By using properties of determinants,show that:
$\left|\begin{array}{ccc}x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4\end{array}\right|=(5x+4)(4-x)^{2}$
$\left|\begin{array}{ccc}x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4\end{array}\right|=(5x+4)(4-x)^{2}$
(A) Let $\Delta = \left|\begin{array}{ccc}x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4\end{array}\right|$.
Applying the row operation $R_{1} \rightarrow R_{1} + R_{2} + R_{3}$,we get:
$\Delta = \left|\begin{array}{ccc}5x+4 & 5x+4 & 5x+4 \\ 2x & x+4 & 2x \\ 2x & 2x & x+4\end{array}\right|$.
Taking $(5x+4)$ common from $R_{1}$:
$\Delta = (5x+4) \left|\begin{array}{ccc}1 & 1 & 1 \\ 2x & x+4 & 2x \\ 2x & 2x & x+4\end{array}\right|$.
Applying column operations $C_{2} \rightarrow C_{2} - C_{1}$ and $C_{3} \rightarrow C_{3} - C_{1}$:
$\Delta = (5x+4) \left|\begin{array}{ccc}1 & 0 & 0 \\ 2x & 4-x & 0 \\ 2x & 0 & 4-x\end{array}\right|$.
Taking $(4-x)$ common from $C_{2}$ and $C_{3}$:
$\Delta = (5x+4)(4-x)(4-x) \left|\begin{array}{ccc}1 & 0 & 0 \\ 2x & 1 & 0 \\ 2x & 0 & 1\end{array}\right|$.
Expanding along $R_{1}$:
$\Delta = (5x+4)(4-x)^{2} [1(1-0)] = (5x+4)(4-x)^{2}$.
Hence,the result is proved.
Applying the row operation $R_{1} \rightarrow R_{1} + R_{2} + R_{3}$,we get:
$\Delta = \left|\begin{array}{ccc}5x+4 & 5x+4 & 5x+4 \\ 2x & x+4 & 2x \\ 2x & 2x & x+4\end{array}\right|$.
Taking $(5x+4)$ common from $R_{1}$:
$\Delta = (5x+4) \left|\begin{array}{ccc}1 & 1 & 1 \\ 2x & x+4 & 2x \\ 2x & 2x & x+4\end{array}\right|$.
Applying column operations $C_{2} \rightarrow C_{2} - C_{1}$ and $C_{3} \rightarrow C_{3} - C_{1}$:
$\Delta = (5x+4) \left|\begin{array}{ccc}1 & 0 & 0 \\ 2x & 4-x & 0 \\ 2x & 0 & 4-x\end{array}\right|$.
Taking $(4-x)$ common from $C_{2}$ and $C_{3}$:
$\Delta = (5x+4)(4-x)(4-x) \left|\begin{array}{ccc}1 & 0 & 0 \\ 2x & 1 & 0 \\ 2x & 0 & 1\end{array}\right|$.
Expanding along $R_{1}$:
$\Delta = (5x+4)(4-x)^{2} [1(1-0)] = (5x+4)(4-x)^{2}$.
Hence,the result is proved.
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