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Question #41212

Difficulty: 8/10 | 🟢 Free

The Question

Let $t_n$ be a geometric progression of real numbers with first term $t_1 = 1$ and a non-zero common ratio $r$ . For each integer $k \ge 1$ , define the $3 \times 3$ matrix $M_k$ as: $M_k = \begin{pmatrix} t_k & t_{k+1} & t_{k+2} \\ -t_{k+1} & t_k & t_{k+1} \\ -t_{k+2} & -t_{k+1} & t_k \end{pmatrix}$ If $r$ satisfies the inequality $\det( - (M_1 M_2)^T ) \ge -16 t_4$ , find the sum of all possible integer values of $r$ in the interval $[-3, 3]$ .

Options

A.
-5
B.
6
C.
0
D.
-6

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Transpose of Product|Selection of GP Terms|Sign Flipping Rule

Why This Question Is Tricky

Common Mistakes

The problem tests the 'Sign Flipping Rule' in the context of matrix determinants and inequalities.

Students often forget that for an $n \times n$ matrix $A$ , $\det(-A) = (-1)^n \det(A)$ .

Here, $n=3$ (odd parity), so $\det(-(M_1 M_2)^T) = -\det(M_1 M_2)$ .

Missing this negative sign inverts the inequality from $-r^3(1+r^2)^4 \ge -16r^3$ to $r^3(1+r^2)^4 \ge -16r^3$ .

When dividing by $r^3$ , the sign of $r$ dictates whether the inequality flips.

The missing negative sign leads to the erroneous conclusion $r > 0$ , yielding a sum of 6.

Correctly applying the odd parity and analyzing cases for $r > 0$ and $r < 0$ yields the asymmetric valid domain $r \in (-\infty, -1] \cup (0, 1]$ , giving the correct sum of -5.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Matrices - Class 12
Concept	Transpose of Product
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free