By using properties of determinants, show that:
$\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|=(5 x+4)(4-x)^{2}$
By using properties of determinants, show that:
$\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|=(5 x+4)(4-x)^{2}$
$\Delta=\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|$
Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3},$ we have:
$\Delta=\left|\begin{array}{ccc}5 x+4 & 5 x+4 & 5 x+4 \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|$
$=(5 x+4)\left|\begin{array}{ccc}1 & 0 & 0 \\ 2 x & x+4 & 0 \\ 2 x & 0 & x+4\end{array}\right|$
Applying $\mathrm{C}_{2} \rightarrow C_{2}-C_{1}, \mathrm{C}_{3} \rightarrow C_{3}-C_{1},$ we have
$\Delta=(5 x+4)\left|\begin{array}{ccc}1 & 0 & 0 \\ 2 x & -x+4 & 0 \\ 2 x & 0 & -x+4\end{array}\right|$
$=(5 x+4)(4-x)(4-x)\left|\begin{array}{ccc}1 & 0 & 0 \\ 2 x & 1 & 0 \\ 2 x & 0 & 1\end{array}\right|$
Expanding along $C_{3},$ we have:
$\Delta=(5 x+4)(4-x)^{2}\left|\begin{array}{cc}1 & 0 \\ 2 x & 1\end{array}\right|$
$=(5 x+4)(4-x)^{2}$
Hence, the given result is proved.
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