Consider the system of linear equations:-x+y+ | vedclass

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(C) The determinant of the coefficient matrix is $\Delta = \begin{vmatrix} -1 & 1 & 2 \ 3 & -a & 5 \ 2 & -2 & -a \end{vmatrix}$.Expanding along the

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$n(S_{1})=2, n(S_{2})=0$

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(C) The determinant of the coefficient matrix is $\Delta = \begin{vmatrix} -1 & 1 & 2 \ 3 & -a & 5 \ 2 & -2 & -a \end{vmatrix}$.Expanding along the first row:$\Delta = -1(a^2 + 10) - 1(-3a - 10) + 2(-6 + 2a)$$= -a^2 - 10 + 3a + 10 - 12 + 4a = -a^2 + 7a - 12 = -(a-3)(a-4)$.For the system to be inconsistent or have infinitely many solutions,we must have $\Delta = 0$,which gives $a = 3$ or $a = 4$.Now,calculate $\Delta_1 = \begin{vmatrix} 0 & 1 & 2 \ 1 & -a & 5 \ 7 & -2 & -a \end{vmatrix}$.$\Delta_1 = 0(a^2 + 10) - 1(-a - 35) + 2(-2 + 7a) = a + 35 - 4 + 14a = 15a + 31$.For $a = 3$,$\Delta_1 = 15(3) + 31 = 76 
eq 0$.For $a = 4$,$\Delta_1 = 15(4) + 31 = 91 
eq 0$.Since $\Delta = 0$ and $\Delta_1 
eq 0$ for both $a=3$ and $a=4$,the system is inconsistent for these values. Thus,$S_1 = \{3, 4\}$ and $n(S_1) = 2$.For infinitely many solutions,we require $\Delta = 0$ and $\Delta_1 = \Delta_2 = \Delta_3 = 0$. Since $\Delta_1 
eq 0$ for $a=3$ and $a=4$,there are no values of $a$ for which the system has infinitely many solutions. Thus,$S_2 = \emptyset$ and $n(S_2) = 0$.

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