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\(\left| {\begin{array}{*{20}{c}} {41}&1&5\\ {79}&7&9\\ {29}&5&3 \end{array}} \right|\) = 
1. 89
2. 10
3. 0
4. None of these
5. 65

1 Answer

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Best answer
Correct Answer - Option 3 : 0

Concept:

  • Elementary transformations do not change the value of the determinant of a matrix.
  • If any two rows (columns) of a matrix are the same then the value of the determinant is zero
  • If all the elements of one row (column) are multiplied by the same quantity say k, then the value of the new determinant is k times the value of the original determinant.

Calculation:

\(\left| {\begin{array}{*{20}{c}} {41}&1&5\\ {79}&7&9\\ {29}&5&3 \end{array}} \right|\)

Apply C1 ↔ C1 - 8C3 on the above determinant, we get

\(\Rightarrow \left| {\begin{array}{*{20}{c}} {41}&1&5\\ {79}&7&9\\ {29}&5&3 \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {1}&1&5\\ {7}&7&9\\ {5}&5&3 \end{array}} \right|\)

As we know that, if any two rows (columns) of a matrix are same then the value of the determinant is zero.

\(\Rightarrow \left| {\begin{array}{*{20}{c}} {41}&1&5\\ {79}&7&9\\ {29}&5&3 \end{array}} \right| = 0\)

So, the value of the determinant is zero.

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