The question you're asking deals with matrices and seems to be asking for the value of $d$, based on the given matrix equations. Let's break it down:

Problem Setup:

You are given three matrices $A$, $B$, and $C$:

$A = \begin{pmatrix} 1 & a + 6 \\ 6 & c \end{pmatrix}, \quad B = \begin{pmatrix} a - 1 & 0 \\ -c & d \end{pmatrix}, \quad C = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$

The equation provided is: $A + B = C$

We need to find the value of $d$.

Step 1: Add the matrices $A$ and $B$

$A + B = \begin{pmatrix} 1 & a + 6 \\ 6 & c \end{pmatrix} + \begin{pmatrix} a - 1 & 0 \\ -c & d \end{pmatrix}$

$A + B = \begin{pmatrix} (1 + (a - 1)) & ((a + 6) + 0) \\ (6 + (-c)) & (c + d) \end{pmatrix}$

Simplifying this:

$A + B = \begin{pmatrix} a & a + 6 \\ 6 - c & c + d \end{pmatrix}$

Step 2: Set the resulting matrix equal to matrix $C$:

$\begin{pmatrix} a & a + 6 \\ 6 - c & c + d \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$

This gives us a system of equations by comparing corresponding elements:

1. $a = 1$
2. $a + 6 = 0$
3. $6 - c = 1$
4. $c + d = 1$

Step 3: Solve the system of equations

From the second equation: $a + 6 = 0 \implies a = -6$

From the third equation: $6 - c = 1 \implies c = 5$

Finally, using the value of $c$ in the fourth equation: $c + d = 1 \implies 5 + d = 1 \implies d = 1 - 5 = -4$

Final Answer:

The value of $d$ is $\boxed{-4}$.

Would you like further clarification or details on any step?

Here are 5 related questions for you to explore:

1. What happens if the matrix $B$ has different dimensions than $A$?
2. How can we use matrix multiplication instead of addition in a similar problem?
3. What if the matrix $C$ had a different dimension (e.g., a 3x3 matrix)?
4. How does matrix inversion work, and can it be applied here?
5. Can you find the determinant of matrix $A$ and $B$?

Tip: Always make sure the matrices are of the same dimension when adding or subtracting them!