Let a, lambda, mu in mathbb{R}. Consider the | vedclass

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(C) The given system of equations is:$ax + 2y = \lambda$$3x - 2y = \mu$The determinant of the coefficient matrix is:$\Delta = \begin{vmatrix} a & 2 \

Vedclass · Accepted Answer

$B, C, D$

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(C) The given system of equations is:$ax + 2y = \lambda$$3x - 2y = \mu$The determinant of the coefficient matrix is:$\Delta = \begin{vmatrix} a & 2 \ 3 & -2 \end{vmatrix} = -2a - 6 = -2(a + 3)$.For a unique solution,$\Delta 
eq 0$,which implies $a 
eq -3$. Thus,statement $(B)$ is correct.If $a = -3$,then $\Delta = 0$. The system will have either no solution or infinitely many solutions.We calculate $\Delta_1 = \begin{vmatrix} \lambda & 2 \ \mu & -2 \end{vmatrix} = -2\lambda - 2\mu = -2(\lambda + \mu)$.If $\lambda + \mu = 0$,then $\Delta_1 = 0$. Since $\Delta = 0$ and $\Delta_1 = 0$,the system has infinitely many solutions. Thus,statement $(C)$ is correct.If $\lambda + \mu 
eq 0$,then $\Delta_1 
eq 0$. Since $\Delta = 0$ and $\Delta_1 
eq 0$,the system has no solution. Thus,statement $(D)$ is correct.Therefore,statements $(B)$,$(C)$,and $(D)$ are correct.

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