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

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.