The number of real values of alpha for which | vedclass

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(A) The given system of equations can be rewritten as:$(1-\alpha)x + 3y + 5z = 0$$5x + (1-\alpha)y + 3z = 0$$3x + 5y + (1-\alpha)z = 0$For the system

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$1$

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(A) The given system of equations can be rewritten as:$(1-\alpha)x + 3y + 5z = 0$$5x + (1-\alpha)y + 3z = 0$$3x + 5y + (1-\alpha)z = 0$For the system to have infinite solutions,the determinant of the coefficient matrix must be zero:$\left|\begin{array}{ccc} 1-\alpha & 3 & 5 \ 5 & 1-\alpha & 3 \ 3 & 5 & 1-\alpha \end{array}ight| = 0$Applying $C_1 ightarrow C_1 + C_2 + C_3$,we get:$\left|\begin{array}{ccc} 9-\alpha & 3 & 5 \ 9-\alpha & 1-\alpha & 3 \ 9-\alpha & 5 & 1-\alpha \end{array}ight| = 0$Taking $(9-\alpha)$ common from the first column:$(9-\alpha) \left|\begin{array}{ccc} 1 & 3 & 5 \ 1 & 1-\alpha & 3 \ 1 & 5 & 1-\alpha \end{array}ight| = 0$Performing $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_1$:$(9-\alpha) \left|\begin{array}{ccc} 1 & 3 & 5 \ 0 & -\alpha-2 & -2 \ 0 & 2 & -\alpha-4 \end{array}ight| = 0$$(9-\alpha) [(-\alpha-2)(-\alpha-4) - (-4)] = 0$$(9-\alpha) [\alpha^2 + 6\alpha + 8 + 4] = 0$$(9-\alpha) (\alpha^2 + 6\alpha + 12) = 0$For $\alpha^2 + 6\alpha + 12 = 0$,the discriminant $D = 6^2 - 4(1)(12) = 36 - 48 = -12 Therefore,the only real value is $\alpha = 9$. The number of real values is $1$.

The number of real values of $\alpha$ for which the system of equations
$x+3y+5z=\alpha x$
$5x+y+3z=\alpha y$
$3x+5y+z=\alpha z$
has infinite number of solutions is

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The number of real values of $\alpha$ for which the system of equations$x+3y+5z=\alpha x$$5x+y+3z=\alpha y$$3x+5y+z=\alpha z$has infinite number of solutions is