有限数学示例

$[\begin{array}{c} a + 7 & 12 z + 1 & 8 m \\ 12 k & 1 & 5 \end{array}] + [\begin{array}{c} 12 a & 9 z & 10 m \\ 6 k & 4 & 5 \end{array}] = [\begin{array}{c} 33 & - 104 & 108 \\ 54 & 5 & 10 \end{array}]$

解题步骤 1

加上相应元素。

$[\begin{array}{c} a + 7 + 12 a & 12 z + 1 + 9 z & 8 m + 10 m \\ 12 k + 6 k & 1 + 4 & 5 + 5 \end{array}] = [\begin{array}{c} 33 & - 104 & 108 \\ 54 & 5 & 10 \end{array}]$

解题步骤 2

Simplify each element.

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解题步骤 2.1

将 $a$ 和 $12 a$ 相加。

$[\begin{array}{c} 13 a + 7 & 12 z + 1 + 9 z & 8 m + 10 m \\ 12 k + 6 k & 1 + 4 & 5 + 5 \end{array}] = [\begin{array}{c} 33 & - 104 & 108 \\ 54 & 5 & 10 \end{array}]$

解题步骤 2.2

将 $12 z$ 和 $9 z$ 相加。

$[\begin{array}{c} 13 a + 7 & 21 z + 1 & 8 m + 10 m \\ 12 k + 6 k & 1 + 4 & 5 + 5 \end{array}] = [\begin{array}{c} 33 & - 104 & 108 \\ 54 & 5 & 10 \end{array}]$

解题步骤 2.3

将 $8 m$ 和 $10 m$ 相加。

$[\begin{array}{c} 13 a + 7 & 21 z + 1 & 18 m \\ 12 k + 6 k & 1 + 4 & 5 + 5 \end{array}] = [\begin{array}{c} 33 & - 104 & 108 \\ 54 & 5 & 10 \end{array}]$

解题步骤 2.4

将 $12 k$ 和 $6 k$ 相加。

$[\begin{array}{c} 13 a + 7 & 21 z + 1 & 18 m \\ 18 k & 1 + 4 & 5 + 5 \end{array}] = [\begin{array}{c} 33 & - 104 & 108 \\ 54 & 5 & 10 \end{array}]$

解题步骤 2.5

将 $1$ 和 $4$ 相加。

$[\begin{array}{c} 13 a + 7 & 21 z + 1 & 18 m \\ 18 k & 5 & 5 + 5 \end{array}] = [\begin{array}{c} 33 & - 104 & 108 \\ 54 & 5 & 10 \end{array}]$

解题步骤 2.6

将 $5$ 和 $5$ 相加。

$[\begin{array}{c} 13 a + 7 & 21 z + 1 & 18 m \\ 18 k & 5 & 10 \end{array}] = [\begin{array}{c} 33 & - 104 & 108 \\ 54 & 5 & 10 \end{array}]$

解题步骤 3

Write as a linear system of equations.

$13 a + 7 = 33$

$21 z + 1 = - 104$

$18 m = 108$

$18 k = 54$

$5 = 5$

$10 = 10$

解题步骤 4

Since $10 = 10$ is always true, the matrix equation has infinite solutions.

Infinite solutions satisfying the system.

有限数学 示例

有限数学示例