Linear Equations

Step 1: For a non-trivial solution, the determinant of coefficients must be zero.

Step 2: Expand along : . . .
Step 3: Factoring gives . This means either or . Checking options: (D) states .

Step 1: Find the determinant of the coefficient matrix. The coefficient matrix is: Expanding the determinant: Thus,
Step 2: Solve for using . For , we substitute the first column of the matrix with the constants. The matrix is: Expanding the determinant:
Step 3: Calculate . Substitute and :

The determinant is given by:


Simplifying the determinant expression:

Solving for :

Therefore, the possible values for are:

Now, calculating the sum of angles:

Step 1: Solve the given condition.
- Substitute : This represents a complex number equation where is expressed as .
Step 2: Solve for and .
- Expand the magnitudes: Now, equate the real and imaginary parts of both sides to solve for and : - After solving, we find: Step 3: Find the quadratic equation.
- The roots are and . Thus, the quadratic equation is: Final Answer: The quadratic equation is .

Step 1: Representing the system of equations as an augmented matrix.

The system can be written as:

Step 2: Determining the determinant of the coefficient matrix.

We denote the determinant of the coefficient matrix as . Using the determinant formula for a 3x3 matrix:

Expanding along the first row:

Simplifying each determinant:

Step 3: Solving for .

To find the values of that make the determinant zero, solve:

Using the quadratic formula:

Thus, the two roots are and .

Step 4: Analyzing the conditions for solutions.

If , substitute into the matrix:

So, the system has a unique solution for , and the second equation must hold for to ensure consistency.

Step 5: Conclusion.

From the above analysis, the correct answer is option (2), as it claims that there is no solution for and for all , which is incorrect. The system has a solution for , but only when .

We are given the equation:

Solving for :

Now, substitute into the equation for :

Simplifying:

Now, differentiate with respect to :

Simplifying:

For to be strictly increasing, we need , which implies:

Solving the inequality , we find that the critical points are:

Thus, is strictly increasing in the intervals:

To solve this problem, we need to analyze the given system of equations and determine when it has a unique solution, infinite solutions, or is inconsistent. The system of equations is:

We'll examine each condition provided in the options by understanding the different scenarios arising from and .

Step 1: Analyze the system.

The system can be represented in matrix form as , where:

Step 2: Check for infinite solutions.

Infinite solutions occur when the system has dependent equations, meaning the rank of matrix is less than the number of variables (here, 3). An easy approach is to check if any of the equations are multiples or linear combinations of each other.

Let's substitute and into the system:

The system simplifies to a form where equations become linearly dependent. They do not provide unique values for each variable, leading to infinite solutions.

Step 3: Check for inconsistency.

An inconsistent system does not have any solutions. This can happen if, after row reduction, an equation like (where ) appears.

For and , the system would reduce to an inconsistency since modifying the constant term results in a contradiction after solving.

Step 4: Check for consistency when .

For , the coefficient matrix does not lead to linear dependency, as . For , the matrix is potentially consistent depending on .

Step 5: Check for uniqueness when and .

If neither value causes the system to degenerate or contradict, altering both and implies differentiability and a unique solution set, as all equations can be differentiated to provide specific values for and .

Conclusion:

The correct answer is: System has a unique solution if and . This statement is NOT correct based on our analysis.

For the system to have infinitely many solutions, the determinant of the coefficient matrix must be zero, as this will indicate linear dependence. The coefficient matrix is:

We compute the determinant of the matrix and solve the equation for the values of and that make the determinant equal to zero. This yields the values for and .

Finally, using these values, we calculate .