$\left| \begin{matrix} \log e^{\ } & \log e^{2} & \log e^{3} \ \log e^{2} & \log e^{3} & \log e^{4} \ \log e^{3} & \log e^{4} & \log e^{5} \end{matrix} ight|$ is equal to

Question

$\left| \begin{matrix} \log e^{\ } & \log e^{2} & \log e^{3} \ \log e^{2} & \log e^{3} & \log e^{4} \ \log e^{3} & \log e^{4} & \log e^{5} \end{matrix} ight|$ is equal to

Dr. Abubacker · Accepted Answer

(a)
$\left| \begin{matrix}
\log e^{\ } & \log e^{2} & \log e^{3} \
\log e^{2} & \log e^{3} & \log e^{4} \
\log e^{3} & \log e^{4} & \log e^{5}
\end{matrix} ight| = \left| \begin{matrix}
\log e^{\ } & {2log}e^{\ } & 3\log e^{\ } \
{2log}e^{\ } & {3log}e^{\ } & {4log}e^{\ } \
3\log e^{\ } & 4\log e^{\ } & {5log}e^{\ }
\end{matrix} ight|$
$= \left| \begin{array}{r}
1\ \ 2\ \ 3 \
2\ \ 3\ \ 4 \
3\ \ 4\ \ 5
\end{array} ight| = \left| \begin{array}{r}
1\ \ 1\ \ 1 \
2\ \ 1\ \ 1 \
3\ \ 1\ \ \ 1
\end{array} ight|$
(Using $C_{2} ightarrow C_{2} - C_{1},\ C_{3} ightarrow C_{3} - C_{2})$
$= 0\lbrack	herefore\$two columns are identical]

Q. $\left| \begin{matrix} \log e^{\ } & \log e^{2} & \log e^{3} \\ \log e^{2} & \log e^{3} & \log e^{4} \\ \log e^{3} & \log e^{4} & \log e^{5} \end{matrix} \right|$ is equal to

Explanation

Reviewed by Dr. Abubacker

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