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3 and 4 .Determinants and Matrices
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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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