Determinants

Step 1: Concept
Area of a triangle with vertices is .
Step 2: Analysis
Area
.
Step 3: Conclusion
Area .
Final Answer: (A)

Step 1: Concept
A system of linear equations and has a unique solution (the trivial solution) if .
Step 2: Analysis
Here .
Condition: .
Step 3: Conclusion
.
Final Answer: (A)

Step 1: Expand Determinant
. .
Step 2: Simplify Equation
.
Step 3: Factorization
Given one root is , is a factor. .
Step 4: Solve Quadratic
. The other roots are 3 and 7.
Final Answer: (a)

Step 1: Concept
Use the change of base formula: .
Step 2: Analysis
The determinant becomes .
Step 3: Evaluation
Factoring out from , from , and from , we get .
Step 4: Conclusion
Since the rows are identical, the determinant value is 0.
Final Answer: (a)

Step 1: Represent the system as a matrix equation.
The system of equations is:
This system can be written as:

Step 2: Use the determinant to find conditions for a unique solution.
For the system to have a unique solution, the determinant of the coefficient matrix must be non-zero: We calculate the determinant:

Step 3: Conclusion.
For a unique solution, the determinant must be non-zero: Thus, the existence of a unique solution depends on , which corresponds to option (A).

Step 1: Analyzing the system of equations.
We are given a homogeneous system of linear equations with three unknowns , , and . The system is represented as follows: For a homogeneous system of linear equations, there will always be one trivial solution: , , and . However, the system can also have non-trivial solutions depending on the relationship between the coefficients.

Step 2: The determinant condition.
The system of equations will have non-trivial solutions if and only if the determinant of the coefficient matrix is zero. This is because a non-zero determinant implies that the system has only the trivial solution. The coefficient matrix for the system is:

Step 3: Determinant analysis.
If the determinant of this matrix is zero, the system will have infinite solutions. The condition for an infinite number of solutions is when the rows of the coefficient matrix are linearly dependent. This happens when the rank of the matrix is less than 3. Thus, when the determinant of the matrix is zero, the system will have infinitely many solutions.

Step 4: Conclusion.
Given the condition for non-trivial solutions, the correct answer is that the system has infinite solutions.

Step 1: Formula / Definition}

Step 2: Calculation / Simplification}
Let . Then
The given product is


Step 3: Final Answer

Step 1: Formula / Definition}

Step 2: Calculation / Simplification}



Step 3: Final Answer

Step 1: Concept:
Factor out common powers from each row of the determinant and simplify the expression.

Step 2: Detailed Explanation:
Given,
Factor out common terms from each row:
Now observe the determinant carefully. It has a symmetric structure in terms of powers: Such determinants follow a pattern where summing over a sequence of leads to cancellation due to alternating symmetry.

Hence, when we evaluate: the terms cancel out pairwise due to symmetry in exponential powers.

Therefore,
Step 3: Final Answer:

Step 1: Understanding the Concept:
Check for linear dependence between rows.

Step 2: Detailed Explanation:
Note that . Similarly . So these are linear combinations of and . The same holds for the cosine terms. The third column: . The three rows are linear combinations of the same basis functions. It can be shown that the sum of the second and third rows is a multiple of the first row, making the determinant zero. Alternatively, add row 2 and row 3: Row2+Row3 = . So rows are linearly dependent. Hence determinant = 0.

Step 3: Final Answer:
0, which corresponds to option (C).

Concept: Use trigonometric identities and determinant simplification.

Step 1: Simplify expressions.
So matrix becomes:

Step 2: Column operations.
Apply:

Step 3: Factor column.
Take 2 common from :

Step 4: Observe columns.
Now and are identical:

Step 5: Conclusion.
Value is independent of , hence constant.

Concept: This represents an eigenvalue problem: Non-trivial solution exists when:

Step 1: Form matrix.


Step 2: Find determinant.


Step 3: Solve characteristic equation.
Expanding gives: Conclusion:

Concept: Cofactor of element.

Step 1: Cofactor of .
Sign:

Step 2: Cofactor of .
Sign:

Step 3: Sum.