矩阵方程组AX=B的解判定及伴随矩阵与B乘积的作用疑问

Great question—let’s break this down using core properties of the adjugate matrix, since that’s the key to understanding why this test works.

First, remember the fundamental identity for any square matrix $A$ and its adjugate $\operatorname{adj} A$:
$$A \cdot \operatorname{adj} A = \operatorname{adj} A \cdot A = |A|I$$
where $I$ is the identity matrix. When $|A|=0$, this simplifies to $A \cdot \operatorname{adj} A = O$ (the zero matrix).

Case 1: $(\operatorname{adj} A) \cdot B \neq O$ → No solution (inconsistent system) #

Suppose, for contradiction, that $AX=B$ has a solution $X$. If we left-multiply both sides of $AX=B$ by $\operatorname{adj} A$, we get:
$$\operatorname{adj} A \cdot A \cdot X = \operatorname{adj} A \cdot B$$
But from the identity above, $\operatorname{adj} A \cdot A = |A|I = O$. So the left side becomes $O \cdot X = O$. This would force $\operatorname{adj} A \cdot B = O$, which contradicts our initial condition $(\operatorname{adj} A) \cdot B \neq O$.

In short: if there was a solution, we’d get a contradiction. Therefore, no solution exists when $(\operatorname{adj} A) \cdot B \neq O$.

Case 2: $(\operatorname{adj} A) \cdot B = O$ → May have infinitely many solutions or no solution #

This condition is necessary but not sufficient for the system to have a solution. Let’s see why:

• We already know that if a solution exists, $\operatorname{adj} A \cdot B$ must be $O$ (from the earlier logic). But the reverse isn’t always true—because the adjugate matrix’s behavior depends on the rank of $A$:
  • If $\text{rank}(A) = n-1$ (where $n$ is the size of $A$): The adjugate matrix has rank 1. Here, $\operatorname{adj} A \cdot B = O$ tells us $B$ aligns with the column space of $A$ (a requirement for solvability). For example, take $A = \begin{bmatrix}1&1\2&2\end{bmatrix}$ (rank 1, $n=2$ so $n-1=1$). $\operatorname{adj} A = \begin{bmatrix}2&-1\-2&1\end{bmatrix}$. If $B = \begin{bmatrix}3\6\end{bmatrix}$, $\operatorname{adj} A \cdot B = \begin{bmatrix}0\0\end{bmatrix}$, and the system $x+y=3, 2x+2y=6$ has infinitely many solutions.
  • If $\text{rank}(A) < n-1$: The adjugate matrix is the zero matrix ($\operatorname{adj} A = O$), so $\operatorname{adj} A \cdot B = O$ is always true—regardless of whether $B$ is in the column space of $A$. For example, take $A = \begin{bmatrix}1&0&0\1&0&0\1&0&0\end{bmatrix}$ (rank 1 < $3-1=2$). $\operatorname{adj} A = O$, so $\operatorname{adj} A \cdot B = O$ for any $B$. But if $B = \begin{bmatrix}0\0\1\end{bmatrix}$, the system $AX=B$ has no solution; if $B = \begin{bmatrix}2\2\2\end{bmatrix}$, it has infinitely many solutions.

So in this case, $\operatorname{adj} A \cdot B = O$ only tells us we haven’t ruled out a solution—we still need to check if $\text{rank}(A) = \text{rank}([A|B])$ (the rank of the augmented matrix) to confirm solvability.

内容的提问来源于stack exchange，提问作者SOORAJ SOMAN