求满足AE=B的初等矩阵E

Hey there! Let's work through this problem step by step, since you mentioned you're stuck on figuring out the right elementary matrix. First, let's clear up a key point:

• When you multiply matrix $A$ by $E$ on the right ($AE$), this corresponds to performing elementary column operations on $A$, not row operations. Row operations would require multiplying $E$ on the left ($EA$). Since your equation is $A*E = B$, we're focused on a single column transformation here.

Step 1: Compare Columns of A and B #

Let's look at your given matrices:
$$
A = \begin{bmatrix} 2 & 4 \ 1 & 6 \end{bmatrix}, \quad B = \begin{bmatrix} 2 & -2 \ 1 & 3 \end{bmatrix}
$$

• The first column of $A$ is identical to the first column of $B$ ($\begin{bmatrix}2\1\end{bmatrix}$), so we don't need to modify this column.
• The second column changes from $\begin{bmatrix}4\6\end{bmatrix}$ (in $A$) to $\begin{bmatrix}-2\3\end{bmatrix}$ (in $B$). Let's calculate the transformation:
  • Top entry: $4 + (-3)*2 = -2$ (using the first column's top value, 2)
  • Bottom entry: $6 + (-3)*1 = 3$ (using the first column's bottom value, 1)

This tells us the elementary operation is: Add -3 times the first column to the second column (or equivalently, subtract 3 times the first column from the second column).

Step 2: Construct the Elementary Matrix E #

Elementary matrices are derived from applying the same operation to the identity matrix $I$:

1. Start with the 2x2 identity matrix:
   $$
   I = \begin{bmatrix}1&0\0&1\end{bmatrix}
   $$
2. Apply our column operation to $I$: add -3 times the first column to the second column. The second column of $I$ becomes $\begin{bmatrix}0 + (-3)*1 \ 1 + (-3)*0\end{bmatrix} = \begin{bmatrix}-3\1\end{bmatrix}$
3. The resulting matrix is our $E$:
   $$
   E = \begin{bmatrix}1&-3\0&1\end{bmatrix}
   $$

Step 3: Verify the Result #

Let's multiply $A$ by $E$ to confirm we get $B$:
$$
AE = \begin{bmatrix}2&4\1&6\end{bmatrix} * \begin{bmatrix}1&-3\0&1\end{bmatrix} = \begin{bmatrix}21 + 40 & 2(-3) + 41 \ 11 + 60 & 1(-3) + 6*1\end{bmatrix} = \begin{bmatrix}2&-2\1&3\end{bmatrix} = B
$$

Perfect, that checks out! The confusion about row vs column operations was probably the main hurdle here—remember right-multiplication = column operations, left-multiplication = row operations.

内容的提问来源于stack exchange，提问作者mathguy