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If $\left|\begin{array}{ccc}y+z & z & y \\ z & z+x & x \\ y & x & x+y\end{array}\right|=4 x y z$
Solution:
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Verified Answer
$$
\begin{aligned}
&\text { LHS }=\left|\begin{array}{ccc}
y+z & z & y \\
z & z+x & x \\
y & x & x+y
\end{array}\right| \\
&=\left|\begin{array}{ccc}
y+z+z+y & z & y \\
z+z+x+x & z+x & x \\
y+x+x+y & x & x+y
\end{array}\right| \quad\left[\because C_1 \rightarrow C_1+C_2+C_3\right] \\
&=2\left|\begin{array}{ccc}
(y+z) & z & y \\
(z+x) & z+x & x \\
(x+y) & x & x+y
\end{array}\right|
\end{aligned}
$$
$$
\begin{aligned}
&=2\left|\begin{array}{ccc}
y & z & y \\
0 & z+x & x \\
y & x & x+y
\end{array}\right|\\
&=2\left|\begin{array}{ccc}
0 & z-x & -x \\
0 & z+x & x \\
y & x & x+y
\end{array}\right|\\
&\left[\because C_1 \rightarrow C_1-C_2\right]\\
&=2\left[y\left(x z-x^2+x z+x^2\right)\right]\\
&=4 x y z=\text { RHS }\\
&\left[\because \mathrm{R}_1 \rightarrow \mathrm{R}_1-\mathrm{R}_3\right]
\end{aligned}
$$
\begin{aligned}
&\text { LHS }=\left|\begin{array}{ccc}
y+z & z & y \\
z & z+x & x \\
y & x & x+y
\end{array}\right| \\
&=\left|\begin{array}{ccc}
y+z+z+y & z & y \\
z+z+x+x & z+x & x \\
y+x+x+y & x & x+y
\end{array}\right| \quad\left[\because C_1 \rightarrow C_1+C_2+C_3\right] \\
&=2\left|\begin{array}{ccc}
(y+z) & z & y \\
(z+x) & z+x & x \\
(x+y) & x & x+y
\end{array}\right|
\end{aligned}
$$
$$
\begin{aligned}
&=2\left|\begin{array}{ccc}
y & z & y \\
0 & z+x & x \\
y & x & x+y
\end{array}\right|\\
&=2\left|\begin{array}{ccc}
0 & z-x & -x \\
0 & z+x & x \\
y & x & x+y
\end{array}\right|\\
&\left[\because C_1 \rightarrow C_1-C_2\right]\\
&=2\left[y\left(x z-x^2+x z+x^2\right)\right]\\
&=4 x y z=\text { RHS }\\
&\left[\because \mathrm{R}_1 \rightarrow \mathrm{R}_1-\mathrm{R}_3\right]
\end{aligned}
$$
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