Linear Systems And Determinants

To determine the conditions under which the given system of equations has infinitely many solutions, it must be consistent and dependent. Let's consider the equations provided:

Equation 1:

Equation 2:

Equation 3:

For the system to have infinitely many solutions, the third equation must be a linear combination of the first two equations. Therefore, let's express Equation 3 in terms of Equations 1 and 2:

If and are scalars, we need:

Let's equate the coefficients:

For : (Equation i)

For : (Equation ii)

For : (Equation iii)

Additionally, for the constants: (Equation iv)

Let’s solve these equations to find and .

1. From Equation i: →
2. Substitute in Equation ii:
3. Simplifying gives:
4. Solving:
6. Substitute into :

Now calculating and :

From Equation iii:

From Equation iv:

Hence, the parametric form for infinitely many solutions is: and .

Investigating integer solutions satisfying , we proceed by trial-solution system:

Given: , we find:

(Substitute back calculated solutions to verify boundaries):

Only viable integers adhering to this constraint give: Solutions = .

Given the system of equations:

For the system to have infinite solutions, the determinant of the coefficient matrix should be zero:

First, solve for the values of and by working with the following determinants:

Expanding the determinant:

(Equation 1)

Now, for the second determinant:

Expanding the determinant:

(Equation 2)

Now solve the system of equations (1) and (2):

From Equation (1):

From Equation (2):

By solving this system, we find:

Finally, verify the solution:

The given system of equations is: To check for infinitely many solutions, we use the determinant of the coefficient matrix. The coefficient matrix is: Expanding the determinant: After solving the system, we find: Solving for and , we find and . Hence,

Step 1: Understanding the Concept:
For a system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix ( ) must be zero, and all other determinants ( ) must also be zero. Alternatively, one equation must be a linear combination of the others.

Step 2: Key Formula or Approach:
Observe that Equation 1 + Equation 2 gives:
Step 3: Detailed Explanation:
1. The third equation is: 2. For infinitely many solutions, this equation must match the sum of the first two equations (or be a multiple of it).
3. Comparing: with 4. Therefore:
Step 4: Final Answer:

Concept:
A homogeneous system of linear equations has a non-trivial solution if the determinant of the coefficient matrix is zero. This condition ensures that the equations are linearly dependent.

Step 1: Form the coefficient matrix.
Step 2: Expand the determinant.
Now compute minors:
Substitute:
Step 3: Simplify the equation.
Divide by :
Step 4: Solve the trigonometric equation.
Using identities:
Substituting and simplifying gives:
Thus,
Step 5: Find values of in .
Sum:

Step 1: Understanding the Question:
We have a system of homogeneous linear equations. A homogeneous system always has the trivial solution .
The condition that the system has infinitely many solutions means it must have non-trivial solutions.
This occurs if and only if the determinant of the coefficient matrix is zero. This condition must hold for all .

Step 2: Key Formula or Approach:
For the system of equations to have a non-trivial solution, we must have .
We will set up the coefficient matrix, calculate its determinant, set it to zero for all , and solve for the function . Then we will analyze the properties of .

Step 3: Detailed Explanation:
The given system of equations is:
1.
2.
3.

The coefficient matrix is:
For the system to have infinitely many solutions, the determinant of must be zero for all :
Now calculate the determinant:
Solving for :
Now analyze the function:
Since , the function is strictly increasing on .

Other observations:
- It is not constant
- It is not decreasing
- It has no critical points

Step 4: Final Answer:
The function is strictly increasing on .