Matrices & Determinants
System of linear equations
Grade Class 12
Question:
The set of all values of λ for which the system of linear equations :<br>2x<sub>1</sub> - 2x<sub>2</sub> + x<sub>3</sub> = λx<sub>1</sub>, 2x<sub>1</sub> - 3x<sub>2</sub> + 2x<sub>3</sub> = λx<sub>2</sub>, -x<sub>1</sub> + 2x<sub>2</sub> = λx<sub>3</sub><br>has a non-trivial solution
(1) contains two elements
(2) contains more than two elements
(3) is an empty set
(4) is a singleton
Step-by-Step Solution
Key Concept: For a homogeneous system of linear equations to have a non-trivial solution, the determinant of the coefficient matrix must be zero.
The given system can be rewritten as:<br>(2-\lambda)x<sub>1</sub> - 2x<sub>2</sub> + x<sub>3</sub> = 0<br>2x<sub>1</sub> - (3+\lambda)x<sub>2</sub> + 2x<sub>3</sub> = 0<br>-x<sub>1</sub> + 2x<sub>2</sub> - \lambda x<sub>3</sub> = 0<br>For a non-trivial solution, the determinant of the coefficient matrix must be zero:<br>| 2-\lambda -2 1 |<br>| 2 -(3+\lambda) 2 | = 0<br>| -1 2 -\lambda |<br>Expanding the determinant:<br>(2-\lambda)((3+\lambda)\lambda - 4) + 2(-2\lambda + 2) + 1(4 - (3+\lambda)) = 0<br>(2-\lambda)(\lambda^2 + 3\lambda - 4) + 2(-2\lambda + 2) + (1 - \lambda) = 0<br>(2-\lambda)(\lambda+4)(\lambda-1) - 4(\lambda-1) - (\lambda-1) = 0<br>(\lambda-1) [ (2-\lambda)(\lambda+4) - 4 - 1 ] = 0<br>(\lambda-1) [ -\lambda^2 - 2\lambda + 8 - 5 ] = 0<br>(\lambda-1) [ -\lambda^2 - 2\lambda + 3 ] = 0<br>-(\lambda-1)(\lambda^2+2\lambda-3) = 0<br>-(\lambda-1)(\lambda+3)(\lambda-1) = 0<br>-(\lambda-1)^2(\lambda+3) = 0<br>The values of \lambda are 1 and -3. Thus, the set contains two elements.
Correct Answer: 1