If $A=\left[\begin{array}{cc}a & b \\ c & -a\end{array}\right]$ such that $A^2=I$ then which of the following is true?

$1-a^2-b c=0$

The correct answer is Option (3) → $1-a^2-b c=0$

$A = \begin{bmatrix} a & b \\ c & -a \end{bmatrix}$

$A^2 = \begin{bmatrix} a & b \\ c & -a \end{bmatrix} \begin{bmatrix} a & b \\ c & -a \end{bmatrix}$

$= \begin{bmatrix} a^2 + bc & ab - ab \\ ca - ca & cb + a^2 \end{bmatrix}$

$= \begin{bmatrix} a^2 + bc & 0 \\ 0 & a^2 + bc \end{bmatrix}$

$A^2 = (a^2 + bc)I$

$A^2 = I \Rightarrow a^2 + bc = 1$

$1 - a^2 - bc = 0$

The correct relation is $1 - a^2 - bc = 0$.