Matrices & Determinants
System of linear equations
Grade Class 12
Question:
<p>Let <i>p, q, r</i> be nonzero real numbers that are, respectively, the 10<sup>th</sup>, 100<sup>th</sup> and 1000<sup>th</sup> terms of a harmonic progression. Consider the system of linear equations</p><p><i>x + y + z = 1</i></p><p><i>10x + 100y + 1000z = 0</i></p><p><i>qrx + pry + pqz = 0</i></p><p>Match List-I with List-II.</p>
(A) (I) → (T); (II) → (R); (III) → (S); (IV) → (T)
(B) (I) → (Q); (II) → (S); (III) → (S); (IV) → (R)
(C) (I) → (Q); (II) → (R); (III) → (P); (IV) → (R)
(D) (I) → (T); (II) → (S); (III) → (P); (IV) → (T)
Step-by-Step Solution
Key Concept: The terms of a harmonic progression are reciprocals of an arithmetic progression. Let the AP be a + (n-1)d. Then p = 1/(a+9d), q = 1/(a+99d), r = 1/(a+999d). The system of equations can be analyzed using the determinant of the coefficient matrix and Cramer's rule.
<p>The system is <i>x + y + z = 1</i>, <i>10x + 100y + 1000z = 0</i>, <i>qrx + pry + pqz = 0</i>. Dividing the third equation by <i>pqr</i>, we get <i>x/p + y/q + z/r = 0</i>. Since <i>p, q, r</i> are in HP, <i>1/p, 1/q, 1/r</i> are in AP. Let <i>1/p = a + 9d</i>, <i>1/q = a + 99d</i>, <i>1/r = a + 999d</i>. The system can be solved by checking the determinant of the coefficient matrix and applying conditions for consistency.</p>
Correct Answer: B