Investigate the values of lambda and mu for t | vedclass

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Question

(A) Given system of linear equations is:$x+2y+3z=6$$x+3y+5z=9$$2x+5y+\lambda z=\mu$The determinant of the coefficient matrix is:$\Delta = \begin{vmatr

Vedclass · Accepted Answer

$A-2, B-3, C-1$

Vedclass · Answer

(A) Given system of linear equations is:$x+2y+3z=6$$x+3y+5z=9$$2x+5y+\lambda z=\mu$The determinant of the coefficient matrix is:$\Delta = \begin{vmatrix} 1 & 2 & 3 \ 1 & 3 & 5 \ 2 & 5 & \lambda \end{vmatrix} = 1(3\lambda - 25) - 2(\lambda - 10) + 3(5 - 6) = 3\lambda - 25 - 2\lambda + 20 - 3 = \lambda - 8$.For unique solution,$\Delta 
eq 0$,so $\lambda 
eq 8$. Thus,$(B)$ matches with $3$.Now,consider $\Delta_1, \Delta_2, \Delta_3$ for $\lambda = 8$:$\Delta_1 = \begin{vmatrix} 6 & 2 & 3 \ 9 & 3 & 5 \ \mu & 5 & 8 \end{vmatrix} = 6(24-25) - 2(72-5\mu) + 3(45-3\mu) = -6 - 144 + 10\mu + 135 - 9\mu = \mu - 15$.If $\lambda = 8$ and $\mu 
eq 15$,then $\Delta_1 
eq 0$,which implies the system has no solution. Thus,$(A)$ matches with $2$.If $\lambda = 8$ and $\mu = 15$,then $\Delta = 0, \Delta_1 = 0, \Delta_2 = 0, \Delta_3 = 0$. The system has infinitely many solutions. Thus,$(C)$ matches with $1$.Therefore,the correct matching is $A-2, B-3, C-1$.

List-$I$	List-$II$
$(A)$ $\lambda=8, \mu \neq 15$	$1$. Infinitely many solutions
$(B)$ $\lambda \neq 8, \mu \in R$	$2$. No solution
$(C)$ $\lambda=8, \mu=15$	$3$. Unique solution

List-$I$	List-$II$
$(P)$ If $\beta=\frac{1}{2}(7\alpha-3)$ and $\gamma=28$,then the system has	$(1)$ a unique solution
$(Q)$ If $\beta=\frac{1}{2}(7\alpha-3)$ and $\gamma \neq 28$,then the system has	$(2)$ no solution
$(R)$ If $\beta \neq \frac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma \neq 28$,then the system has	$(3)$ infinitely many solutions
$(S)$ If $\beta \neq \frac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma=28$,then the system has	$(4)$ $x=11, y=-2$ and $z=0$ as a solution
	$(5)$ $x=-15, y=4$ and $z=0$ as a solution

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