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

Hint:

For any matrix \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), the Cayley-Hamilton theorem states \( A^2 - (Tr(A))A + |A|I = 0 \).
If \( Tr(A) = 0 \), then \( A^2 = -|A|I \).
For \( A^2 = I \), we need \( -|A| = 1 \), which means \( |A| = -1 \).
The determinant \( |A| = -a^2 - bc \). Setting \( -a^2 - bc = -1 \) gives \( a^2 + bc = 1 \).

Answer 1

The Correct Option is B

Step 1: Write out the square.
We are told $A^2 = I$. So let us multiply $A$ by itself and see what comes out.
\[ A^2 = \begin{bmatrix} a & b \\ c & -a \end{bmatrix}\begin{bmatrix} a & b \\ c & -a \end{bmatrix} \]

Step 2: Fill in each entry.
The top left is $a\cdot a + b\cdot c = a^2 + bc$. The top right is $ab - ab = 0$. The bottom left is $ca - ac = 0$. The bottom right is $cb + a^2 = a^2 + bc$.
\[ A^2 = \begin{bmatrix} a^2+bc & 0 \\ 0 & a^2+bc \end{bmatrix} \]

Step 3: Match with the identity.
The off diagonal terms are already zero, just like in $I$. So we only need the diagonal terms to equal $1$.
\[ a^2 + bc = 1 \]

Step 4: Shape it like the options.
Move both terms across so the right side is zero.
\[ 1 - a^2 - bc = 0 \]
That is option 2.
\[ \boxed{1 - a^2 - bc = 0} \]