如何根据四元方程组求线性变换矩阵？给定具体变换方程组求解

First, let's recap the core idea: For a linear transformation that maps an n-dimensional input vector $\mathbf{x} = (x_1, x_2, ..., x_n)$ to an m-dimensional output vector $\mathbf{y} = (y_1, y_2, ..., y_m)$, the corresponding matrix is an $m \times n$ matrix. Each row of this matrix directly corresponds to the coefficients of the input variables in the equation for the corresponding output component.

Here's the step-by-step process:

• Step 1: Determine the dimensions of the matrix. The number of rows equals the number of output variables (y's), and the number of columns equals the number of input variables (x's).
• Step 2: For each output equation $y_i = a_{i1}x_1 + a_{i2}x_2 + ... + a_{in}x_n$, take the coefficients $a_{i1}, a_{i2}, ..., a_{in}$ and place them as the $i$-th row of the matrix.
• Step 3: Assemble all the rows together to form the complete transformation matrix.

Applying This to Your Given System #

Your linear transformation is defined by:
$$
\begin{align*}
y_1 &= 9x_1 + 3x_2 - 3x_3 \
y_2 &= 2x_1 - 9x_2 + x_3 \
y_3 &= 4x_1 - 9x_2 - 2x_3 \
y_4 &= 5x_1 + x_2 + 5x_3
\end{align*}
$$

Let's break this down:

• We have 4 output variables ($y_1$ to $y_4$) and 3 input variables ($x_1$ to $x_3$), so our matrix will be $4 \times 3$.
• Extract each row from the equations:
  • Row 1 (for $y_1$): $[9, 3, -3]$
  • Row 2 (for $y_2$): $[2, -9, 1]$
  • Row 3 (for $y_3$): $[4, -9, -2]$
  • Row 4 (for $y_4$): $[5, 1, 5]$

Putting it all together, the matrix representing this linear transformation is:
$$
A = \begin{pmatrix}
9 & 3 & -3 \
2 & -9 & 1 \
4 & -9 & -2 \
5 & 1 & 5
\end{pmatrix}
$$

Just to confirm: When you multiply this matrix by the input vector $\mathbf{x} = \begin{pmatrix}x_1 \ x_2 \ x_3\end{pmatrix}$, you'll get exactly the output vector $\mathbf{y} = \begin{pmatrix}y_1 \ y_2 \ y_3 \ y_4\end{pmatrix}$ as defined by your equations.

内容的提问来源于stack exchange，提问作者user490308