Linear Algebra - Matrices (Multiple Choice Question) (1st Year College)

The answer is e). It is pretty easy to check that $i^2 = j^2 = k^2 = -1$ (1 being the identity matrix), and that $ij = -ji = k$, $jk = -kj = i$, $ki=-ik=j$. Since you only asked for the answer to the multiple-choice question, I assume you already checked this by yourself. If not, just follow what happens when you apply these matrices to the elements of the canonical basis.

With these rules, \[ (a1 + bi + cj + dk)(a1 - bi - cj - dk) = (a^2 + b^2 + c^2 + d^2) 1 + (-ab-cd+ab+cd)i + (-ac-bd+ac+bd)j + (-ad-bc+ad+bc)k = (a^2 + b^2 + c^2 + d^2) 1 \] where the coefficient of $1$ is given by multiplying the terms with the same matrix and the rest by the mixed products, that all appear twice with opposite signs and cancel. To get the inverse you can simply divide by $a^2+ b^2 + c^2 + d^2$, so the answer is e).