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Question: Answered & Verified by Expert
By using properties of determinants, show that
(a) $\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|=(a+b+c)^3$
(b) $\left|\begin{array}{ccc}x+y+2 z & x & y \\ z & y+z+2 x & y \\ z & x & z+x+2 y\end{array}\right|=2(x+y+z)^3$
MathematicsDeterminants
Solution:
2796 Upvotes Verified Answer
(a) L.H.S. $=\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$
$=(a+b+c)\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$
$=(a+b+c)\left|\begin{array}{ccc}0 & 1 & 1 \\ b+c+a & b-c-a & 2 b \\ 0 & 2 c & c-a-b\end{array}\right|$
$C_1-C_2 \Rightarrow C_1$
$=(a+b+c)\left[-(b+c+a)\left|\begin{array}{cc}1 & 1 \\ 2 c & c-a-b\end{array}\right|\right]$
$=(a+b+c)^3=$ R.H.S.
(b) Let $\Delta=\left|\begin{array}{ccc}\mathrm{x}+\mathrm{y}+2 \mathrm{z} & \mathrm{x} & \mathrm{y} \\ \mathrm{z} & \mathrm{y}+\mathrm{z}+2 \mathrm{x} & \mathrm{y} \\ \mathrm{z} & \mathrm{x} & \mathrm{z}+\mathrm{x}+2 \mathrm{y}\end{array}\right|$
Applying $\mathrm{C}_1 \rightarrow \mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3$
$\Delta=\left|\begin{array}{ccc}2(x+y+2 z) & x & y \\ 2(x+y+z) & y+z+2 x & y \\ 2(x+y+z) & x & z+x+2 y\end{array}\right|$
$=2(x+y+z)\left|\begin{array}{ccc}
1 & x & y \\
1 & y+z+2 x & y \\
1 & x & z+x+2 y
\end{array}\right|$
Applying $\mathrm{R}_1 \rightarrow \mathrm{R}_1-\mathrm{R}_2, \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_3$
$\begin{aligned}
&\Delta=2(x+y+z)\left|\begin{array}{ccc}
0 & -(x+y+z) & y-z \\
0 & x+y+z & -(x+2 y) \\
1 & x & z+x+2 y
\end{array}\right| \\
&=2(x+y+z)^2[(x+2 y)-(y-z)] ; \\
&=2(x+y+z)^3
\end{aligned}$

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