Problem 123
Question
The system of linear equations $$ \begin{aligned} &x+\lambda y-z=0 \\\&\lambda x-y-z=0 \\ &x+y-\lambda z=0 \end{aligned} $$ has a non-trivial solution for: (A) exactly three values of \(\lambda\). (B) infinitely many values of \(\lambda\). (C) exactly one value of \(\lambda\). (D) Exactly two values of \(\lambda\).
Step-by-Step Solution
Verified Answer
The system has a non-trivial solution for exactly one value of \(\lambda\), (C).
1Step 1: Formulate the System Matrix
Write the system of equations in matrix form as \(A\vec{x} = \vec{0}\), where \(A\) is the coefficient matrix and \(\vec{x} = [x, y, z]^T\). We have:\[A = \begin{bmatrix} 1 & \lambda & -1 \\lambda & -1 & -1 \1 & 1 & -\lambda \end{bmatrix}\]
2Step 2: Determine the Condition for Non-Trivial Solution
For the system to have a non-trivial solution, the determinant of matrix \(A\) must be zero. This leads us to:\[\text{det}(A) = \begin{vmatrix} 1 & \lambda & -1 \\lambda & -1 & -1 \1 & 1 & -\lambda \end{vmatrix} = 0\]
3Step 3: Calculate the Determinant
Compute the determinant of \(A\):\[\text{det}(A) = 1(\lambda - \lambda^2) + \lambda(\lambda - 1) - 1(\-\lambda - 1)\]Simplify the equation:\[\text{det}(A) = \lambda - \lambda^2 + \lambda^2 - \lambda + \lambda + 1\] Where:\[= 1 + \lambda\]
4Step 4: Find Values of \(\lambda\)
To find when the determinant equals zero:\[1 + \lambda = 0\]Solving for \(\lambda\), we get:\[\lambda = -1\]
5Step 5: Conclude the Number of Solutions
Since the only value for \(\lambda\) that satisfies the condition is \(-1\), the system has a non-trivial solution for exactly one value of \(\lambda\).
Key Concepts
System of Linear EquationsCoefficient MatrixNon-Trivial Solution
System of Linear Equations
A system of linear equations is a collection of one or more linear equations involving the same set of variables. These equations are solved simultaneously to find the values of the variables that make each equation true. In its simplest form, a system with two variables, such as \(x\) and \(y\), can be visualized as two lines on a plane.
When you solve a system of linear equations, you look for solutions that satisfy all the equations. This solution can be:
When you solve a system of linear equations, you look for solutions that satisfy all the equations. This solution can be:
- A single point where all lines or planes intersect (one unique solution).
- No solutions if the lines are parallel and never intersect.
- Infinitely many solutions if the lines are exactly the same, meaning they lie on top of each other.
Coefficient Matrix
The coefficient matrix is a crucial concept in solving systems of linear equations using matrices. It consists of the coefficients of the variables from the system. For example, in the system given:\[\begin{aligned}&x+\lambda y-z = 0 \&\lambda x-y-z = 0 \&x+y-\lambda z = 0\end{aligned}\]The coefficient matrix \(A\) is formed by arranging the coefficients of \(x\), \(y\), and \(z\) into a matrix:\[A = \begin{bmatrix} 1 & \lambda & -1 \\lambda & -1 & -1 \1 & 1 & -\lambda \end{bmatrix}\]
This representation helps in leveraging several matrix operations to solve the system. It allows us to use methods like Gaussian elimination or, in this case, find the determinant to determine the existence or uniqueness of solutions.
This representation helps in leveraging several matrix operations to solve the system. It allows us to use methods like Gaussian elimination or, in this case, find the determinant to determine the existence or uniqueness of solutions.
Non-Trivial Solution
In the context of linear equations, a non-trivial solution refers to any solution other than the zero solution (where all variables equal zero). A non-trivial solution indicates that there are meaningful values for the variables that satisfy the system.
For a homogeneous system of linear equations—where all constant terms are zero, like the system in consideration—finding a non-trivial solution means that there is more than just the "boring" solution where every variable equals zero. To determine if such solutions exist, we often look at the determinant of the coefficient matrix. If the determinant is zero, the system may have non-trivial solutions.
In our exercise, the determinant condition \(\text{det}(A) = 0\) led us to solve for \(\lambda\) and find that a non-trivial solution exists only for \(\lambda = -1\). This means, at this specific value of \(\lambda\), there exists a solution to the system other than \([0, 0, 0]^T\).
For a homogeneous system of linear equations—where all constant terms are zero, like the system in consideration—finding a non-trivial solution means that there is more than just the "boring" solution where every variable equals zero. To determine if such solutions exist, we often look at the determinant of the coefficient matrix. If the determinant is zero, the system may have non-trivial solutions.
In our exercise, the determinant condition \(\text{det}(A) = 0\) led us to solve for \(\lambda\) and find that a non-trivial solution exists only for \(\lambda = -1\). This means, at this specific value of \(\lambda\), there exists a solution to the system other than \([0, 0, 0]^T\).
Other exercises in this chapter
Problem 119
Let \(P\) and \(Q\) be 3 by 3 matrices with \(P \neq Q\). If \(P^{3}=\) \(Q^{3}\) and \(P^{2} Q=Q^{2} P\), then determinant of \(\left(P^{2}+Q^{2}\right)\) is e
View solution Problem 120
If \(P=\left(\begin{array}{lll}1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4\end{array}\right]\) is the adjoint of a \(3 \times 3\) matrix \(A\) and \(|A|=4\), then
View solution Problem 118
The number of values of \(\mathrm{k}\) for which the homogeneous system of linear equations \(4 x+k y+2 z=0 ; k x+4 y+z=0 ; 2 x+2 y+z=0\) possess a non- zero so
View solution