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Question #40320

Difficulty: 8/10 | 🟢 Free

The Question

Let $P, Q,$ and $R$ be three finite sets such that $|P \cup Q \cup R| = 40$ , $|P| = 20$ , $|Q| = 22$ , $|R| = 25$ , $|P \cap Q| = 8$ , $|Q \cap R| = 7$ , and $|P \cap R| = 15$ . Let $B = \begin{bmatrix} |P \cap Q| \\ |Q \cap R| \\ |P \cap Q \cap R| \end{bmatrix}$ and $A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 1 \\ 2 & 3 & 1 \end{bmatrix}$ . If $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is the unique solution to the matrix equation $AX = B$ , then the value of $\det(A^3) + x + y + z$ is:

Options

A.
-8
B.
8
C.
-6
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

Cardinality Inclusion-Exclusion|System of Linear Equations (Cramer/Inverse)|Row/Column Exchange Parity

Why This Question Is Tricky

Common Mistakes

This question integrates set cardinality (Inclusion-Exclusion) with a system of linear equations, specifically targeting the 'Row Exchange Parity' trap.

The matrix A is deliberately designed with a zero in the top-left entry ( $a_{11} = 0$ ).

Students using Gaussian elimination to compute the determinant or apply Cramer's rule will naturally swap Row 1 and Row 2.

The classic trap is failing to multiply the determinant by -1 after this single row swap.

If the parity flip is ignored, the student incorrectly computes $\det(A) = 1$ .

This cascades into two possible errors: if using Cramer's rule, the entire solution vector X flips sign, yielding a sum of 7 and a final answer of 8.

If solving by substitution but still miscalculating the determinant, they arrive at -6.

Both distractors are explicitly provided in the options.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Sets - Class 11
Concept	Cardinality Inclusion-Exclusion
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free