Matrices & Determinants
Harmonic Progression
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>
(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 system of equations can be written as a matrix equation AX=B. The harmonic progression terms imply p=1/a_10, q=1/a_100, r=1/a_1000. The third equation qrx + pry + pqz = 0 can be divided by pqr to get x/p + y/q + z/r = 0. Analyze the determinant of the coefficient matrix and use Cramer's rule or consistency conditions.
The system is x+y+z=1, 10x+100y+1000z=0, qrx+pry+pqz=0. Dividing the third equation by pqr gives x/p + y/q + z/r = 0. Since p, q, r are in HP, 1/p, 1/q, 1/r are in AP. Let 1/p = a+9d, 1/q = a+99d, 1/r = a+999d. The determinant of the coefficient matrix is D = |1 1 1; 10 100 1000; 1/p 1/q 1/r|. Since 1/p, 1/q, 1/r are in AP, the third row is a linear combination of the first two rows, so D=0. The system is consistent if the augmented matrix has the same rank. Solving leads to the matching (I)\to (T), (II)\to (R), (III)\to (S), (IV)\to (T).
Correct Answer: A