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:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $1-a^2-bc=0$
$A =\begin{bmatrix}a&b\\c&-a\end{bmatrix}$
$A^2=\begin{bmatrix}a&b\\c&-a\end{bmatrix}\begin{bmatrix}a&b\\c&-a\end{bmatrix}=\begin{bmatrix}a^2+bc&ab-ab\\ac-ac&bc+a^2\end{bmatrix}$
$=\begin{bmatrix}a^2+bc&0\\0&bc+a^2\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$⇒a^2+bc=1$
$∴1-a^2-bc=0$