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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