Question:

<p><strong>Step 1: Understand the definition of the rank of a matrix.</strong></p><p>The rank of a matrix is the maximum number of linearly independent row vectors or column vectors in the matrix. Alternatively, the rank of a matrix is the order of the largest non-singular (invertible) square submatrix.</p><p><strong>Step 2: Relate the determinant of a square matrix to its rank.</strong></p><p>For a square matrix, say $A$ of order $n \times n$, its determinant, denoted as $\det(A)$ or $\Delta$, provides crucial information about its rank and invertibility. A square matrix $A$ is said to be non-singular (or invertible) if and only if its determinant is non-zero, i.e., $\det(A) \neq 0$.</p><p><strong>Step 3: Apply the given condition to determine the rank.</strong></p><p>The problem states that the determinant $\Delta \neq 0$. Since a determinant is defined for a square matrix, let's assume the given matrix is an $n \times n$ square matrix. Because $\Delta \neq 0$, the matrix itself is non-singular. This implies that the entire $n \times n$ matrix is an $n \times n$ non-singular submatrix. Since this is the largest possible square submatrix (as it is the matrix itself), the rank of the matrix must be $n$.</p><p><strong>Step 4: Conclude the answer.</strong></p><p>The rank of the matrix is $n$, which is equal to its order. Therefore, option (B) is the correct answer.</p> <div class="key-concept"><strong>Key Concept:</strong> A square matrix has a non-zero determinant if and only if it is non-singular (invertible), which directly implies that its rank is equal to its order.</div> <div class="trap-box"><strong>Trap:</strong> A common mistake is to confuse the conditions for a matrix to be singular (determinant zero) or to incorrectly apply the definition of rank, especially for non-square matrices, although the question's mention of 'determinant' implies a square matrix.</div>