Understanding Determinant Value Calculation

The problem asks us to find the value of a given 3x3 determinant involving square roots. Calculating the determinant value directly by expanding might be complicated due to the radical terms. We should look for ways to simplify the determinant using properties of determinants, such as row or column operations.

The given determinant is:

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

Simplifying the Determinant using Column Operations

Let's examine the columns. Notice that the second and third columns have \(\sqrt{5}\) as a common factor, either explicitly or implicitly (\(5 = \sqrt{5} \times \sqrt{5}\), \(\sqrt{10} = \sqrt{2} \times \sqrt{5}\), \(\sqrt{15} = \sqrt{3} \times \sqrt{5}\), \(\sqrt{65} = \sqrt{13} \times \sqrt{5}\)). We can factor out \(\sqrt{5}\) from the second column \(C_2\) and another \(\sqrt{5}\) from the third column \(C_3\). This brings out a factor of \((\sqrt{5}) \times (\sqrt{5}) = 5\) outside the determinant.

Factoring out 5:

\( D = 5 \left| {\begin{array}{*{20}{c}} {\sqrt {13} + \sqrt 3 }&{2}&{1}\\ {\sqrt {15} + \sqrt {26} }&{\sqrt 5 }&{\sqrt {2} }\\ {3 + \sqrt {65} }&{\sqrt {3} }&{\sqrt 5 } \end{array}} \right| \)

Let the new matrix inside the determinant be \(A'\). So \(D = 5 \det(A')\). We need to calculate the determinant value of \(A'\):

Now, let's look for relationships between the columns of \(A'\). Consider the first column \(C_1\). Can we simplify it using \(C_2\) and \(C_3\)? Let's try the column operation \(C_1 \to C_1 - \sqrt{3} C_2 - \sqrt{13} C_3\).

Let's apply this operation to the first column elements:

• Row 1: \((\sqrt{13} + \sqrt 3) - \sqrt{3} \times (2) - \sqrt{13} \times (1) = \sqrt{13} + \sqrt 3 - 2\sqrt{3} - \sqrt{13} = -\sqrt{3}\)
• Row 2: \((\sqrt{15} + \sqrt{26}) - \sqrt{3} \times (\sqrt 5) - \sqrt{13} \times (\sqrt 2) = (\sqrt{3}\sqrt{5} + \sqrt{2}\sqrt{13}) - \sqrt{15} - \sqrt{26} = \sqrt{15} + \sqrt{26} - \sqrt{15} - \sqrt{26} = 0\)
• Row 3: \((3 + \sqrt{65}) - \sqrt{3} \times (\sqrt 3) - \sqrt{13} \times (\sqrt 5) = (3 + \sqrt{5}\sqrt{13}) - 3 - \sqrt{65} = 3 + \sqrt{65} - 3 - \sqrt{65} = 0\)

Applying this column operation \(C_1 \to C_1 - \sqrt{3} C_2 - \sqrt{13} C_3\) transforms the first column into \(\begin{pmatrix} -\sqrt{3} \\ 0 \\ 0 \end{pmatrix}\). This operation does not change the determinant value of the matrix \(A'\).

The determinant of \(A'\) becomes:

\( \det(A') = \left| {\begin{array}{*{20}{c}} {-\sqrt{3} }&{2}&{1}\\ {0}&{\sqrt 5 }&{\sqrt {2} }\\ {0}&{\sqrt {3} }&{\sqrt 5 } \end{array}} \right| \)

Calculating the Determinant Value

Now we can calculate this determinant by expanding along the first column, as it has two zeros:

\( \det(A') = (-\sqrt{3}) \times \left| {\begin{array}{*{20}{c}} {\sqrt 5 }&{\sqrt {2} }\\ {\sqrt {3} }&{\sqrt 5 } \end{array}} \right| - (0) \times \left| {\begin{array}{*{20}{c}} {2}&{1}\\ {\sqrt {3} }&{\sqrt 5 } \end{array}} \right| + (0) \times \left| {\begin{array}{*{20}{c}} {2}&{1}\\ {\sqrt 5 }&{\sqrt {2} } \end{array}} \right| \)

Calculating the 2x2 determinant:

\( \left| {\begin{array}{*{20}{c}} {\sqrt 5 }&{\sqrt {2} }\\ {\sqrt {3} }&{\sqrt 5 } \end{array}} \right| = (\sqrt 5 \times \sqrt 5) - (\sqrt 2 \times \sqrt 3) = 5 - \sqrt 6 \)

Substituting this back into the determinant of \(A'\):

\( \det(A') = (-\sqrt{3}) \times (5 - \sqrt 6) = -5\sqrt{3} + \sqrt{18} = -5\sqrt{3} + \sqrt{9 \times 2} = -5\sqrt{3} + 3\sqrt{2} \)

The value of the original determinant \(D\) is \(5 \times \det(A')\):

\( D = 5 \times (-5\sqrt{3} + 3\sqrt{2}) = -25\sqrt{3} + 15\sqrt{2} \)

Matching the Result with Options

Let's compare this result with the given options. The desired format in the options is \(5\sqrt{3} (\dots)\). Let's factor out \(5\sqrt{3}\) from our result \(-25\sqrt{3} + 15\sqrt{2}\).

\( -25\sqrt{3} + 15\sqrt{2} = 5\sqrt{3} \times (-5) + 5\sqrt{3} \times \left(\frac{15\sqrt{2}}{5\sqrt{3}}\right) \)

Let's simplify the second term inside the bracket:

\( \frac{15\sqrt{2}}{5\sqrt{3}} = \frac{3\sqrt{2}}{\sqrt{3}} = \frac{3\sqrt{2} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{3\sqrt{6}}{3} = \sqrt{6} \)

So, the expression becomes:

\( -25\sqrt{3} + 15\sqrt{2} = 5\sqrt{3} \times (-5) + 5\sqrt{3} \times (\sqrt{6}) = 5\sqrt{3} (-5 + \sqrt{6}) = 5\sqrt{3} (\sqrt{6} - 5) \)

This calculation confirms the Determinant Value Calculation matches one of the options.

The final determinant value is \(5\sqrt{3}(\sqrt{6} - 5)\). This detailed step-by-step process using column operations helps in finding the Determinant Value Calculation for this complex 3x3 determinant.

Understanding matrix properties and column operations is crucial for solving such problems in mathematics and algebra. This example illustrates how strategic operations can simplify a complex Determinant Value Calculation problem involving square root terms, making it easier to solve during exam preparation.