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The value of \(\left| {\begin{array}{*{20}{c}} 1&1&1\\ 1&{1 + {\rm{x}}}&1\\ 1&1&{1 + {\rm{y}}} \end{array}} \right|\) is

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  1. x + y
  2. x – y
  3. xy
  4. 1 + x + y

Answer (Detailed Solution Below)

Option 3 : xy
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Detailed Solution

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Concept:

Elementary row or column transformations do not change the value of the determinant of a matrix.

Calculation:

\(\left| {\begin{array}{*{20}{c}} 1&1&1\\ 1&{1 + {\rm{x}}}&1\\ 1&1&{1 + {\rm{y}}} \end{array}} \right|\)

Applying R2 → R2 – R1, R3 → R3 – R1, we get

\(= \left| {\begin{array}{*{20}{c}} 1&1&1\\ 0&{\rm{x}}&0\\ 0&0&{\rm{y}} \end{array}} \right|\)

Now, Expanding along C1

= 1 (xy – 0) – 0 + 0 = xy

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