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Question #41738

Difficulty: 8/10 | 🟢 Free

The Question

Consider the system of linear equations AX=BAX = B, where A=(21−11−12α11),X=(xyz),B=(34β)A = \begin{pmatrix} 2 & 1 & -1 \\ 1 & -1 & 2 \\ \alpha & 1 & 1 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 3 \\ 4 \\ \beta \end{pmatrix} The equations represent three planes in R3\mathbb{R}^3. It is given that the line of intersection of the first two planes is strictly parallel to the third plane (i.e., it does not lie in the third plane). Let C=adj(A)C = \text{adj}(A). Which of the following is a possible value for the column vector CBCB?

Options

  • A.

    (2−10−6)\begin{pmatrix} 2 \\ -10 \\ -6 \end{pmatrix}

  • B.

    (000)\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}

  • C.

    (2106)\begin{pmatrix} 2 \\ 10 \\ 6 \end{pmatrix}

  • D.

    (−210−6)\begin{pmatrix} -2 \\ 10 \\ -6 \end{pmatrix}

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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

Line lying in a plane|Collinearity and Coplanarity|Adjoint-Determinant Relation

Why This Question Is Tricky

Common Mistakes

This question tests the geometric interpretation of systems of linear equations and the Adjoint-Determinant relation.

A common trap is assuming that ∣A∣=0|A|=0 is the only condition needed for parallel planes, or confusing the condition for a line lying in a plane ((adjA)B=0(\text{adj} A)B = 0, infinite solutions) with the line being strictly parallel ((adjA)B≠0(\text{adj} A)B \neq 0, no solution).

Students falling for the trap will calculate CBCB and incorrectly equate it to the zero vector, choosing the null vector option.

The problem forces them to recognize that strict parallelism requires inconsistency, hence a non-zero vector proportional to the adjoint's column space.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterThree-Dimensional Geometry - Class 12
ConceptLine lying in a plane
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree