Question

For \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \), if \( A = \begin{bmatrix} a & b \\ c & -a \end{bmatrix} \) be 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. \(1 + a^2 - bc = 0\)

Answer 1

The Correct Option is B

Given \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \) and \( A = \begin{bmatrix} a & b \\ c & -a \end{bmatrix} \), we need to find the condition for \( A^2 = I \). Calculating \( A^2 \):  

\[ A^2 = \begin{bmatrix} a & b \\ c & -a \end{bmatrix} \begin{bmatrix} a & b \\ c & -a \end{bmatrix} = \begin{bmatrix} a^2 + bc & ab - ab \\ ac - ac & bc + a^2 \end{bmatrix} = \begin{bmatrix} a^2 + bc & 0 \\ 0 & a^2 + bc \end{bmatrix} \]

For \( A^2 = I \):

\[ \begin{bmatrix} a^2 + bc & 0 \\ 0 & a^2 + bc \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]

This implies:

\( a^2 + bc = 1 \)

Rearranging gives:

\( 1 - a^2 - bc = 0 \)

Answer 2

The matrix A is given as:

\[ A = \begin{bmatrix} a & b \\ c & -a \end{bmatrix}. \]

To compute \( A^2 \):

\[ A^2 = \begin{bmatrix} a & b \\ c & -a \end{bmatrix} \cdot \begin{bmatrix} a & b \\ c & -a \end{bmatrix}. \]

\[ A^2 = \begin{bmatrix} a^2 + bc & ab - ab \\ ac - ac & bc + (-a)^2 \end{bmatrix} = \begin{bmatrix} a^2 + bc & 0 \\ 0 & bc + a^2 \end{bmatrix}. \]

Since \( A^2 = I \), we have:

\[ \begin{bmatrix} a^2 + bc & 0 \\ 0 & bc + a^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. \]

Equating the diagonal elements:

\[ a^2 + bc = 1. \]

Rewriting this equation:

\[ 1 - a^2 - bc = 0. \]

Thus, the correct answer is Option (B).