Question

The value of the determinant \(\left| {\begin{array}{*{20}{c}} {\sqrt {13} + \sqrt 3 }&{2\sqrt 5 }&{\sqrt 5 }\\ {\sqrt {15} + \sqrt {26} }&5&{\sqrt {10} }\\ {3 + \sqrt {65} }&{\sqrt {15} }&5 \end{array}} \right|\) is
1. 5
2. \(5\sqrt 3 \left( {\sqrt 6 - 5} \right)\)
3. \(5\sqrt 3\)
4. \(5\sqrt 3 \left( {\sqrt 6 + 5} \right)\)

Answer 1

Correct Answer - Option 2 : \(5\sqrt 3 \left( {\sqrt 6 - 5} \right)\)

Concept:

Any square matrix says A, is said to be invertible if the determinant of A i.e |A| ≠ 0 else it is said to be a singular matrix.

If \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right]\) then determinant of A is given by: |A| = (a­11 × a22) – (a12 – a21).

If \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right]\) then determinant of A is given by:

|A| = a11 × {(a22 × a33) – (a23 × a32)} - a12 × {(a21 × a33) – (a23 × a31)} + a13 × {(a21 × a32) – (a22 × a31)}

Calculation:

Determinant is given by:

\(\left| {\begin{array}{*{20}{c}} {\sqrt {13} + \sqrt 3 }&{2\sqrt 5 }&{\sqrt 5 }\\ {\sqrt {15} + \sqrt {26} }&5&{\sqrt {10} }\\ {3 + \sqrt {65} }&{\sqrt {15} }&5 \end{array}} \right|\)

\(\Rightarrow 25\sqrt {13} + 25\sqrt 3 - 5\sqrt {78} - 15\sqrt 2 - 50\sqrt 3 + 30\sqrt 2 + 5\sqrt {78} - 25\sqrt {13} \)

\(\Rightarrow - 25\sqrt 3 + 15\sqrt 2\)

\( \Rightarrow 5\left[ { - 5\sqrt 3 + 3\sqrt 2 } \right]\)

\(\Rightarrow 5\sqrt 3 \left[ { - 5 + \sqrt 3 \sqrt 2 } \right]\)

\(5\sqrt 3 \left[ {\sqrt 6 - 5} \right]\)