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STD 12 - 3 and 4 . determinant and metrices — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3 and 4 . determinant and metrices4 Marks
MCQ
If $\left| {\begin{array}{*{20}{c}}1&a&{{a^2}}\\1&x&{{x^2}}\\{{b^2}}&{ab}&{{a^2}} \end{array}} \right|$ $= 0$ , then :
  • A
    $x = a$
  • B
    $x = b$
  • C
    $x = \frac{a}{b}$
  • ✓
    both $(A)$ and $(C)$
✓

Answer

Correct option: D.
both $(A)$ and $(C)$
d
$R_2 \rightarrow R_2 -R_1$ and $R_3 \rightarrow R_3 - R_1$ gives
$(x - a) (b - 1)$ $\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2}}\\0&1&{x + a}\\ {b + 1}&a&0\end{array}\,} \right|$ open by $c_1$ and get the value of $x = a/b, x = a$

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