有限数学示例

$[\begin{array}{c} a & 15 \\ 2 & b \end{array}] - 2 [\begin{array}{c} 5 & 8 \\ 7 & 10 \end{array}] = [\begin{array}{c} - 3 & - 1 \\ - 16 & 2 \end{array}]$

解题步骤 1

化简每一项。

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解题步骤 1.1

将 $- 2$ 乘以矩阵中的每一个元素。

$[\begin{array}{c} a & 15 \\ 2 & b \end{array}] + [\begin{array}{c} - 2 \cdot 5 & - 2 \cdot 8 \\ - 2 \cdot 7 & - 2 \cdot 10 \end{array}] = [\begin{array}{c} - 3 & - 1 \\ - 16 & 2 \end{array}]$

解题步骤 1.2

化简矩阵中的每一个元素。

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解题步骤 1.2.1

将 $- 2$ 乘以 $5$ 。

$[\begin{array}{c} a & 15 \\ 2 & b \end{array}] + [\begin{array}{c} - 10 & - 2 \cdot 8 \\ - 2 \cdot 7 & - 2 \cdot 10 \end{array}] = [\begin{array}{c} - 3 & - 1 \\ - 16 & 2 \end{array}]$

解题步骤 1.2.2

将 $- 2$ 乘以 $8$ 。

$[\begin{array}{c} a & 15 \\ 2 & b \end{array}] + [\begin{array}{c} - 10 & - 16 \\ - 2 \cdot 7 & - 2 \cdot 10 \end{array}] = [\begin{array}{c} - 3 & - 1 \\ - 16 & 2 \end{array}]$

解题步骤 1.2.3

将 $- 2$ 乘以 $7$ 。

$[\begin{array}{c} a & 15 \\ 2 & b \end{array}] + [\begin{array}{c} - 10 & - 16 \\ - 14 & - 2 \cdot 10 \end{array}] = [\begin{array}{c} - 3 & - 1 \\ - 16 & 2 \end{array}]$

解题步骤 1.2.4

将 $- 2$ 乘以 $10$ 。

$[\begin{array}{c} a & 15 \\ 2 & b \end{array}] + [\begin{array}{c} - 10 & - 16 \\ - 14 & - 20 \end{array}] = [\begin{array}{c} - 3 & - 1 \\ - 16 & 2 \end{array}]$

解题步骤 2

加上相应元素。

$[\begin{array}{c} a - 10 & 15 - 16 \\ 2 - 14 & b - 20 \end{array}] = [\begin{array}{c} - 3 & - 1 \\ - 16 & 2 \end{array}]$

解题步骤 3

Simplify each element.

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解题步骤 3.1

从 $15$ 中减去 $16$ 。

$[\begin{array}{c} a - 10 & - 1 \\ 2 - 14 & b - 20 \end{array}] = [\begin{array}{c} - 3 & - 1 \\ - 16 & 2 \end{array}]$

解题步骤 3.2

从 $2$ 中减去 $14$ 。

$[\begin{array}{c} a - 10 & - 1 \\ - 12 & b - 20 \end{array}] = [\begin{array}{c} - 3 & - 1 \\ - 16 & 2 \end{array}]$

解题步骤 4

Write as a linear system of equations.

$a - 10 = - 3$

$- 1 = - 1$

$- 12 = - 16$

$b - 20 = 2$

解题步骤 5

Since $- 12 \neq - 16$ , the matrix equation has no solution.

无解

有限数学 示例

有限数学示例