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

Question

Question

Answer 1

Concept:

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

Calculation:

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

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

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

Now, Expanding along C₃

= 1 (0 – ab) – 0 + 0 = -ab