Question

Without expanding, prove that:
$\left| {\begin{array}{*{20}{c}}
  {x + y}&{y + z}&{z + x} \\
  z&x&y \\
  1&1&1
\end{array}} \right| = 0$

Answer 1

Hint: Here, we will apply the properties of determinants to simplify the given determinant.

Complete step-by-step answer:
Given: $\left| {\begin{array}{*{20}{c}}
  {x + y}&{y + z}&{z + x} \\
  z&x&y \\
  1&1&1
\end{array}} \right| = 0$
L.H.S=
$\left| {\begin{array}{*{20}{c}}
  {x + y}&{y + z}&{z + x} \\
  z&x&y \\
  1&1&1
\end{array}} \right|$
Applying row operations;
i.e., ${R_1} \to {R_1} + {R_2}$
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {x + y + z}&{x + y + z}&{y + z + x} \\
  z&x&y \\
  1&1&1
\end{array}} \right|$
Take $(x + y + z)$ common from first row
\[ \Rightarrow (x + y + z)\left| {\begin{array}{*{20}{c}}
  1&1&1 \\
  z&x&y \\
  1&1&1
\end{array}} \right|\]
As you see, row 1 and row 3 are the same.
We know from properties of determinants that in the determinant if two rows are identical, then the value of the determinant is zero.
\[ \Rightarrow (x + y + z)\left| {\begin{array}{*{20}{c}}
  1&1&1 \\
  z&x&y \\
  1&1&1
\end{array}} \right| = 0 = R.H.S\]
Hence Proved.

Note: In these types of questions, application of determinant properties is always preferable rather than directly solving to ease the process of arriving at the solution.