To solve this problem, we need to interpret the given notation. The expression "If $x + iy \begin{bmatrix} \dots \end{bmatrix}$" implies that the complex number $x+iy$ is equal to the determinant of the given matrix. Step 1: Calculate the determinant of the matrix. Let the given matrix be $A$: $$ A = \begin{bmatrix} 6i & -3i & 1 \\ 4 & 3i & -1 \\ 20 & 3 & i \end{bmatrix} $$ The determinant of a $3 \times 3$ matrix $\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$ is given by $a(ei - fh) - b(di - fg) + c(dh - eg)$. Applying this formula to our matrix: $$ \det(A) = 6i \left| \begin{matrix} 3i & -1 \\ 3 & i \end{matrix} \right| - (-3i) \left| \begin{matrix} 4 & -1 \\ 20 & i \end{matrix} \right| + 1 \left| \begin{matrix} 4 & 3i \\ 20 & 3 \end{matrix} \right| $$ Step 2: Calculate the $2 \times 2$ determinants. 1. The first minor: $$ \left| \begin{matrix} 3i & -1 \\ 3 & i \end{matrix} \right| = (3i)(i) - (-1)(3) = 3i^2 + 3 $$ Since $i^2 = -1$: $$ 3(-1) + 3 = -3 + 3 = 0 $$ 2. The second minor: $$ \left| \begin{matrix} 4 & -1 \\ 20 & i \end{matrix} \right| = (4)(i) - (-1)(20) = 4i + 20 $$ 3. The third minor: $$ \left| \begin{matrix} 4 & 3i \\ 20 & 3 \end{matrix} \right| = (4)(3) - (3i)(20) = 12 - 60i $$ Step 3: Substitute the minor determinants back into the main determinant formula. $$ x + iy = 6i(0) - (-3i)(4i + 20) + 1(12 - 60i) $$ $$ x + iy = 0 + 3i(4i + 20) + 12 - 60i $$ Distribute $3i$: $$ x + iy = 12i^2 + 60i + 12 - 60i $$ Substitute $i^2 = -1$: $$ x + iy = 12(-1) + 60i + 12 - 60i $$ $$ x + iy = -12 + 60i + 12 - 60i $$ Combine the real and imaginary parts: $$ x + iy = (-12 + 12) + (60i - 60i) $$ $$ x + iy = 0 + 0i $$ $$ x + iy = 0 $$ Step 4: Determine $x$ and $y$, then find $x - iy$. From $x + iy = 0$, we can equate the real and imaginary parts: $x = 0$ $y = 0$ Now, we need to find $x - iy$: $$ x - iy = 0 - i(0) $$ $$ x - iy = 0 $$ Comparing this result with the given options: a. 0 b. $1 + 3i$ c. $3i$ d. $3 + i$ The calculated value matches option a. The final answer is $\boxed{\text{a. 0}}$.