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By using properties of determinants, show that
(a) $\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|=(5 x+4)(4-x)^2$
(b) $\left|\begin{array}{ccc}y+x & y & y \\ y & y+k & y \\ y & y & y+k\end{array}\right|=k^2(3 y+k)$
(a) $\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|=(5 x+4)(4-x)^2$
(b) $\left|\begin{array}{ccc}y+x & y & y \\ y & y+k & y \\ y & y & y+k\end{array}\right|=k^2(3 y+k)$
Solution:
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Verified Answer
(a) L.H.S. $=\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|$
$\begin{aligned}
&=\left|\begin{array}{ccc}
5 x+4 & 2 x & 2 x \\
5 x+4 & x+4 & 2 x \\
5 x+4 & 2 x & x+4
\end{array}\right| ; \mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3 \Rightarrow \mathrm{C}_1 \\
&=(5 x+4)\left|\begin{array}{ccc}
1 & 2 x & 2 x \\
1 & x+4 & 2 x \\
1 & 2 x & x+4
\end{array}\right|
\end{aligned}$
$=(5 x+4)\left|\begin{array}{ccc}0 & x-4 & 0 \\ 1 & x+4 & 2 x \\ 1 & 2 x & x+4\end{array}\right| ; \mathrm{R}_1-\mathrm{R}_2 \Rightarrow \mathrm{R}_1$
$=(5 x+4)\left[-(x-4)\left|\begin{array}{cc}1 & 2 x \\ 1 & x+4\end{array}\right|\right]$
$=(5 x+4)(x-4)(x-4)=(5 x+4)(x-4)^2$
$=$ R.H.S.
(b) L.H.S. $=\left|\begin{array}{ccc}y+k & y & y \\ y & y+k & y \\ y & y & y+k\end{array}\right|$
$=\left|\begin{array}{ccc}3 y+k & y & y \\ 3 y+k & y+k & y \\ 3 y+k & y & y+k\end{array}\right| ; \mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3 \Rightarrow \mathrm{C}_1$
$=(3 y+k)\left[k\left|\begin{array}{cc}1 & y \\ 1 & y+k\end{array}\right|\right]$
$=(3 y+k) \cdot k[y+k-y]=(3 y+k) k^2$
$=$ R.H.S.
$\begin{aligned}
&=\left|\begin{array}{ccc}
5 x+4 & 2 x & 2 x \\
5 x+4 & x+4 & 2 x \\
5 x+4 & 2 x & x+4
\end{array}\right| ; \mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3 \Rightarrow \mathrm{C}_1 \\
&=(5 x+4)\left|\begin{array}{ccc}
1 & 2 x & 2 x \\
1 & x+4 & 2 x \\
1 & 2 x & x+4
\end{array}\right|
\end{aligned}$
$=(5 x+4)\left|\begin{array}{ccc}0 & x-4 & 0 \\ 1 & x+4 & 2 x \\ 1 & 2 x & x+4\end{array}\right| ; \mathrm{R}_1-\mathrm{R}_2 \Rightarrow \mathrm{R}_1$
$=(5 x+4)\left[-(x-4)\left|\begin{array}{cc}1 & 2 x \\ 1 & x+4\end{array}\right|\right]$
$=(5 x+4)(x-4)(x-4)=(5 x+4)(x-4)^2$
$=$ R.H.S.
(b) L.H.S. $=\left|\begin{array}{ccc}y+k & y & y \\ y & y+k & y \\ y & y & y+k\end{array}\right|$
$=\left|\begin{array}{ccc}3 y+k & y & y \\ 3 y+k & y+k & y \\ 3 y+k & y & y+k\end{array}\right| ; \mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3 \Rightarrow \mathrm{C}_1$
$=(3 y+k)\left[k\left|\begin{array}{cc}1 & y \\ 1 & y+k\end{array}\right|\right]$
$=(3 y+k) \cdot k[y+k-y]=(3 y+k) k^2$
$=$ R.H.S.
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