We are given the equation:

$(1 + i)x + 2u = v$

Where:

• $u = [8, 1 + 3i, 8 - 1]$ (a $1 \times 3$ matrix),
• $v = [4, 6 + 6i, 2i - 6]$ (a $1 \times 3$ matrix),
• $x$ is the unknown matrix we need to solve for.

Steps:

1. Isolate $x$:

   $(1 + i)x = v - 2u$

   So, we subtract $2u$ from $v$:

   [ v - 2u = [4, 6 + 6i, 2i - 6] - 2 \times [8, 1 + 3i, 8 - 1] ]

2. Multiply $u$ by 2:

   $2u = [16, 2 + 6i, 16 - 2]$

   Now, subtract $2u$ from $v$:

   [ v - 2u = [4, 6 + 6i, 2i - 6] - [16, 2 + 6i, 16 - 2] ]

3. Perform the subtraction:

   $v - 2u = [4 - 16, (6 + 6i) - (2 + 6i), (2i - 6) - (16 - 2)]$

   Simplifying the components:

   $v - 2u = [-12, 4 + 6i, 2i - 6 - 16 + 2]$

   $v - 2u = [-12, 4 + 6i, 2i - 20]$

4. Now, solve for $x$:

   $x = \frac{v - 2u}{1 + i}$

   To divide by $1 + i$, multiply the numerator and denominator by the complex conjugate of $1 + i$, which is $1 - i$:

   $x = \frac{[-12, 4 + 6i, 2i - 20]}{1 + i} \times \frac{1 - i}{1 - i}$

   Simplifying the denominator:

   $(1 + i)(1 - i) = 1^2 - i^2 = 1 + 1 = 2$

   So:

   $x = \frac{1 - i}{2} \times [-12, 4 + 6i, 2i - 20]$

   Now, distribute $\frac{1 - i}{2}$ across each element of the matrix:

   $x = \left[ \frac{(-12)(1 - i)}{2}, \frac{(4 + 6i)(1 - i)}{2}, \frac{(2i - 20)(1 - i)}{2} \right]$

Let's compute each term.

First term:

$\frac{(-12)(1 - i)}{2} = \frac{-12 + 12i}{2} = -6 + 6i$

Second term:

$\frac{(4 + 6i)(1 - i)}{2}$ Expand using distributive property:

= 4 - 4i + 6i - 6 = -2 + 2i$$ Now divide by 2: $$\frac{-2 + 2i}{2} = -1 + i$$ #### Third term: $$\frac{(2i - 20)(1 - i)}{2}$$ Expand using distributive property: $$(2i - 20)(1 - i) = 2i(1) - 2i(i) - 20(1) + 20(i) = 2i + 2 - 20 + 20i = -18 + 22i$$ Now divide by 2: $$\frac{-18 + 22i}{2} = -9 + 11i$$ ### Final result for $$x$$: $$x = [-6 + 6i, -1 + i, -9 + 11i]$$ --- Would you like more details or any clarifications? Here are some related questions: 1. How can we verify this result using matrix multiplication? 2. What happens if we change the values of $$u$$ and $$v$$? 3. How do complex conjugates help in dividing complex numbers? 4. How would the solution change if we had a $$2 \times 2$$ matrix instead? 5. What other matrix operations can we perform on $$u$$ and $$v$$? **Tip:** Always use the complex conjugate to simplify division by a complex number, as it eliminates the imaginary part in the denominator.