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  • The number of values of for which the system of equations: <img alt="\mathrm{x

The number of values of \mathrm{\alpha } for which the system of equations:
\mathrm{x+y+z= \alpha }
\mathrm{\alpha x+2\alpha y+3z= -1 }
\mathrm{ x+3\alpha y+5z= 4 }
is inconsistent, is

Option: 1

0


Option: 2

1


Option: 3

2


Option: 4

3


Answers (1)

best_answer

\mathrm{x+y+z= \alpha }
\mathrm{\alpha x+2\, \alpha y+3z= -1 }
\mathrm{x+3\, \alpha y+5z= 4 }

For inconsistant system,

\mathrm{D= 0} & atleast one of \mathrm{D_{1},\, D_{2},\, D_{3}\neq 0}

\mathrm{D= \begin{vmatrix} 1 & 1 &1 \\ \alpha &2 \alpha & 3\\ 1& 3\alpha& 5 \end{vmatrix}= \alpha+3-5\alpha+3\alpha^{2}-2\alpha}
                                   \mathrm{= 3\alpha ^{2}-6\alpha+3}
                                   \mathrm{= 3\left ( \alpha-1 \right )^{2}}


\mathrm{D_{1}= \begin{vmatrix} \alpha &1 &1 \\ -1& 2\alpha& 3\\ 4&3\alpha & 5 \end{vmatrix}= \alpha^{2}+17-3\alpha-8\alpha}\
                                        \mathrm{= \alpha^{2}-11\alpha+17}

\mathrm{D_{2}= \begin{vmatrix} 1 & \alpha &1 \\ \alpha & -1 &3 \\ 1 & 4 & 5 \end{vmatrix}=-17+3\alpha-5\alpha^{2}+4\alpha+1 }
                                     \mathrm{= -5\alpha^{2}+7\alpha-16}\


\mathrm{D_{3}=\begin{vmatrix} 1 &1 &\alpha \\ \alpha&2\alpha &-1 \\ 1& 3\alpha & 4 \end{vmatrix}= 11\alpha-1-4\alpha+3\alpha^{3}-2\alpha^{2}}
                                         \mathrm{=3\alpha^{3}-2\alpha^{2}+7\alpha-1}

Hence inconsistant solutions for \mathrm{\alpha=1}
The correct answer is option (B)

Posted by

Shailly goel

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