求满足矩阵方程AB=C的矩阵A（B不可逆时的解法咨询）

Hey there! Let's work through this problem step by step.

First, you're right that matrix ( B = \begin{bmatrix} 1 & 2 & 3 \ 1 & 1 & 1 \ 1 & 3 & 5 \end{bmatrix} ) is singular (its determinant is 0, as you calculated), so it doesn't have an inverse. That means we can't use the formula ( A = C B^{-1} ) directly—but there are still ways to find all possible matrices ( A ) that satisfy ( AB = C ).

Step 1: Break the matrix equation into linear systems #

Let's denote ( A ) as a 3x3 matrix:
$$
A = \begin{bmatrix}
a_{11} & a_{12} & a_{13} \
a_{21} & a_{22} & a_{23} \
a_{31} & a_{32} & a_{33}
\end{bmatrix}
$$
The equation ( AB = C ) translates to each row of ( A ) multiplying ( B ) gives the corresponding row of ( C ). For example:

• The first row of ( A ) times ( B ) equals the first row of ( C ): ( [a_{11}, a_{12}, a_{13}] \cdot B = [1, 3, 5] )
• The second row of ( A ) times ( B ) equals the second row of ( C ): ( [a_{21}, a_{22}, a_{23}] \cdot B = [1, 2, 3] )
• The third row of ( A ) times ( B ) equals the third row of ( C ): ( [a_{31}, a_{32}, a_{33}] \cdot B = [1, 1, 1] )

Each of these is a linear system we can solve independently.

Step 2: Solve each linear system #

Let's start with the first system (for the first row of ( A )):
$$
\begin{cases}
a_{11} + a_{12} + a_{13} = 1 \
2a_{11} + a_{12} + 3a_{13} = 3 \
3a_{11} + a_{12} + 5a_{13} = 5
\end{cases}
$$
Write the augmented matrix and perform row operations:
$$
\left[\begin{array}{ccc|c}
1 & 1 & 1 & 1 \
2 & 1 & 3 & 3 \
3 & 1 & 5 & 5
\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}
1 & 1 & 1 & 1 \
0 & -1 & 1 & 1 \
0 & 0 & 0 & 0
\end{array}\right]
$$
The system has infinitely many solutions. Let ( a_{13} = t ) (a free variable). Then:

• From the second row: ( -a_{12} + t = 1 \implies a_{12} = t - 1 )
• From the first row: ( a_{11} + (t-1) + t = 1 \implies a_{11} = 2 - 2t )

So the first row of ( A ) is ( [2-2t, t-1, t] ) where ( t \in \mathbb{R} ).

Next, the second system (second row of ( A )):
$$
\begin{cases}
a_{21} + a_{22} + a_{23} = 1 \
2a_{21} + a_{22} + 3a_{23} = 2 \
3a_{21} + a_{22} + 5a_{23} = 3
\end{cases}
$$
Augmented matrix and row ops:
$$
\left[\begin{array}{ccc|c}
1 & 1 & 1 & 1 \
2 & 1 & 3 & 2 \
3 & 1 & 5 & 3
\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}
1 & 1 & 1 & 1 \
0 & -1 & 1 & 0 \
0 & 0 & 0 & 0
\end{array}\right]
$$
Let ( a_{23} = s ) (free variable):

• ( -a_{22} + s = 0 \implies a_{22} = s )
• ( a_{21} + s + s = 1 \implies a_{21} = 1 - 2s )

Second row of ( A ): ( [1-2s, s, s] ) where ( s \in \mathbb{R} ).

Third system (third row of ( A )):
$$
\begin{cases}
a_{31} + a_{32} + a_{33} = 1 \
2a_{31} + a_{32} + 3a_{33} = 1 \
3a_{31} + a_{32} + 5a_{33} = 1
\end{cases}
$$
Augmented matrix and row ops:
$$
\left[\begin{array}{ccc|c}
1 & 1 & 1 & 1 \
2 & 1 & 3 & 1 \
3 & 1 & 5 & 1
\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}
1 & 1 & 1 & 1 \
0 & -1 & 1 & -1 \
0 & 0 & 0 & 0
\end{array}\right]
$$
Let ( a_{33} = u ) (free variable):

• ( -a_{32} + u = -1 \implies a_{32} = u + 1 )
• ( a_{31} + (u+1) + u = 1 \implies a_{31} = -2u )

Third row of ( A ): ( [-2u, u+1, u] ) where ( u \in \mathbb{R} ).

Step 3: Combine the solutions #

Putting it all together, the general solution for ( A ) is:
$$
A = \begin{bmatrix}
2-2t & t-1 & t \
1-2s & s & s \
-2u & u+1 & u
\end{bmatrix}
$$
where ( t, s, u ) are any real numbers.

Key takeaway #

When ( B ) is singular (no inverse), we can't use the inverse formula. Instead, we break the matrix equation into linear systems for each row of ( A ). First, we need to check if solutions exist: this happens when each row of ( C ) is a linear combination of the columns of ( B ) (equivalently, the rank of ( B ) equals the rank of the augmented matrix ( [B^T | C_i^T] ) for each row ( C_i ) of ( C )). In your case, all systems are consistent, so we get infinitely many solutions parameterized by free variables.

内容的提问来源于stack exchange，提问作者Herman