Matrices & Determinants
System of linear equations — matrix method
Grade Class 12
Question:
<p>Let <em>a</em>, <em>λ</em>, <em>μ</em> ∈ R. Consider the system of linear equations</p><p><em>ax</em> + 2<em>y</em> = <em>λ</em></p><p>3<em>x</em> - 2<em>y</em> = <em>μ</em></p><p>Which of the following statement(s) is(are) correct?</p><p>(A) if <em>a</em> = -3, then the system has infinitely many solutions for all values of <em>λ</em> and <em>μ</em></p><p>(B) if <em>a</em> ≠ -3, then the system has a a unique solution for all values of <em>λ</em> and <em>μ</em></p><p>(C) if <em>λ</em> + <em>μ</em> = 0, then the system has infinitely many solutions for <em>a</em> = -3</p><p>(D) if <em>λ</em> + <em>μ</em> ≠ 0, then the system has no solution for <em>a</em> = -3</p>
(A) if <em>a</em> = -3, then the system has infinitely many solutions for all values of <em>λ</em> and <em>μ</em>
(B) if <em>a</em> ≠ -3, then the system has a a unique solution for all values of <em>λ</em> and <em>μ</em>
(C) if <em>λ</em> + <em>μ</em> = 0, then the system has infinitely many solutions for <em>a</em> = -3
(D) if <em>λ</em> + <em>μ</em> ≠ 0, then the system has no solution for <em>a</em> = -3
Step-by-Step Solution
Key Concept: The system of equations is ax + 2y = \lambda and 3x - 2y = \mu. The determinant of the coefficient matrix is D = |a 2; 3 -2| = -2a - 6 = -2(a + 3). If a \neq -3, D \neq 0, so the system has a unique solution. If a = -3, the equations are -3x + 2y = \lambda and 3x - 2y = \mu. Adding them gives 0 = \lambda + \mu. If \lambda + \mu = 0, there are infinitely many solutions. If \lambda + \mu \neq 0, there is no solution.
<p>The system is:</p><p><em>ax</em> + 2<em>y</em> = <em>\lambda</em></p><p>3<em>x</em> - 2<em>y</em> = <em>\mu</em></p><p>The determinant of the coefficient matrix is <em>D</em> = (-2<em>a</em>) - (6) = -2(<em>a</em> + 3).</p><p>If <em>a</em> \neq -3, <em>D</em> \neq 0, so the system has a unique solution for all <em>\lambda</em>, <em>\mu</em>. Thus (B) is correct.</p><p>If <em>a</em> = -3, the system becomes:</p><p>-3<em>x</em> + 2<em>y</em> = <em>\lambda</em></p><p>3<em>x</em> - 2<em>y</em> = <em>\mu</em></p><p>Adding the two equations gives 0 = <em>\lambda</em> + <em>\mu</em>.</p><p>If <em>\lambda</em> + <em>\mu</em> = 0, the equations are consistent and dependent, leading to infinitely many solutions. Thus (C) is correct.</p><p>If <em>\lambda</em> + <em>\mu</em> \neq 0, the system is inconsistent, leading to no solution. Thus (D) is correct.</p><p>(A) is incorrect because it requires <em>\lambda</em> + <em>\mu</em> = 0 for infinitely many solutions.</p>
Correct Answer: 2, 3, 4