Step 1: Understanding the Question:
We use the properties of the rank of matrices in systems of linear equations (Rouche-Capelli Theorem).
For a system \( MX = Y \) with \( M \) being \( 3 \times 3 \):
1. Unique Solution: \( \text{rank}(M) = \text{rank}[M:Y] = 3 \) (Full rank).
2. Infinite Solutions: \( \text{rank}(M) = \text{rank}[M:Y]<3 \).
Step 2: Understanding the the Data:
* System \( AX = B \): Has a unique solution.
Since \( A \) is \( 3 \times 3 \), this implies \( \text{rank}
(A) = 3 \).
Consequently, the augmented matrix \( [A:D] \) (which is \( 3 \times 4 \)) must have a rank of 3 because it contains a \( 3 \times 3 \) submatrix \( A \) with rank 3, and the maximum possible rank for a matrix with 3 rows is 3.
So, \( \text{rank}[A:D] = 3 \).
* System \( CX = D \): Has infinite solutions.
Since \( C \) is \( 3 \times 3 \), this implies \( \text{rank}(C)<3 \).
Also, \( \text{rank}(C) = \text{rank}[C:D] = k \), where \( k \in \{1, 2\} \).
Step 3: Evaluating Options:
Now we evaluate \( \text{rank}[C:B] \). This is an augmented matrix of size \( 3 \times 4 \). Its rank can be at most 3 (limited by the number of rows).
So, \( \text{rank}[C:B] \le 3 \).
Comparison:
We established that \( \text{rank}[A:D] = 3 \).
We know that \( \text{rank}[C:B] \le 3 \).
Therefore, \( \text{rank}[A:D] \ge \text{rank}[C:B] \) is always true (since \( 3 \ge \text{anything} \le 3 \)).
Let's check other options:
(A) \( \text{rank}[A:D] = \text{rank}[C:B] \): Not necessarily true. \( [C:B] \) could have rank 2.
(B) \( \text{rank}
(A) = \text{rank}(C) \): False. \( 3 \neq <3 \).
(C) \( \text{rank}[A:B]<\text{rank}[B:D] \): \( \text{rank}[A:B]=3 \). Rank of \( [B:D] \) is max 3 (as it has 3 rows or assuming it's formed by columns, wait \( B, D \) are \( 3 \times 1 \), so \( [B:D] \) is \( 3 \times 2 \), max rank 2). So \( 3<2 \) is False.
Step 4: Required Answer:
Option (D) is the correct statement.