证明：满足det(A+B)=det(A-B)的3阶实矩阵B可逆当且仅当b₁₁≠b₂₁

Given matrix (A = \begin{pmatrix} 0 & 1 & 2 \ 0 & 1 & 2 \ 0 & 2 & 3 \end{pmatrix}) and (B = [b_{ij}] \in \mathbb{R}^{3 \times 3}) satisfying (\det(A+B) = \det(A-B)), we need to show (B) is invertible if and only if (b_{11} \neq b_{21}).

Step 1: Rewrite the determinant condition using column-wise expansion #

First, note (A) can be written as (A = [\mathbf{0}, \mathbf{c}_2, \mathbf{c}_3]), where (\mathbf{c}_2 = (1,1,2)^T) and (\mathbf{c}_3 = (2,2,3)^T). Let the columns of (B) be (\mathbf{B}_1, \mathbf{B}2, \mathbf{B}3) (so (\mathbf{B}1 = (b{11}, b{21}, b{31})^T)).

We can express the combined matrices as:

• (A+B = [\mathbf{B}_1, \mathbf{c}_2+\mathbf{B}_2, \mathbf{c}_3+\mathbf{B}_3])
• (A-B = [-\mathbf{B}_1, \mathbf{c}_2-\mathbf{B}_2, \mathbf{c}_3-\mathbf{B}_3])

Using the multilinearity and alternating properties of determinants, expand both:

• (\det(A+B) = \det(\mathbf{B}_1, \mathbf{c}_2, \mathbf{c}_3) + \det(\mathbf{B}_1, \mathbf{c}_2, \mathbf{B}_3) + \det(\mathbf{B}_1, \mathbf{B}_2, \mathbf{c}_3) + \det(B))
• (\det(A-B) = -\det(\mathbf{B}_1, \mathbf{c}_2-\mathbf{B}_2, \mathbf{c}_3-\mathbf{B}_3)) (factoring out (-1) from the first column)
  Expanding the second determinant further gives:
  (\det(A-B) = -\left[\det(\mathbf{B}_1, \mathbf{c}_2, \mathbf{c}_3) - \det(\mathbf{B}_1, \mathbf{c}_2, \mathbf{B}_3) - \det(\mathbf{B}_1, \mathbf{B}_2, \mathbf{c}_3) + \det(B)\right])

Step 2: Set determinants equal and simplify #

Set (\det(A+B) = \det(A-B)) and rearrange all terms to one side:
[
\begin{align*}
&\det(\mathbf{B}_1, \mathbf{c}_2, \mathbf{c}_3) + \det(\mathbf{B}_1, \mathbf{c}_2, \mathbf{B}_3) + \det(\mathbf{B}_1, \mathbf{B}_2, \mathbf{c}_3) + \det(B) \
&+ \det(\mathbf{B}_1, \mathbf{c}_2, \mathbf{c}_3) - \det(\mathbf{B}_1, \mathbf{c}_2, \mathbf{B}_3) - \det(\mathbf{B}_1, \mathbf{B}_2, \mathbf{c}_3) + \det(B) = 0
\end{align*}
]
Most terms cancel out, leaving:
[2\det(\mathbf{B}_1, \mathbf{c}_2, \mathbf{c}_3) + 2\det(B) = 0 \implies \det(B) = -\det(\mathbf{B}_1, \mathbf{c}_2, \mathbf{c}_3)]

Step 3: Compute (\det(\mathbf{B}_1, \mathbf{c}_2, \mathbf{c}_3)) #

Calculate the determinant of the matrix formed by (\mathbf{B}1, \mathbf{c}2, \mathbf{c}3):
[
\det\begin{pmatrix} b{11} & 1 & 2 \ b{21} & 1 & 2 \ b{31} & 2 & 3 \end{pmatrix} = b_{11}(1\cdot3 - 2\cdot2) - 1(b_{21}\cdot3 - 2\cdot b_{31}) + 2(b_{21}\cdot2 -1\cdot b_{31})
]
Simplify each term:
[
= b_{11}(-1) - (3b_{21} - 2b_{31}) + 2(2b_{21} - b_{31}) = -b_{11} -3b_{21} +2b_{31} +4b_{21} -2b_{31}
]
Cancel like terms ((2b_{31}) and (-2b_{31}), (-3b_{21}) and (4b_{21})):
[
= b_{21} - b_{11}
]

Step 4: Relate determinant of B to (b_{11}) and (b_{21}) #

From Step 2, we have:
[
\det(B) = -(b_{21} - b_{11}) = b_{11} - b_{21}
]

Conclusion #

A matrix is invertible if and only if its determinant is non-zero. So putting it all together:

• (B) is invertible exactly when (\det(B) \neq 0), which from our calculation is equivalent to (b_{11} - b_{21} \neq 0), or simply (b_{11} \neq b_{21}).

This completes the proof.

内容的提问来源于stack exchange，提问作者solomeo paredes