The problem asks us to show that the product \( C = A \cdot B \) is a singular matrix, given that \( A \) is a \( 3 \times 2 \) matrix and \( B \) is a \( 2 \times 3 \) matrix. A singular matrix is one that does not have an inverse, which means its determinant is zero.
First, let’s find the dimensions of the resulting matrix \( C \) when \( A \) and \( B \) are multiplied.
The dimensions of \( C \) can be determined by the outer dimensions of \( A \) and \( B \). In this case, \( A \) is \( 3 \times 2 \) and \( B \) is \( 2 \times 3 \), so \( C \) will be \( 3 \times 3 \).
The rank of \( C \) is limited by the smaller of the two ranks of \( A \) and \( B \). Since \( A \) is \( 3 \times 2 \), its rank can be at most 2. Similarly, \( B \) is \( 2 \times 3 \), so its rank can also be at most 2.
Therefore, the rank of \( C \) can be at most 2.
\[ \text{Rank}(C) \leq \min(\text{Rank}(A), \text{Rank}(B)) \leq 2 \]
For a \( 3 \times 3 \) matrix to be invertible (non-singular), its rank must be 3. However, we’ve established that the rank of \( C \) can be at most 2. Therefore, \( C \) must be singular.
To confirm, the determinant of a singular matrix is zero:
\[ \text{Det}(C) = 0 \]
We have shown that the matrix \( C = A \cdot B \) will be a \( 3 \times 3 \) matrix with a rank of at most 2. Since the rank is less than 3, \( C \) is a singular matrix, and its determinant is zero. Therefore, \( C = A \cdot B \) is indeed a singular matrix.