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ધારો કે સુરેખ સમીકરણ સંહતિ
$x+y+\alpha z=2$
$3 x+y+z=4$
$x+2 z=1$
ને અનન્ય ઉએેલ $\left( x ^{*}, y ^{*}, z ^{*}\right)$ છે. જો $\left(\alpha, x ^{*}\right),\left( y ^{*}, \alpha\right)$ અને $\left( x ^{*},- y ^{*}\right)$ તો $\alpha$સમરેખ બિંદુઓ હોય. તો $\alpha$ ની તમામ શક્ય કિંમતોનાં નિરપેક્ષ મૂલ્યોનો સરવાળો ........ છે.
$4$
$3$
$2$
$1$
Solution
$\Delta=\left|\begin{array}{lll}1 & 1 & \alpha \\ 3 & 1 & 1 \\ 1 & 0 & 2\end{array}\right|=-(\alpha+3)$
$\Delta_{1}=\left|\begin{array}{lll}2 & 1 & \alpha \\ 4 & 1 & 1 \\ 1 & 0 & 2\end{array}\right|=-(3+\alpha)$
$\Delta_{2}=\left|\begin{array}{lll}1 & 2 & \alpha \\ 3 & 4 & 1 \\ 1 & 1 & 2\end{array}\right|=-(\alpha+3)$
$\Delta_{3}=\left|\begin{array}{lll}1 & 1 & 2 \\ 3 & 1 & 4 \\ 1 & 0 & 1\end{array}\right|=0$
$\alpha \neq-3, x=1, y=1, z=0$,
Now points $(\alpha, 1),(1, \alpha) \ and (1,-1)$ are collinear
$\left|\begin{array}{lcc}\alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & -1 & 1\end{array}\right|=0$
$\Rightarrow \alpha(\alpha+1)-1(1-1)+1(-1-\alpha)=0$
$\alpha^{2}+\alpha-1-\alpha=0$
