Correct Answer - Option 1 : 0
Concept:
The determinant of Matrix \(M = {\left[ {\begin{array}{*{20}{c}} a\\ b\\ :\\ d \end{array}} \right]_{n \times 1}}{\left[ {p\;\;\;q\;\;\;..\;\;s} \right]_{1 \times n\;}}\) is 0, when two of either row or column will be the same.
Calculation:
\(A = \left[ {\begin{array}{*{20}{c}} 3\\ { - 2}\\ 4 \end{array}} \right]\left[ {2\;\;\;5\;\;\;-2} \right]\)
\(A = \;{\left[ {\begin{array}{*{20}{c}} 6&{15}&{-6}\\ { - 4}&{ - 10}&{ -4}\\ 8&{20}&{-8} \end{array}} \right]_{3 \times 3}}\)
\(\left| A \right| = \;\left| {\begin{array}{*{20}{c}} 6&{15}&{-6}\\ { - 4}&{ -10}&{ 4}\\ 8&{20}&{-8} \end{array}} \right|\)
\(\left| A \right| = \;3 \times - 2 \times 4\left| {\begin{array}{*{20}{c}} 2&5&- 2\\ 2&5&- 2\\ 2&5&- 2 \end{array}} \right|\)
Since 2 rows are identical determinant is 0
\(\left| A \right| = \;2 \times - 4 \times 7 \times 0 = 0\)
The determinant of A is equal to 0
