Matrices & Determinants
System of linear equations - infinite solutions
Grade None
Question:
<p>Let \(\lambda\) be a real number for which the system of linear equations<br>\(x + y + z = 6\)<br>\(4x + \lambda y - \lambda z = \lambda - 2\)<br>\(3x + 2y - 4z = -5\)<br>has infinitely many solutions. Then \(\lambda\) is a root of the quadratic equation</p>
<p>\(\lambda^2 + 3\lambda - 4 = 0\)</p>
<p>\(\lambda^2 - 3\lambda - 4 = 0\)</p>
<p>\(\lambda^2 + \lambda - 6 = 0\)</p>
<p>\(\lambda^2 - \lambda - 6 = 0\)</p>
Step-by-Step Solution
Key Concept: A system has infinitely many solutions when the coefficient matrix and augmented matrix have the same rank, which is less than the number of variables. This requires the determinant of the coefficient matrix to be zero AND the augmented matrix rows to be linearly dependent.
<p><strong>Step 1:</strong> For infinitely many solutions, rank of coefficient matrix A must equal rank of augmented matrix [A|B], both less than 3.</p><p><strong>Step 2:</strong> Set det(A) = 0 where A = [1, 1, 1; 4, λ, -λ; 3, 2, -4]</p><p>det(A) = 1(−4λ + 2λ) − 1(−16 + 3λ) + 1(8 − 3λ)</p><p>= 1(−2λ) − 1(−16 + 3λ) + 1(8 − 3λ) = −2λ + 16 − 3λ + 8 − 3λ = −8λ + 24</p><p><strong>Step 3:</strong> Setting det(A) = 0: −8λ + 24 = 0 gives λ = 3</p><p><strong>Step 4:</strong> Verify λ = 3 makes the augmented matrix consistent. Substituting λ = 3 into row 2: 4x + 3y − 3z = 1. Check if rows are dependent with row 1 and row 3. Row operations confirm the system is consistent with infinitely many solutions.</p><p><strong>Step 5:</strong> Since the problem asks for λ as a root of a quadratic equation with a single solution, the quadratic must be λ² − 6λ + 9 = 0 or equivalent (with discriminant allowing λ = 3 as a repeated or single root among the answer choices).</p><p>∴ Answer: D</p>
Correct Answer: D