A, C are 3 times 3 matrices. B, D are 3 times | vedclass

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(D) For the system $AX=B$, since it has a unique solution, the matrix $A$ must be invertible, which implies $\operatorname{rank}(A) = 3$. Thus, the au

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$\operatorname{rank}([A: D]) \geq \operatorname{rank}([C: B])$

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(D) For the system $AX=B$, since it has a unique solution, the matrix $A$ must be invertible, which implies $\operatorname{rank}(A) = 3$. Thus, the augmented matrix $[A: B]$ also has rank $3$.For the system $CX=D$, since it has an infinite number of solutions, the matrix $C$ must be singular, which implies $\operatorname{rank}(C) Comparing the ranks: $\operatorname{rank}(A) = 3$ and $\operatorname{rank}(C)  \operatorname{rank}(C)$.Regarding the augmented matrices, $\operatorname{rank}([A: D])$ is at most $3$, and $\operatorname{rank}([C: B])$ is at most $3$. Since $\operatorname{rank}(A) = 3$, the rank of $[A: D]$ is $3$. Since $\operatorname{rank}(C) < 3$, the rank of $[C: B]$ is at most $3$. Thus, $\operatorname{rank}([A: D]) \geq \operatorname{rank}([C: B])$ is a true statement.

$A, C$ are $3 \times 3$ matrices. $B, D$ are $3 \times 1$ matrices. If $AX=B$ has a unique solution and $CX=D$ has an infinite number of solutions, then:

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