Question

Find the value with out expanding:

$\left|\begin{array}{ccc}1& \log x& 3\\ 2& \log x^2& 4\\ 3& \log x^3& 5\end{array}\right|$.

Answer 1

Now,

$\left|\begin{array}{ccc}1& \log x& 3\\ 2& \log x^2& 4\\ 3& \log x^3& 5\end{array}\right|$

$=\left|\begin{array}{ccc}1& \log x& 3\\ 2& 2\log x& 4\\ 3& 3\log x& 5\end{array}\right|$

Now, $R_2^{\prime }=R_2-R_1$ and $R_3^{\prime }=R_3-R_1$

$=\left|\begin{array}{ccc}1& \log x& 3\\ 1& \log x& 1\\ 1& \log x& 1\end{array}\right|$

$=0$ [ Since $R_2=R_1$]