To determine the nature of the solution for a system of linear equations, we first calculate the determinant of the coefficient matrix, denoted by $\Delta$. 
$\Delta = \det(A) = 2(2 \cdot 4 - 2 \cdot (-1)) - 3(3 \cdot 4 - 2 \cdot 1) + 2(3 \cdot (-1) - 2 \cdot 1)$.
$\Delta = 2(8 + 2) - 3(12 - 2) + 2(-3 - 2)$.
$\Delta = 2(10) - 3(10) + 2(-5)$.
$\Delta = 20 - 30 - 10 = -20$.
Since the determinant $\Delta$ is non-zero ($\Delta = -20 \neq 0$), the system of linear equations is consistent and has a unique solution.
Therefore, option (B) is the correct statement.
Let $$ B = \begin{bmatrix} 1 & 3 \\ 1 & 5 \end{bmatrix} $$ and $A$ be a $2 \times 2$ matrix such that $$ AB^{-1} = A^{-1}. $$ If $BCB^{-1} = A$ and $$ C^4 + \alpha C^2 + \beta I = O, $$ then $2\beta - \alpha$ is equal to:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)