & - 5 \end{matrix} ∣ ∣ ∣ + 0

$- 3 (0 - 1) + 3 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0$

解题步骤 3.4.2.2

从

0

$0$ 中减去

1

$1$ 。

- 3 \cdot - 1 + 3 ∣ ∣ ∣ \begin{matrix} 1 & - 1 1 & - 5 \end{matrix} ∣ ∣ ∣ + 0

$- 3 \cdot - 1 + 3 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0$

- 3 \cdot - 1 + 3 ∣ ∣ ∣ \begin{matrix} 1 & - 1 1 & - 5 \end{matrix} ∣ ∣ ∣ + 0

$- 3 \cdot - 1 + 3 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0$

- 3 \cdot - 1 + 3 ∣ ∣ ∣ \begin{matrix} 1 & - 1 1 & - 5 \end{matrix} ∣ ∣ ∣ + 0

$- 3 \cdot - 1 + 3 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0$

解题步骤 3.5

计算

∣ ∣ ∣ \begin{matrix} 1 & - 1 1 & - 5 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} |$ 。

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解题步骤 3.5.1

可以使用公式

∣ ∣ ∣ \begin{matrix} a & b c & d \end{matrix} ∣ ∣ ∣ = a d - c b

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ 求

2 \times 2

$2 \times 2$ 矩阵的行列式。

- 3 \cdot - 1 + 3 (1 \cdot - 5 - 1 \cdot - 1) + 0

$- 3 \cdot - 1 + 3 (1 \cdot - 5 - 1 \cdot - 1) + 0$

解题步骤 3.5.2

化简行列式。

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解题步骤 3.5.2.1

化简每一项。

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解题步骤 3.5.2.1.1

将

- 5

$- 5$ 乘以

1

$1$ 。

- 3 \cdot - 1 + 3 (- 5 - 1 \cdot - 1) + 0

$- 3 \cdot - 1 + 3 (- 5 - 1 \cdot - 1) + 0$

解题步骤 3.5.2.1.2

将

- 1

$- 1$ 乘以

- 1

$- 1$ 。

- 3 \cdot - 1 + 3 (- 5 + 1) + 0

$- 3 \cdot - 1 + 3 (- 5 + 1) + 0$

- 3 \cdot - 1 + 3 (- 5 + 1) + 0

$- 3 \cdot - 1 + 3 (- 5 + 1) + 0$

解题步骤 3.5.2.2

将

- 5

$- 5$ 和

1

$1$ 相加。

- 3 \cdot - 1 + 3 \cdot - 4 + 0

$- 3 \cdot - 1 + 3 \cdot - 4 + 0$

- 3 \cdot - 1 + 3 \cdot - 4 + 0

$- 3 \cdot - 1 + 3 \cdot - 4 + 0$

- 3 \cdot - 1 + 3 \cdot - 4 + 0

$- 3 \cdot - 1 + 3 \cdot - 4 + 0$

解题步骤 3.6

化简行列式。

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解题步骤 3.6.1

化简每一项。

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解题步骤 3.6.1.1

将

- 3

$- 3$ 乘以

- 1

$- 1$ 。

3 + 3 \cdot - 4 + 0

$3 + 3 \cdot - 4 + 0$

解题步骤 3.6.1.2

将

3

$3$ 乘以

- 4

$- 4$ 。

3 - 12 + 0

$3 - 12 + 0$

3 - 12 + 0

$3 - 12 + 0$

解题步骤 3.6.2

从

3

$3$ 中减去

12

$12$ 。

- 9 + 0

$- 9 + 0$

解题步骤 3.6.3

将

- 9

$- 9$ 和

0

$0$ 相加。

- 9

$- 9$

- 9

$- 9$

D = - 9

$D = - 9$

解题步骤 4

Since the determinant is not

0

$0$ , the system can be solved using Cramer's Rule.

解题步骤 5

Find the value of

x

$x$ by Cramer's Rule, which states that

x = \frac{D_{x}}{D}

$x = \frac{D_{x}}{D}$ .

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解题步骤 5.1

Replace column

1

$1$ of the coefficient matrix that corresponds to the

x

$x$ -coefficients of the system with

⎡ ⎢ ⎣ \begin{matrix} - 2 40 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} - 2 \\ 4 \\ 0 \end{array}]$ .

∣ ∣ ∣ ∣ \begin{matrix} - 2 & 1 & - 1 4 & 0 & 1 0 & 1 & - 5 \end{matrix} ∣ ∣ ∣ ∣

$| \begin{array}{c} - 2 & 1 & - 1 \\ 4 & 0 & 1 \\ 0 & 1 & - 5 \end{array} |$

解题步骤 5.2

Find the determinant.

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解题步骤 5.2.1

Choose the row or column with the most

0

$0$ elements. If there are no

0

$0$ elements choose any row or column. Multiply every element in column

1

$1$ by its cofactor and add.

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解题步骤 5.2.1.1

Consider the corresponding sign chart.

∣ ∣ ∣ ∣ \begin{matrix} + & - & + - & + & - + & - & + \end{matrix} ∣ ∣ ∣ ∣

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

解题步骤 5.2.1.2

The cofactor is the minor with the sign changed if the indices match a

-

$-$ position on the sign chart.

解题步骤 5.2.1.3

The minor for

a_{11}

$a_{11}$ is the determinant with row

1

$1$ and column

1

$1$ deleted.

∣ ∣ ∣ \begin{matrix} 0 & 1 1 & - 5 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 0 & 1 \\ 1 & - 5 \end{array} |$

解题步骤 5.2.1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

- 2 ∣ ∣ ∣ \begin{matrix} 0 & 1 1 & - 5 \end{matrix} ∣ ∣ ∣

$- 2 | \begin{array}{c} 0 & 1 \\ 1 & - 5 \end{array} |$

解题步骤 5.2.1.5

The minor for

a_{21}

$a_{21}$ is the determinant with row

2

$2$ and column

1

$1$ deleted.

∣ ∣ ∣ \begin{matrix} 1 & - 1 1 & - 5 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} |$

解题步骤 5.2.1.6

Multiply element

a_{21}

$a_{21}$ by its cofactor.

- 4 ∣ ∣ ∣ \begin{matrix} 1 & - 1 1 & - 5 \end{matrix} ∣ ∣ ∣

$- 4 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} |$

解题步骤 5.2.1.7

The minor for

a_{31}

$a_{31}$ is the determinant with row

3

$3$ and column

1

$1$ deleted.

∣ ∣ ∣ \begin{matrix} 1 & - 1 0 & 1 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

解题步骤 5.2.1.8

Multiply element

a_{31}

$a_{31}$ by its cofactor.

0 ∣ ∣ ∣ \begin{matrix} 1 & - 1 0 & 1 \end{matrix} ∣ ∣ ∣

$0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

解题步骤 5.2.1.9

Add the terms together.

- 2 ∣ ∣ ∣ \begin{matrix} 0 & 1 1 & - 5 \end{matrix} ∣ ∣ ∣ - 4 ∣ ∣ ∣ \begin{matrix} 1 & - 1 1 & - 5 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 1 & - 1 0 & 1 \end{matrix} ∣ ∣ ∣

$- 2 | \begin{array}{c} 0 & 1 \\ 1 & - 5 \end{array} | - 4 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

- 2 ∣ ∣ ∣ \begin{matrix} 0 & 1 1 & - 5 \end{matrix} ∣ ∣ ∣ - 4 ∣ ∣ ∣ \begin{matrix} 1 & - 1 1 & - 5 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 1 & - 1 0 & 1 \end{matrix} ∣ ∣ ∣

$- 2 | \begin{array}{c} 0 & 1 \\ 1 & - 5 \end{array} | - 4 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

解题步骤 5.2.2

将

0

$0$ 乘以

∣ ∣ ∣ \begin{matrix} 1 & - 1 0 & 1 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$ 。

- 2 ∣ ∣ ∣ \begin{matrix} 0 & 1 1 & - 5 \end{matrix} ∣ ∣ ∣ - 4 ∣ ∣ ∣ \begin{matrix} 1 & - 1 1 & - 5 \end{matrix} ∣ ∣ ∣ + 0

$- 2 | \begin{array}{c} 0 & 1 \\ 1 & - 5 \end{array} | - 4 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0$

解题步骤 5.2.3

计算

∣ ∣ ∣ \begin{matrix} 0 & 1 1 & - 5 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 0 & 1 \\ 1 & - 5 \end{array} |$ 。

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解题步骤 5.2.3.1

可以使用公式

∣ ∣ ∣ \begin{matrix} a & b c & d \end{matrix} ∣ ∣ ∣ = a d - c b

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ 求

2 \times 2

$2 \times 2$ 矩阵的行列式。

- 2 (0 \cdot - 5 - 1 \cdot 1) - 4 ∣ ∣ ∣ \begin{matrix} 1 & - 1 1 & - 5 \end{matrix} ∣ ∣ ∣ + 0

$- 2 (0 \cdot - 5 - 1 \cdot 1) - 4 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0$

解题步骤 5.2.3.2

化简行列式。

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解题步骤 5.2.3.2.1

化简每一项。

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解题步骤 5.2.3.2.1.1

将

0

$0$ 乘以

- 5

$- 5$ 。

$- 2 (0 - 1 \cdot 1) - 4 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0$

解题步骤 5.2.3.2.1.2

将

$- 1$ 乘以

$1$ 。

$- 2 (0 - 1) - 4 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0$

解题步骤 5.2.3.2.2

从

$0$ 中减去

$1$ 。

$- 2 \cdot - 1 - 4 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0$

解题步骤 5.2.4

计算

$| \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} |$ 。

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解题步骤 5.2.4.1

可以使用公式

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ 求

$2 \times 2$ 矩阵的行列式。

$- 2 \cdot - 1 - 4 (1 \cdot - 5 - 1 \cdot - 1) + 0$

解题步骤 5.2.4.2

化简行列式。

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解题步骤 5.2.4.2.1

化简每一项。

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解题步骤 5.2.4.2.1.1

将

$- 5$ 乘以

$1$ 。

$- 2 \cdot - 1 - 4 (- 5 - 1 \cdot - 1) + 0$

解题步骤 5.2.4.2.1.2

将

$- 1$ 乘以

$- 1$ 。

$- 2 \cdot - 1 - 4 (- 5 + 1) + 0$

解题步骤 5.2.4.2.2

将

$- 5$ 和

$1$ 相加。

$- 2 \cdot - 1 - 4 \cdot - 4 + 0$

解题步骤 5.2.5

化简行列式。

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解题步骤 5.2.5.1

化简每一项。

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解题步骤 5.2.5.1.1

将

$- 2$ 乘以

$- 1$ 。

$2 - 4 \cdot - 4 + 0$

解题步骤 5.2.5.1.2

将

$- 4$ 乘以

$- 4$ 。

$2 + 16 + 0$

解题步骤 5.2.5.2

将

$2$ 和

$16$ 相加。

$18 + 0$

解题步骤 5.2.5.3

将

$18$ 和

$0$ 相加。

$18$

$D_{x} = 18$

解题步骤 5.3

Use the formula to solve for

$x$ .

$x = \frac{D_{x}}{D}$

解题步骤 5.4

Substitute

$- 9$ for

$D$ and

$18$ for

$D_{x}$ in the formula.

$x = \frac{18}{- 9}$

解题步骤 5.5

用

$18$ 除以

$- 9$ 。

$x = - 2$

解题步骤 6

Find the value of

$y$ by Cramer's Rule, which states that

$y = \frac{D_{y}}{D}$ .

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解题步骤 6.1

Replace column

$2$ of the coefficient matrix that corresponds to the

$y$ -coefficients of the system with

$[\begin{array}{c} - 2 \\ 4 \\ 0 \end{array}]$ .

$| \begin{array}{c} - 3 & - 2 & - 1 \\ - 3 & 4 & 1 \\ 0 & 0 & - 5 \end{array} |$

解题步骤 6.2

Find the determinant.

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解题步骤 6.2.1

Choose the row or column with the most

$0$ elements. If there are no

$0$ elements choose any row or column. Multiply every element in row

$3$ by its cofactor and add.

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解题步骤 6.2.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

解题步骤 6.2.1.2

The cofactor is the minor with the sign changed if the indices match a

$-$ position on the sign chart.

解题步骤 6.2.1.3

The minor for

$a_{31}$ is the determinant with row

$3$ and column

$1$ deleted.

$| \begin{array}{c} - 2 & - 1 \\ 4 & 1 \end{array} |$

解题步骤 6.2.1.4

Multiply element

$a_{31}$ by its cofactor.

$0 | \begin{array}{c} - 2 & - 1 \\ 4 & 1 \end{array} |$

解题步骤 6.2.1.5

The minor for

$a_{32}$ is the determinant with row

$3$ and column

$2$ deleted.

$| \begin{array}{c} - 3 & - 1 \\ - 3 & 1 \end{array} |$

解题步骤 6.2.1.6

Multiply element

$a_{32}$ by its cofactor.

$0 | \begin{array}{c} - 3 & - 1 \\ - 3 & 1 \end{array} |$

解题步骤 6.2.1.7

The minor for

$a_{33}$ is the determinant with row

$3$ and column

$3$ deleted.

$| \begin{array}{c} - 3 & - 2 \\ - 3 & 4 \end{array} |$

解题步骤 6.2.1.8

Multiply element

$a_{33}$ by its cofactor.

$- 5 | \begin{array}{c} - 3 & - 2 \\ - 3 & 4 \end{array} |$

解题步骤 6.2.1.9

Add the terms together.

$0 | \begin{array}{c} - 2 & - 1 \\ 4 & 1 \end{array} | + 0 | \begin{array}{c} - 3 & - 1 \\ - 3 & 1 \end{array} | - 5 | \begin{array}{c} - 3 & - 2 \\ - 3 & 4 \end{array} |$

解题步骤 6.2.2

将

$0$ 乘以

$| \begin{array}{c} - 2 & - 1 \\ 4 & 1 \end{array} |$ 。

$0 + 0 | \begin{array}{c} - 3 & - 1 \\ - 3 & 1 \end{array} | - 5 | \begin{array}{c} - 3 & - 2 \\ - 3 & 4 \end{array} |$

解题步骤 6.2.3

将

$0$ 乘以

$| \begin{array}{c} - 3 & - 1 \\ - 3 & 1 \end{array} |$ 。

$0 + 0 - 5 | \begin{array}{c} - 3 & - 2 \\ - 3 & 4 \end{array} |$

解题步骤 6.2.4

计算

$| \begin{array}{c} - 3 & - 2 \\ - 3 & 4 \end{array} |$ 。

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解题步骤 6.2.4.1

可以使用公式

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ 求

$2 \times 2$ 矩阵的行列式。

$0 + 0 - 5 (- 3 \cdot 4 - (- 3 \cdot - 2))$

解题步骤 6.2.4.2

化简行列式。

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解题步骤 6.2.4.2.1

化简每一项。

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解题步骤 6.2.4.2.1.1

将

$- 3$ 乘以

$4$ 。

$0 + 0 - 5 (- 12 - (- 3 \cdot - 2))$

解题步骤 6.2.4.2.1.2

乘以

$- (- 3 \cdot - 2)$ 。

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解题步骤 6.2.4.2.1.2.1

将

$- 3$ 乘以

$- 2$ 。

$0 + 0 - 5 (- 12 - 1 \cdot 6)$

解题步骤 6.2.4.2.1.2.2

将

$- 1$ 乘以

$6$ 。

$0 + 0 - 5 (- 12 - 6)$

解题步骤 6.2.4.2.2

从

$- 12$ 中减去

$6$ 。

$0 + 0 - 5 \cdot - 18$

解题步骤 6.2.5

化简行列式。

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解题步骤 6.2.5.1

将

$- 5$ 乘以

$- 18$ 。

$0 + 0 + 90$

解题步骤 6.2.5.2

将

$0$ 和

$0$ 相加。

$0 + 90$

解题步骤 6.2.5.3

将

$0$ 和

$90$ 相加。

$90$

$D_{y} = 90$

解题步骤 6.3

Use the formula to solve for

$y$ .

$y = \frac{D_{y}}{D}$

解题步骤 6.4

Substitute

$- 9$ for

$D$ and

$90$ for

$D_{y}$ in the formula.

$y = \frac{90}{- 9}$

解题步骤 6.5

用

$90$ 除以

$- 9$ 。

$y = - 10$

解题步骤 7

Find the value of

$z$ by Cramer's Rule, which states that

$z = \frac{D_{z}}{D}$ .

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解题步骤 7.1

Replace column

$3$ of the coefficient matrix that corresponds to the

$z$ -coefficients of the system with

$[\begin{array}{c} - 2 \\ 4 \\ 0 \end{array}]$ .

$| \begin{array}{c} - 3 & 1 & - 2 \\ - 3 & 0 & 4 \\ 0 & 1 & 0 \end{array} |$

解题步骤 7.2

Find the determinant.

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解题步骤 7.2.1

Choose the row or column with the most

$0$ elements. If there are no

$0$ elements choose any row or column. Multiply every element in row

$3$ by its cofactor and add.

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解题步骤 7.2.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

解题步骤 7.2.1.2

The cofactor is the minor with the sign changed if the indices match a

$-$ position on the sign chart.

解题步骤 7.2.1.3

The minor for

$a_{31}$ is the determinant with row

$3$ and column

$1$ deleted.

$| \begin{array}{c} 1 & - 2 \\ 0 & 4 \end{array} |$

解题步骤 7.2.1.4

Multiply element

$a_{31}$ by its cofactor.

$0 | \begin{array}{c} 1 & - 2 \\ 0 & 4 \end{array} |$

解题步骤 7.2.1.5

The minor for

$a_{32}$ is the determinant with row

$3$ and column

$2$ deleted.

$| \begin{array}{c} - 3 & - 2 \\ - 3 & 4 \end{array} |$

解题步骤 7.2.1.6

Multiply element

$a_{32}$ by its cofactor.

$- 1 | \begin{array}{c} - 3 & - 2 \\ - 3 & 4 \end{array} |$

解题步骤 7.2.1.7

The minor for

$a_{33}$ is the determinant with row

$3$ and column

$3$ deleted.

$| \begin{array}{c} - 3 & 1 \\ - 3 & 0 \end{array} |$

解题步骤 7.2.1.8

Multiply element

$a_{33}$ by its cofactor.

$0 | \begin{array}{c} - 3 & 1 \\ - 3 & 0 \end{array} |$

解题步骤 7.2.1.9

Add the terms together.

$0 | \begin{array}{c} 1 & - 2 \\ 0 & 4 \end{array} | - 1 | \begin{array}{c} - 3 & - 2 \\ - 3 & 4 \end{array} | + 0 | \begin{array}{c} - 3 & 1 \\ - 3 & 0 \end{array} |$

解题步骤 7.2.2

将

$0$ 乘以

$| \begin{array}{c} 1 & - 2 \\ 0 & 4 \end{array} |$ 。

$0 - 1 | \begin{array}{c} - 3 & - 2 \\ - 3 & 4 \end{array} | + 0 | \begin{array}{c} - 3 & 1 \\ - 3 & 0 \end{array} |$

解题步骤 7.2.3

将

$0$ 乘以

$| \begin{array}{c} - 3 & 1 \\ - 3 & 0 \end{array} |$ 。

$0 - 1 | \begin{array}{c} - 3 & - 2 \\ - 3 & 4 \end{array} | + 0$

解题步骤 7.2.4

计算

$| \begin{array}{c} - 3 & - 2 \\ - 3 & 4 \end{array} |$ 。

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解题步骤 7.2.4.1

可以使用公式

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ 求

$2 \times 2$ 矩阵的行列式。

$0 - 1 (- 3 \cdot 4 - (- 3 \cdot - 2)) + 0$

解题步骤 7.2.4.2

化简行列式。

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解题步骤 7.2.4.2.1

化简每一项。

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解题步骤 7.2.4.2.1.1

将

$- 3$ 乘以

$4$ 。

$0 - 1 (- 12 - (- 3 \cdot - 2)) + 0$

解题步骤 7.2.4.2.1.2

乘以

$- (- 3 \cdot - 2)$ 。

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解题步骤 7.2.4.2.1.2.1

将

$- 3$ 乘以

$- 2$ 。

$0 - 1 (- 12 - 1 \cdot 6) + 0$

解题步骤 7.2.4.2.1.2.2

将

$- 1$ 乘以

$6$ 。

$0 - 1 (- 12 - 6) + 0$

解题步骤 7.2.4.2.2

从

$- 12$ 中减去

$6$ 。

$0 - 1 \cdot - 18 + 0$

解题步骤 7.2.5

化简行列式。

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解题步骤 7.2.5.1

将

$- 1$ 乘以

$- 18$ 。

$0 + 18 + 0$

解题步骤 7.2.5.2

将

$0$ 和

$18$ 相加。

$18 + 0$

解题步骤 7.2.5.3

将

$18$ 和

$0$ 相加。

$18$

$D_{z} = 18$

解题步骤 7.3

Use the formula to solve for

$z$ .

$z = \frac{D_{z}}{D}$

解题步骤 7.4

Substitute

$- 9$ for

$D$ and

$18$ for

$D_{z}$ in the formula.

$z = \frac{18}{- 9}$

解题步骤 7.5

用

$18$ 除以

$- 9$ 。

$z = - 2$

解题步骤 8

列出方程组的解。

$x = - 2$

$y = - 10$

$z = - 2$

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