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5. Let λ∈R\lambda \in \mathbb{R}, and consider the system of linear equations in...
May 11, 2024
Solution by Steps
step 1
To row reduce the augmented matrix, we perform elementary row operations to reach row echelon form
step 2
The given matrix is [1amp;0amp;−5amp;−41amp;−λamp;−2amp;21amp;2amp;λamp;2]\begin{bmatrix} 1 & 0 & -5 & -4 \\ 1 & -\lambda & -2 & 2 \\ 1 & 2 & \lambda & 2 \end{bmatrix}. We start by subtracting the first row from the second and third rows to eliminate the xx-terms
step 3
After the row operations, we obtain [1amp;0amp;−5amp;−40amp;1amp;3/2amp;30amp;0amp;0amp;0]\begin{bmatrix} 1 & 0 & -5 & -4 \\ 0 & 1 & 3/2 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix}, assuming λ=−2\lambda = -2. This form shows that for λ=−2\lambda = -2, the system has infinitely many solutions since the last row corresponds to the equation 0x+0y+0z=00x + 0y + 0z = 0
[question 1] Answer
For λ=−2\lambda = -2, the system has infinitely many solutions.
step 4
To find the values of λ\lambda for which the system has no solutions or exactly one solution, we examine the determinants of the submatrices
step 5
The determinant of the submatrix [1amp;−51amp;−2]\begin{bmatrix} 1 & -5 \\ 1 & -2 \end{bmatrix} is −6-6, which is non-zero, so it does not affect the number of solutions
step 6
The determinant of the submatrix [1amp;−51amp;λ]\begin{bmatrix} 1 & -5 \\ 1 & \lambda \end{bmatrix} is 1(−λ)−(−5)(1)=−λ+51(-\lambda) - (-5)(1) = -\lambda + 5. Setting this equal to zero gives λ=5\lambda = 5
step 7
The determinant of the submatrix [−λamp;−22amp;λ]\begin{bmatrix} -\lambda & -2 \\ 2 & \lambda \end{bmatrix} is −λ(λ)−(−2)(2)=−λ2+4-\lambda(\lambda) - (-2)(2) = -\lambda^2 + 4. Setting this equal to zero gives λ2=4\lambda^2 = 4, so λ=±2\lambda = \pm 2
step 8
The system has no solutions when the determinant of the entire matrix is zero, which occurs when λ=5\lambda = 5 or λ=±2\lambda = \pm 2. However, we already know that for λ=−2\lambda = -2, the system has infinitely many solutions. Therefore, the system has no solutions for λ=2\lambda = 2 and λ=5\lambda = 5
step 9
The system has exactly one solution when the determinant of the entire matrix is non-zero, which is for values of λ\lambda other than 22 and 55
[question 2] Answer
The system has no solutions for λ=2\lambda = 2 and λ=5\lambda = 5, exactly one solution for values of λ\lambda other than 22 and 55, and infinitely many solutions for λ=−2\lambda = -2.
Key Concept
Row reduction and determinant analysis
Explanation
Row reduction is used to simplify the system to row echelon form, revealing the number of solutions. Determinant analysis helps determine the values of λ\lambda for which the system has no solutions, exactly one solution, or infinitely many solutions.
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