3 and 4 .Determinants and Matrices
hard

$\alpha $ के किस मान के लिए समीकरण निकाय ${(\alpha  + 1)^3}x + {(\alpha  + 2)^3}y - {(\alpha  + 3)^3} = 0$, $(\alpha  + 1)x + (\alpha  + 2)y - (\alpha  + 3) = 0,$ $x + y - 1 = 0$ संगत है

A

$1$

B

$0$

C

$-3$

D

$-2$

Solution

संगत हल के लिए, $|A| = 0$

अर्थात्, $\left| {\,\begin{array}{*{20}{c}}{{{(\alpha  + 1)}^3}}&{{{(\alpha  + 2)}^3}}&{ – {{(\alpha  + 3)}^3}}\\{\alpha  + 1}&{\alpha  + 2}&{ – (\alpha  + 3)}\\1&1&{ – 1}\end{array}} \right| = 0$

$\Rightarrow$ $6\alpha  + 12 = 0 \Rightarrow \alpha  =  – 2$.

Standard 12
Mathematics

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