Class CBSE Class 12 Mathematics Matrices and Determinants Q #1684
COMPETENCY BASED
APPLY
1 Marks 2026 AISSCE(Board Exam) MCQ SINGLE
If matrix $A=\begin{bmatrix}-p&q\\r&p\end{bmatrix}$ is such that $A^{2}=I$ then :
(A) $1+p^{2}+qr=0$
(B) $1-p^{2}-qr=0$
(C) $1-p^{2}+qr=0$
(D) $1+p^{2}-qr=0$
Correct Answer: B

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Detailed Solution

Step 1: Compute $A^{2}$

Given $A = \begin{bmatrix}-p & q \\ r & p\end{bmatrix}$. We calculate $A^{2} = A \times A$:

$$A^{2} = \begin{bmatrix}-p & q \\ r & p\end{bmatrix} \begin{bmatrix}-p & q \\ r & p\end{bmatrix}$$ $$A^{2} = \begin{bmatrix}(-p)(-p) + (q)(r) & (-p)(q) + (q)(p) \\ (r)(-p) + (p)(r) & (r)(q) + (p)(p)\end{bmatrix}$$ $$A^{2} = \begin{bmatrix}p^{2} + qr & -pq + pq \\ -pr + pr & qr + p^{2}\end{bmatrix} = \begin{bmatrix}p^{2} + qr & 0 \\ 0 & p^{2} + qr\end{bmatrix}$$

Step 2: Equate to Identity Matrix $I$

We are given $A^{2} = I$, where $I = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$. Therefore:

$$\begin{bmatrix}p^{2} + qr & 0 \\ 0 & p^{2} + qr\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$$

Step 3: Solve for the relation

By comparing the corresponding elements, we get:

$$p^{2} + qr = 1$$ $$p^{2} + qr - 1 = 0$$

Multiplying by -1 to match the options:

$$1 - p^{2} - qr = 0$$

Final Answer: $1-p^{2}-qr=0$

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student must utilize the definition of matrix multiplication and the property of the identity matrix to derive a specific algebraic relationship.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the execution of a series of steps (matrix multiplication followed by algebraic manipulation) to reach the solution.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. It tests the student's ability to handle symbolic matrix operations rather than just numerical ones, which is a core competency in the Matrices chapter.