有限数学示例

$[\begin{array}{c} \frac{1}{7} & \frac{2}{7} \\ \frac{3}{7} & \frac{- 1}{7} \end{array}] [\begin{array}{c} 1 & 2 \\ 3 & - 1 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{1}{7} & \frac{2}{7} \\ \frac{3}{7} & - \frac{1}{7} \end{array}] [\begin{array}{c} - 1 \\ 4 \end{array}]$

解题步骤 1

将负号移到分数的前面。

$[\begin{array}{c} \frac{1}{7} & \frac{2}{7} \\ \frac{3}{7} & - \frac{1}{7} \end{array}] [\begin{array}{c} 1 & 2 \\ 3 & - 1 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{1}{7} & \frac{2}{7} \\ \frac{3}{7} & - \frac{1}{7} \end{array}] [\begin{array}{c} - 1 \\ 4 \end{array}]$

解题步骤 2

乘以 $[\begin{array}{c} \frac{1}{7} & \frac{2}{7} \\ \frac{3}{7} & - \frac{1}{7} \end{array}] [\begin{array}{c} 1 & 2 \\ 3 & - 1 \end{array}]$ 。

点击获取更多步骤...

解题步骤 2.1

Two matrices can be multiplied if and only if the number of columns in the first matrix is equal to the number of rows in the second matrix. In this case, the first matrix is $2 \times 2$ and the second matrix is $2 \times 2$ .

解题步骤 2.2

将第一个矩阵中的每一行乘以第二个矩阵中的每一列。

$[\begin{array}{c} \frac{1}{7} \cdot 1 + \frac{2}{7} \cdot 3 & \frac{1}{7} \cdot 2 + \frac{2}{7} \cdot - 1 \\ \frac{3}{7} \cdot 1 - \frac{1}{7} \cdot 3 & \frac{3}{7} \cdot 2 - \frac{1}{7} \cdot - 1 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{1}{7} & \frac{2}{7} \\ \frac{3}{7} & - \frac{1}{7} \end{array}] [\begin{array}{c} - 1 \\ 4 \end{array}]$

解题步骤 2.3

通过展开所有表达式化简矩阵的每一个元素。

$[\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{1}{7} & \frac{2}{7} \\ \frac{3}{7} & - \frac{1}{7} \end{array}] [\begin{array}{c} - 1 \\ 4 \end{array}]$

解题步骤 3

Multiplying any matrix by an identity matrix is the matrix itself.

点击获取更多步骤...

解题步骤 3.1

解题步骤 3.2

将第一个矩阵中的每一行乘以第二个矩阵中的每一列。

$[\begin{array}{c} 1 x + 0 y \\ 0 x + 1 y \end{array}] = [\begin{array}{c} \frac{1}{7} & \frac{2}{7} \\ \frac{3}{7} & - \frac{1}{7} \end{array}] [\begin{array}{c} - 1 \\ 4 \end{array}]$

解题步骤 3.3

通过展开所有表达式化简矩阵的每一个元素。

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{1}{7} & \frac{2}{7} \\ \frac{3}{7} & - \frac{1}{7} \end{array}] [\begin{array}{c} - 1 \\ 4 \end{array}]$

解题步骤 4

乘以 $[\begin{array}{c} \frac{1}{7} & \frac{2}{7} \\ \frac{3}{7} & - \frac{1}{7} \end{array}] [\begin{array}{c} - 1 \\ 4 \end{array}]$ 。

点击获取更多步骤...

解题步骤 4.1

解题步骤 4.2

将第一个矩阵中的每一行乘以第二个矩阵中的每一列。

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{1}{7} \cdot - 1 + \frac{2}{7} \cdot 4 \\ \frac{3}{7} \cdot - 1 - \frac{1}{7} \cdot 4 \end{array}]$

解题步骤 4.3

通过展开所有表达式化简矩阵的每一个元素。

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 1 \\ - 1 \end{array}]$

解题步骤 5

Write as a linear system of equations.

$x = 1$

$y = - 1$

解题步骤 6

这个方程组已经解出来了。

$x = 1, y = - 1$

有限数学 示例

有限数学示例