 Solution

Let $B=\left(\begin{array}{cc}0& b\\ c& 0\end{array}\right)$ be the lower block. ThenThe defining relation $a^4=I$ requires $T(a{)}^4=I_3$, so for real constants $bc=\pm 1$. Both possibilities ensure invertibility.

If $bc=-1$, the generator matrix has order four, because its square is not the identity while its fourth power is. Its four powers are distinct, giving a faithful representation. If $bc=1$, its square is the identity but it is not itself the identity; its lower diagonal entries are zero. The image then has order two and the kernel of a group homomorphism is $\{I,a^2\}$. Thus the faithfulness criterion for a block representation of the cyclic group of order four givesOther parameter values give no representation of $Z_4$, not additional unfaithful cases; $bc=0$ also makes the assigned matrix singular. No orthogonality of these matrices was required.

For the final construction takeMultiplication adds exponents, so $D(a^r)D(a^s)=D(a^{r+s})$ and $D(a{)}^4=I_2$. The representing matrix is the identity exactly for even exponents. Therefore this is a two-dimensional representation with kernel $\{I,a^2\}$, as required. It factors through the quotient cyclic group of order two.