Let A=[a_ij]_22, where a_ij0 for all i,j and A^2=I.

Question

Let $A=[a_{ij}]_{2	imes2}$, where $a_{ij}
e0$ for all $i,j$ and $A^2=I$. Let $a$ be the sum of all diagonal elements of $A$ and $b=|A|$. Then $3a^2+4b^2$ is equal to:

Accepted Answer

$A^2=I\Rightarrow|A|^2=1\Rightarrow b=|A|=\pm1$.

Writing $A=\begin{pmatrix}\alpha&\beta\\gamma&\delta\end{pmatrix}$ with $A^2=I$: from the off-diagonal entries of $A^2$, $(\alpha+\delta)\beta=0$ and $(\alpha+\delta)\gamma=0$. Since $\beta,\gamma
e0$, $\alpha+\delta=0$, i.e. $a=\operatorname{trace}(A)=0$.

$3a^2+4b^2=3(0)^2+4(1)^2=4$.

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