Step 1: Understanding the Concept:
This question deals with linear algebra and the properties of square matrices.
Specifically, it tests the fundamental relationship between a matrix's invertibility, its singularity, and its determinant.
Step 2: Detailed Explanation:
Let \( A \) be a square matrix of size \( n \times n \).
- By definition, a square matrix \( A \) is said to be invertible (or non-singular) if there exists another square matrix \( B \) of the same size such that:
\[ AB = BA = I \]
where \( I \) represents the identity matrix of size \( n \times n \).
- According to the properties of determinants, the determinant of the product of two matrices is the product of their individual determinants:
\[ \det(AB) = \det(A) \cdot \det(B) \]
- Taking the determinant on both sides of the invertibility equation:
\[ \det(AB) = \det(I) \implies \det(A) \cdot \det(B) = 1 \]
- For the product of two real numbers \( \det(A) \) and \( \det(B) \) to equal 1, neither of the numbers can be zero.
Therefore:
\[ \det(A) \neq 0 \]
This proves that if a matrix is invertible, its determinant must be non-zero.
Conversely, if \( \det(A) \neq 0 \), we can construct the inverse of the matrix using the adjugate matrix formula:
\[ A^{-1} = \frac{1}{\det(A)} \text{adj}(A) \]
which is well-defined only when the determinant is non-zero.
Let us evaluate the other options to confirm why they are incorrect:
- Option (A): By definition, a singular matrix is one whose determinant is equal to zero. Thus, the statement is false.
- Option (B): A non-singular matrix is one whose determinant is non-zero. Thus, this statement is false.
- Option (D): If a matrix is not invertible, its determinant must be exactly equal to zero. Thus, this statement is false.
Step 3: Final Answer:
The statement "If a matrix is invertible, then determinant is non-zero" is always true.
Therefore, the correct choice is Option (C).