Question

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. \( 1 + p^2 - qr = 0 \)

Hint:

Always double-check signs during matrix multiplication, especially when dealing with variables like \(-p\).
For a \( 2 \times 2 \) matrix with trace zero, \( A^2 \) will always be a scalar multiple of the identity matrix.

Answer 1

The Correct Option is B

Concept:
Matrix multiplication is performed by multiplying the rows of the first matrix with the columns of the second matrix.
For a \(2\times2\) matrix, the identity matrix is: \[ I= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] Two matrices are equal when their corresponding elements are equal. 
Step 1: Calculate \(A^2\)
Given: \[ A= \begin{bmatrix} -p & q \\ r & p \end{bmatrix} \] Therefore: \[ A^2= \begin{bmatrix} -p & q \\ r & p \end{bmatrix} \begin{bmatrix} -p & q \\ r & p \end{bmatrix} \] \[ A^2= \begin{bmatrix} (-p)(-p)+qr & (-p)q+qp \\ r(-p)+pr & rq+p^2 \end{bmatrix} \] \[ A^2= \begin{bmatrix} p^2+qr & 0 \\ 0 & p^2+qr \end{bmatrix} \] 
Step 2: Apply the condition \(A^2=I\)
Given: \[ A^2=I \] Therefore: \[ \begin{bmatrix} p^2+qr & 0 \\ 0 & p^2+qr \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] Comparing corresponding elements: \[ p^2+qr=1 \] 
Step 3: Rearrange the equation
From: \[ p^2+qr=1 \] we get: \[ 1-p^2-qr=0 \] 
Final Answer:
Therefore, \[ \boxed{1-p^2-qr=0} \] Hence, the correct answer is Option (B).