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The value of k for which the system of equations x + ky + 3z = 0, 4x + 3y + kz = 0, 2x + y + 2z = 0 has non-trivial solution is
- a)k = 0 or 9/2
- b)k = 10
- c)k < 9
- d)k > 0
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The value of k for which the system of equations x + ky + 3z = 0,4x + ...
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 x1, x2, …, xn. 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.
- Where, Ax, Ay, Az is the coefficient matrix of the given system of equations after replacing the first, second, and third columns from the constant term column which will be having all the entries as 0.
- In the case of homogeneous equations, the determinants of, Ax, Ay, Az will be 0 definitely.
- So, for the system of homogeneous equations having the the-trivial solution, the determinant of A should be zero.
- The system of homogeneous equations has a unique solution (trivial solution) if and only if the determinant of A is non-zero.
Calculation:
For non - trivial solution, the |A| = 0
For non - trivial solution, the |A| = 0
⇒ 1(6 - K) - K(8 - 2K) + 3(4 - 6) = 0
⇒ 9K - 2K2 = 0
⇒ k = 0 or 9/2
Most Upvoted Answer
The value of k for which the system of equations x + ky + 3z = 0,4x + ...
To find the value of k for which the system of equations has a non-trivial solution, we need to find when the determinant of the coefficient matrix is equal to zero.
The coefficient matrix of the system is:
| 1 k 3 |
| 4 3 k |
| 2 1 2 |
To find the determinant, we can use Laplace expansion along the first row:
det = 1 * det (3 k)
(1 2)
- k * det (4 k)
(2 2)
+ 3 * det (4 3)
(2 1)
Expanding each determinant:
det (3 k)
(1 2) = 3*2 - k*1 = 6 - k
det (4 k)
(2 2) = 4*2 - k*2 = 8 - 2k
det (4 3)
(2 1) = 4*1 - 3*2 = 4 - 6 = -2
Substituting these determinants back into the expression for the determinant:
det = 1 * (6 - k) - k * (8 - 2k) + 3 * (-2)
= 6 - k - 8k + 2k^2 - 6
= 2k^2 - 9k
To find when the determinant is equal to zero, we solve the equation:
2k^2 - 9k = 0
Factoring out k:
k(2k - 9) = 0
Setting each factor equal to zero:
k = 0 or 2k - 9 = 0
If k = 0, the system of equations becomes:
x 0y 3z = 0
4x 3y 0z = 0
2x y 2z = 0
This system has a non-trivial solution when x = 0, y = 0, and z ≠ 0.
If 2k - 9 = 0, then k = 9/2. The system of equations becomes:
x (9/2)y 3z = 0
4x 3y (9/2)z = 0
2x y 2z = 0
This system has a non-trivial solution when x = 0, y = 0, and z ≠ 0.
Therefore, the value of k for which the system of equations has a non-trivial solution is k = 0 or k = 9/2.
So the correct answer is a) k = 0 or 9/2.
The coefficient matrix of the system is:
| 1 k 3 |
| 4 3 k |
| 2 1 2 |
To find the determinant, we can use Laplace expansion along the first row:
det = 1 * det (3 k)
(1 2)
- k * det (4 k)
(2 2)
+ 3 * det (4 3)
(2 1)
Expanding each determinant:
det (3 k)
(1 2) = 3*2 - k*1 = 6 - k
det (4 k)
(2 2) = 4*2 - k*2 = 8 - 2k
det (4 3)
(2 1) = 4*1 - 3*2 = 4 - 6 = -2
Substituting these determinants back into the expression for the determinant:
det = 1 * (6 - k) - k * (8 - 2k) + 3 * (-2)
= 6 - k - 8k + 2k^2 - 6
= 2k^2 - 9k
To find when the determinant is equal to zero, we solve the equation:
2k^2 - 9k = 0
Factoring out k:
k(2k - 9) = 0
Setting each factor equal to zero:
k = 0 or 2k - 9 = 0
If k = 0, the system of equations becomes:
x 0y 3z = 0
4x 3y 0z = 0
2x y 2z = 0
This system has a non-trivial solution when x = 0, y = 0, and z ≠ 0.
If 2k - 9 = 0, then k = 9/2. The system of equations becomes:
x (9/2)y 3z = 0
4x 3y (9/2)z = 0
2x y 2z = 0
This system has a non-trivial solution when x = 0, y = 0, and z ≠ 0.
Therefore, the value of k for which the system of equations has a non-trivial solution is k = 0 or k = 9/2.
So the correct answer is a) k = 0 or 9/2.
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Community Answer
The value of k for which the system of equations x + ky + 3z = 0,4x + ...
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 x1, x2, …, xn. 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.
- Where, Ax, Ay, Az is the coefficient matrix of the given system of equations after replacing the first, second, and third columns from the constant term column which will be having all the entries as 0.
- In the case of homogeneous equations, the determinants of, Ax, Ay, Az will be 0 definitely.
- So, for the system of homogeneous equations having the the-trivial solution, the determinant of A should be zero.
- The system of homogeneous equations has a unique solution (trivial solution) if and only if the determinant of A is non-zero.
Calculation:
For non - trivial solution, the |A| = 0
For non - trivial solution, the |A| = 0
⇒ 1(6 - K) - K(8 - 2K) + 3(4 - 6) = 0
⇒ 9K - 2K2 = 0
⇒ k = 0 or 9/2
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The value of k for which the system of equations x + ky + 3z = 0,4x + 3y + kz = 0, 2x + y + 2z = 0 has non-trivial solution isa)k = 0 or 9/2b)k = 10c)k < 9d)k > 0Correct answer is option 'A'. Can you explain this answer?
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