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For what value of λ, the system of equations 3x - y + z = 0, 15x - 6y + 5z = 0, x - 2y + 2z = 0 has a non-zero solution?
- a)1
- b)- 2
- c)- 3
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
For what value of λ, the system of equations 3x - y + z = 0, 15...
Analysis:
To find the value of λ for which the given system of equations has a non-zero solution, we can rewrite the system in matrix form and then find the determinant of the coefficient matrix.
Matrix Form:
The given system of equations can be represented in matrix form as:
\[ \begin{bmatrix} 3 & -1 & 1 \\ 15 & -6 & 5 \\ 1 & -2 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]
Finding Determinant:
To have a non-zero solution, the determinant of the coefficient matrix must be zero. Therefore, we calculate the determinant of the coefficient matrix as follows:
\[ \text{det} \begin{bmatrix} 3 & -1 & 1 \\ 15 & -6 & 5 \\ 1 & -2 & 2 \end{bmatrix} = 3(-6 \times 2 - 5 \times (-2)) - (-1)(15 \times 2 - 5 \times 1) + 1(15 \times (-2) - (-6) \times 1) \]
\[ = 3(-12 + 10) - (-15 + 5) + (30 + 6) = 3(-2) - (-10) + 36 = -6 + 10 + 36 = 40 \]
The determinant of the coefficient matrix is 40, which is non-zero. Hence, the system of equations has a non-zero solution for all values of λ.
Conclusion:
Therefore, the correct answer is option 'D' - None of these.
To find the value of λ for which the given system of equations has a non-zero solution, we can rewrite the system in matrix form and then find the determinant of the coefficient matrix.
Matrix Form:
The given system of equations can be represented in matrix form as:
\[ \begin{bmatrix} 3 & -1 & 1 \\ 15 & -6 & 5 \\ 1 & -2 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]
Finding Determinant:
To have a non-zero solution, the determinant of the coefficient matrix must be zero. Therefore, we calculate the determinant of the coefficient matrix as follows:
\[ \text{det} \begin{bmatrix} 3 & -1 & 1 \\ 15 & -6 & 5 \\ 1 & -2 & 2 \end{bmatrix} = 3(-6 \times 2 - 5 \times (-2)) - (-1)(15 \times 2 - 5 \times 1) + 1(15 \times (-2) - (-6) \times 1) \]
\[ = 3(-12 + 10) - (-15 + 5) + (30 + 6) = 3(-2) - (-10) + 36 = -6 + 10 + 36 = 40 \]
The determinant of the coefficient matrix is 40, which is non-zero. Hence, the system of equations has a non-zero solution for all values of λ.
Conclusion:
Therefore, the correct answer is option 'D' - None of these.
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