线性代数示例

$[\begin{array}{c} \cos (3 x) & \sin (3 x) \\ - \sin (3 x) & \cos (3 x) \end{array}] \cdot [\begin{array}{c} b \\ d \end{array}] = [\begin{array}{c} 0 \\ \tan (3 x) \end{array}]$

解题步骤 1

化简。

点击获取更多步骤...

解题步骤 1.1

乘以 $[\begin{array}{c} \cos (3 x) & \sin (3 x) \\ - \sin (3 x) & \cos (3 x) \end{array}] [\begin{array}{c} b \\ d \end{array}]$ 。

点击获取更多步骤...

解题步骤 1.1.1

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 $2 \times 1$ .

解题步骤 1.1.2

将第一个矩阵中的每一行乘以第二个矩阵中的每一列。

$[\begin{array}{c} \cos (3 x) b + \sin (3 x) d \\ - \sin (3 x) b + \cos (3 x) d \end{array}] = [\begin{array}{c} 0 \\ \tan (3 x) \end{array}]$

解题步骤 1.1.3

通过展开所有表达式化简矩阵的每一个元素。

$[\begin{array}{c} b \cos (3 x) + d \sin (3 x) \\ - b \sin (3 x) + d \cos (3 x) \end{array}] = [\begin{array}{c} 0 \\ \tan (3 x) \end{array}]$

解题步骤 1.2

将 $\tan (3 x)$ 重写为正弦和余弦形式。

$[\begin{array}{c} b \cos (3 x) + d \sin (3 x) \\ - b \sin (3 x) + d \cos (3 x) \end{array}] = [\begin{array}{c} 0 \\ \frac{\sin (3 x)}{\cos (3 x)} \end{array}]$

解题步骤 1.3

将 $\frac{\sin (3 x)}{\cos (3 x)}$ 转换成 $\tan (3 x)$ 。

$[\begin{array}{c} b \cos (3 x) + d \sin (3 x) \\ - b \sin (3 x) + d \cos (3 x) \end{array}] = [\begin{array}{c} 0 \\ \tan (3 x) \end{array}]$

解题步骤 2

矩阵方程可表示为一组方程。

$b \cos (3 x) + d \sin (3 x) = 0$

$- b \sin (3 x) + d \cos (3 x) = \tan (3 x)$

线性代数 示例

线性代数示例