Since the only matrix in view here is the identity matrix, there is no linear algebra. The identity matrix can be factored out, and the only interest is in the scalar factor multiplying it. Doing this, you want the expectation of $$ (b^2 -2bY +Y^2) \mathcal{I}(Y \ge b) + \\ 2(ab -(a+b) +Y^2) \mathcal{I}(b \le Y \le a) + \\ (a^2 -2aY +Y^2) \mathcal{I}(Y\le a) $$ WE have used that the indicator function, as a zero-one function, is its own square, and that product of two indicator function is the indicator function on the intersection of the two sets.

I will not give all the details, only some examples of the calculations involved. You put it all together yourself.

Expectation of first line above will involve $$ \DeclareMathOperator{\E}{\mathbb{E}} \E\{ Y^2 \mathcal{I}(Y\ge b )\} = \int_b^\infty y^2 \phi(y)\; dy = \\ \frac{\sqrt{2}b e^{-b^2/2}-2\Phi(b)\sqrt{\pi}+2\sqrt{\pi}}{2\sqrt\pi} $$ and there will be other similar integrals. Here $\phi, \Phi$ is standard normal density and CDF, respectively.