有限数学示例

$[\begin{array}{c} a + 8 & 4 z + 1 & 8 m \\ 7 k & 4 & 0 \end{array}] + [\begin{array}{c} 10 a & 10 z & 7 m \\ 8 k & 5 & 0 \end{array}] = [\begin{array}{c} 41 & - 69 & 105 \\ 45 & 9 & 0 \end{array}]$

解题步骤 1

加上相应元素。

$[\begin{array}{c} a + 8 + 10 a & 4 z + 1 + 10 z & 8 m + 7 m \\ 7 k + 8 k & 4 + 5 & 0 + 0 \end{array}] = [\begin{array}{c} 41 & - 69 & 105 \\ 45 & 9 & 0 \end{array}]$

解题步骤 2

Simplify each element.

点击获取更多步骤...

解题步骤 2.1

将 $a$ 和 $10 a$ 相加。

$[\begin{array}{c} 11 a + 8 & 4 z + 1 + 10 z & 8 m + 7 m \\ 7 k + 8 k & 4 + 5 & 0 + 0 \end{array}] = [\begin{array}{c} 41 & - 69 & 105 \\ 45 & 9 & 0 \end{array}]$

解题步骤 2.2

将 $4 z$ 和 $10 z$ 相加。

$[\begin{array}{c} 11 a + 8 & 14 z + 1 & 8 m + 7 m \\ 7 k + 8 k & 4 + 5 & 0 + 0 \end{array}] = [\begin{array}{c} 41 & - 69 & 105 \\ 45 & 9 & 0 \end{array}]$

解题步骤 2.3

将 $8 m$ 和 $7 m$ 相加。

$[\begin{array}{c} 11 a + 8 & 14 z + 1 & 15 m \\ 7 k + 8 k & 4 + 5 & 0 + 0 \end{array}] = [\begin{array}{c} 41 & - 69 & 105 \\ 45 & 9 & 0 \end{array}]$

解题步骤 2.4

将 $7 k$ 和 $8 k$ 相加。

$[\begin{array}{c} 11 a + 8 & 14 z + 1 & 15 m \\ 15 k & 4 + 5 & 0 + 0 \end{array}] = [\begin{array}{c} 41 & - 69 & 105 \\ 45 & 9 & 0 \end{array}]$

解题步骤 2.5

将 $4$ 和 $5$ 相加。

$[\begin{array}{c} 11 a + 8 & 14 z + 1 & 15 m \\ 15 k & 9 & 0 + 0 \end{array}] = [\begin{array}{c} 41 & - 69 & 105 \\ 45 & 9 & 0 \end{array}]$

解题步骤 2.6

将 $0$ 和 $0$ 相加。

$[\begin{array}{c} 11 a + 8 & 14 z + 1 & 15 m \\ 15 k & 9 & 0 \end{array}] = [\begin{array}{c} 41 & - 69 & 105 \\ 45 & 9 & 0 \end{array}]$

解题步骤 3

Write as a linear system of equations.

$11 a + 8 = 41$

$14 z + 1 = - 69$

$15 m = 105$

$15 k = 45$

$9 = 9$

$0 = 0$

解题步骤 4

Since $0 = 0$ 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