Question
Without expanding the determinants, show that
$\left|\begin{array}{lll}l & m & n \\ e & d & f \\ u & v & w\end{array}\right|=\left|\begin{array}{lll}n & f & w \\ l & e & u \\ m & d & v\end{array}\right|$
$\left|\begin{array}{lll}l & m & n \\ e & d & f \\ u & v & w\end{array}\right|=\left|\begin{array}{lll}n & f & w \\ l & e & u \\ m & d & v\end{array}\right|$
Interchanging rows and columns, we get
L.H.S. $=\left|\begin{array}{lll}l & \mathrm{e} & \mathrm{u} \\ \mathrm{m} & \mathrm{d} & \mathrm{v} \\ \mathrm{n} & \mathrm{f} & \mathrm{w}\end{array}\right|$
Applying $\mathrm{R}_2 \leftrightarrow \mathrm{R}_3$, we get
L.H.S. $=-\left|\begin{array}{lll}l & e & u \\ n & f & w \\ m & d & v\end{array}\right|$
Applying $R_1 \leftrightarrow R_2$, we get
$\begin{aligned} \text { L.H.S. } & =\left|\begin{array}{ccc}\mathrm{n} & \mathrm{f} & \mathrm{w} \\ l & \mathrm{e} & \mathrm{u} \\ \mathrm{m} & \mathrm{d} & \mathrm{v}\end{array}\right| \\ & =\text { R.H.S. }\end{aligned}$
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$f(x)=\frac{6 x-7}{3}$