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(A) Given the system of equations:$x+2y+z=7$$x+\alpha z=11$$2x-3y+\beta z=\gamma$The determinant of the coefficient matrix is $\Delta = \begin{vmatrix

Vedclass · Accepted Answer

$(P) \rightarrow (3), (Q) \rightarrow (2), (R) \rightarrow (1), (S) \rightarrow (4)$

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(A) Given the system of equations:$x+2y+z=7$$x+\alpha z=11$$2x-3y+\beta z=\gamma$The determinant of the coefficient matrix is $\Delta = \begin{vmatrix} 1 & 2 & 1 \ 1 & 0 & \alpha \ 2 & -3 & \beta \end{vmatrix} = 1(0 - (-3\alpha)) - 2(\beta - 2\alpha) + 1(-3 - 0) = 3\alpha - 2\beta + 4\alpha - 3 = 7\alpha - 2\beta - 3$.If $\beta = \frac{1}{2}(7\alpha - 3)$,then $\Delta = 0$.For $(P)$: $\beta = \frac{1}{2}(7\alpha - 3)$ and $\gamma = 28$. Calculating $\Delta_x, \Delta_y, \Delta_z$,we find they are all $0$. Thus,the system has infinitely many solutions. $(P ightarrow 3)$.For $(Q)$: $\beta = \frac{1}{2}(7\alpha - 3)$ and $\gamma 
eq 28$. Since $\Delta = 0$ and at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero,the system has no solution. $(Q ightarrow 2)$.For $(R)$: $\beta 
eq \frac{1}{2}(7\alpha - 3)$,so $\Delta 
eq 0$. The system has a unique solution. $(R ightarrow 1)$.For $(S)$: $\beta 
eq \frac{1}{2}(7\alpha - 3)$ and $\alpha = 1, \gamma = 28$. Since $\Delta 
eq 0$,there is a unique solution. Substituting $x=11, y=-2, z=0$ into the equations: $11+2(-2)+0 = 7$ (True),$11+1(0) = 11$ (True),$2(11)-3(-2)+\beta(0) = 22+6 = 28 = \gamma$ (True). Thus,$(x=11, y=-2, z=0)$ is the unique solution. $(S ightarrow 4)$.

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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