Question

Find the value of the determinant:
 \[\det \left( {\begin{array}{*{20}{l}}
  {\sqrt {13} + \sqrt 3 }&{2\sqrt 5 }&{\sqrt 5 } \\
  {\sqrt {15} + \sqrt {26} }&5&{\sqrt {10} } \\
  {3 + \sqrt {65} }&{\sqrt {15} }&5
\end{array}} \right)\]
A) \[15\sqrt 2 - 25\sqrt 3 \]
B) \[15\sqrt 5 - 25\sqrt 3 \]
C) \[25\sqrt 2 + 15\sqrt 3 \]
D) 0

Answer 1

Hint:
We will find the value of determinant of the given matrix.

Complete step by step solution:
Let \[A = \left( {\begin{array}{*{20}{l}}
  {\sqrt {13} + \sqrt 3 }&{2\sqrt 5 }&{\sqrt 5 } \\
  {\sqrt {15} + \sqrt {26} }&5&{\sqrt {10} } \\
  {3 + \sqrt {65} }&{\sqrt {15} }&5
\end{array}} \right)\]
 \[\left| A \right| = (\sqrt {13} + \sqrt 3 )\left( {\begin{array}{*{20}{l}}
  5&{\sqrt {10} } \\
  {\sqrt {15} }&5
\end{array}} \right) - 2\sqrt 5 \left( {\begin{array}{*{20}{l}}
  {\sqrt {15} + \sqrt {26} }&{\sqrt {10} } \\
  {3 + \sqrt {65} }&5
\end{array}} \right) + \sqrt 5 \left( {\begin{array}{*{20}{l}}
  {\sqrt {15} + \sqrt {26} }&5 \\
  {3 + \sqrt {65} }&{\sqrt {15} }
\end{array}} \right)\]
 \[
  \det A = \left( {\sqrt {13} + \sqrt 3 } \right)\left( {5 \times 5 - \sqrt {10} \sqrt {15} } \right) - 2\sqrt 5 \left( {5 \times \left( {\sqrt {15} + \sqrt {26} } \right) - \sqrt {10} \left( {3 + \sqrt {65} } \right)} \right) + \sqrt 5 \left( {\sqrt {15} \left( {\sqrt {15} + \sqrt {26)} } \right) - 5\left( {3 + \sqrt {65} } \right)} \right) \\
   + \sqrt 5 (\sqrt {15} (\sqrt {15} + \sqrt {26)} - 5(3 + \sqrt {65} )) \\
 \]
On simplification, we get
 \[
   = 25\sqrt {13} + 25\sqrt 3 - \sqrt {13} \sqrt {15} \sqrt {10} - \sqrt 3 \sqrt {15} \sqrt {10} - 10\sqrt 5 \sqrt {15} - 10\sqrt 5 \sqrt {26} + 6\sqrt 5 \sqrt {10} + 2\sqrt 5 \sqrt {65} \sqrt {10} \\
   + \sqrt 5 \sqrt {15} \sqrt {15} + \sqrt 5 \sqrt {15} \sqrt {26} - 15\sqrt 5 - 5\sqrt 5 \sqrt {65} \\
 \]
On expanding the factors under root, we get
 \[
   = 25\sqrt {13} + 25\sqrt 3 - \sqrt {13} \sqrt {5 \times 3} \sqrt {5 \times 2} - \sqrt 3 \sqrt {5 \times 3} \sqrt {5 \times 2} - 10\sqrt 5 \sqrt {5 \times 3} - 10\sqrt 5 \sqrt {13 \times 2} + 6\sqrt 5 \sqrt {5 \times 2} \\
   + 2\sqrt 5 \sqrt {13 \times 5} \sqrt {5 \times 2} + 15\sqrt 5 + 5\sqrt 2 \sqrt 3 \sqrt {13} - 15\sqrt 5 - 5\sqrt 5 \sqrt {13 \times 5} \\
 \]
On simplification we get
 \[
   = 25\sqrt {13} + 25\sqrt 3 - 5\sqrt {13} \sqrt 2 \sqrt 3 - 15\sqrt 2 - 50\sqrt 3 - 10\sqrt 5 \sqrt 2 \sqrt {13} + 30\sqrt 2 + 10\sqrt {13} \sqrt 5 \sqrt 2 + 15\sqrt 5 \\
   + 5\sqrt 2 \sqrt 3 \sqrt {13} - 15\sqrt 5 - 25\sqrt {13} \\
 \]
On adding like terms, we get
 \[ = 15\sqrt 2 - 25\sqrt 3 \]

Therefore, \[\left| A \right| = 15\sqrt 2 - 25\sqrt 3 \]
Hence, option (A) is the correct answer.

Note:
For calculating the determinant of any matrix, the matrix should be a square matrix. Determinant of any matrix A can be represented as \[\left| A \right|\]. Determinant of the identity matrix is always 1.