Gujarati
Hindi
3 and 4 .Determinants and Matrices
normal

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

$x = \frac{a}{b}$

D

both $(A)$ and $(C)$

Solution

$R_2 \rightarrow R_2 -R_1$ 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 $x = a/b, x = a$

Standard 12
Mathematics

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