The value of k, for which the following system of linear equations has...
Concept:
Consider the system of m linear equations
a11 x1 + a12 x2 + … + a1n xn = 0
a21 x1 + a22 x2 + … + a2n xn = 0
…
am1 x1 + am2 x2 + … + amn xn = 0
The above equations containing the n unknowns x
1, x
2, …, x
n. To determine whether the above system of equations is consistent or not, we need to find the rank of the following matrix.
A is the coefficient matrix of the given system of equations.
- The system of homogeneous equations has a unique solution (trivial solution) if and only if the determinant of A is non-zero.
- The system of homogeneous equations has a Non - trivial solution if and only if the determinant of A is zero.
Calculation:
Given:
x + 2y – 3z = 0
2x + y + z = 0
x – y + kz = 0
For non-trivial solution the determinant should be zero
∴ 1(k + 1) – 2(2k - 1) – 3(-2 - 1) = 0
∴ k + 1 – 4k + 2 + 9 = 0
∴ 12 = 3k
∴ k = 4
The value of k, for which the following system of linear equations has...
To find the value of k for which the given system of linear equations has a non-trivial solution, we need to analyze the coefficients of the variables in each equation and determine the conditions for a non-trivial solution.
The given system of equations is:
1) x + 2y - 3z = 0
2) 2x + y + z = 0
3) x - y + kz = 0
In order for the system to have a non-trivial solution, there must be dependence between the equations, meaning that one equation can be obtained as a linear combination of the other two.
Let's analyze the coefficients of the variables in the system:
1) The coefficient of x in equation 1 is 1.
2) The coefficient of x in equation 2 is 2.
3) The coefficient of x in equation 3 is 1.
Since the coefficients of x in the three equations are different, there is no linear combination that can eliminate the variable x. Therefore, the value of k does not affect the existence of a non-trivial solution.
Next, let's analyze the coefficients of y in the system:
1) The coefficient of y in equation 1 is 2.
2) The coefficient of y in equation 2 is 1.
3) The coefficient of y in equation 3 is -1.
The coefficients of y in equations 1 and 2 are different, but we can eliminate the variable y by adding equation 1 and equation 3. This gives us:
(x + 2y - 3z) + (x - y + kz) = 0
2x + (y - y) + (k - 3)z = 0
2x + (k - 3)z = 0
Now, let's analyze the coefficients of z in the system:
1) The coefficient of z in equation 1 is -3.
2) The coefficient of z in equation 2 is 1.
3) The coefficient of z in equation 3 is k.
If we want to eliminate the variable z, we need the coefficients in equations 1 and 2 to be equal. Therefore, we have the condition:
-3 = 1
This condition is not satisfied, so we cannot eliminate the variable z.
Therefore, the value of k does not affect the existence of a non-trivial solution. Hence, the correct answer is option A) 4.