Let's tackle the problem step-by-step.

Given matrices: $A = \begin{bmatrix} 2 & 1 & 0 \\ -1 & 4 & 1 \\ 3 & 0 & -1 \end{bmatrix}, \quad B = \begin{bmatrix} -3 & -1 & 2 \\ 1 & -1 & 0 \\ 1 & 1 & -1 \end{bmatrix}, \quad C = \begin{bmatrix} 7 & -2 & 6 \\ -4 & 6 & -1 \\ 12 & -1 & 3 \end{bmatrix}$

Part (a)

1. Find $(A + B)C$:

   • First, calculate $A + B$.
   • Then, multiply $(A + B)$ with $C$.

2. Deduce $(A + B)^{-1}$ (if it exists):

   • To find the inverse, calculate the determinant of $(A + B)$.
   • If the determinant is non-zero, find the inverse.

Part (b)

Use the result from part (a) to solve a system of linear equations (the problem does not specify, but it could mean using $(A + B)^{-1}$ to solve a system).

I'll calculate each step.Here are the solutions:

1. Matrix $A + B$: $A + B = \begin{bmatrix} -1 & 0 & 2 \\ 0 & 3 & 1 \\ 4 & 1 & -2 \end{bmatrix}$

2. Matrix $(A + B)C$: $(A + B)C = \begin{bmatrix} 17 & 0 & 0 \\ 0 & 17 & 0 \\ 0 & 0 & 17 \end{bmatrix}$ This is a scalar matrix, which implies that $(A + B)C = 17I$, where $I$ is the identity matrix.

3. Determinant of $A + B$:
   $\det(A + B) = -17$, which is non-zero, so $A + B$ is invertible.

4. Inverse of $A + B$: $(A + B)^{-1} = \begin{bmatrix} 0.4118 & -0.1176 & 0.3529 \\ -0.2353 & 0.3529 & -0.0588 \\ 0.7059 & -0.0588 & 0.1765 \end{bmatrix}$

Part (b)

Using $(A + B)^{-1}$, we can solve any system of equations involving $(A + B)$ if required.

Would you like more details on any of these steps or need help with the system of equations?

Related Questions:

1. What are the steps to find the inverse of a 3x3 matrix?
2. How does a scalar matrix like $(A + B)C = 17I$ simplify calculations?
3. Why is the determinant important for matrix invertibility?
4. Can you explain the properties of identity and scalar matrices?
5. How would a non-invertible matrix affect a system of equations?

Tip: When multiplying matrices, check if they result in a scalar multiple of the identity matrix. This simplifies finding their inverses or powers.