Question

If \( A = \begin{bmatrix} a & b \\ c & -a \end{bmatrix} \) is such that \( A^2 = I \), then _____

• A. \( 1 + a^2 + bc = 0 \)
• B. \( 1 - a^2 - bc = 0 \)
• C. \( 1 - a^2 + bc = 0 \)
• D. \( 1 + a^2 - bc = 0 \)

Hint:

For \( A^2 = I \), diagonal elements must be 1 and off-diagonal must be 0.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
$A^2 = I$ means the matrix $A$ multiplied by itself results in the identity matrix $\begin{pmatrix} 1 & 0
0 & 1 \end{pmatrix}$.
Step 2: Formula Application:
$$A^2 = \begin{pmatrix} a & b
c & -a \end{pmatrix} \begin{pmatrix} a & b
c & -a \end{pmatrix} = \begin{pmatrix} a^2 + bc & ab - ab
ac - ac & bc + a^2 \end{pmatrix} = \begin{pmatrix} a^2 + bc & 0
0 & a^2 + bc \end{pmatrix}$$
Step 3: Explanation:
Equating $A^2$ to $I$: $\begin{pmatrix} a^2 + bc & 0
0 & a^2 + bc \end{pmatrix} = \begin{pmatrix} 1 & 0
0 & 1 \end{pmatrix}$ This gives $a^2 + bc = 1$, which can be rewritten as $1 - a^2 - bc = 0$.
Step 4: Final Answer:
The correct option is (b).