This problem requires a strong understanding of matrix properties, particularly idempotency, the binomial theorem for matrices, and the properties of the trace operator. The goal is to efficiently calculate the sum of diagonal elements (trace) of a matrix raised to a high power.

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Key Concepts and Formulas Used

1. Idempotent Matrix: A square matrix $M$ is called idempotent if $M^2 = M$. If a matrix is idempotent, then any positive integer power of the matrix is equal to the matrix itself: $M^k = M$ for all integers $k \ge 1$. This property significantly simplifies calculations involving high powers of such matrices.

2. Binomial Theorem for Matrices: For two square matrices $X$ and $Y$ of the same order that commute (i.e., $XY = YX$), the binomial expansion is given by: $(X+Y)^n = \sum_{k=0}^n {^nC_k} X^{n-k} Y^k$ A crucial special case is when one of the matrices is the identity matrix $I$. The identity matrix always commutes with any square matrix $A$ ($AI = IA = A$). Thus, we can write: $(A+I)^n = \sum_{k=0}^n {^nC_k} A^{n-k} I^k$ Remember that $I^k = I$ for any positive integer $k$, and $A^0 = I$ by convention in such expansions.

3. Trace of a Matrix: The trace of a square matrix $M$, denoted as $\text{Tr}(M)$, is the sum of its diagonal elements. For a matrix $M = [m_{ij}]$, $\text{Tr}(M) = \sum_{i} m_{ii}$.

4. Linearity of the Trace Operator: The trace operator is linear. For any scalar constants $a, b$ and matrices $X, Y$ of the same dimension: $\text{Tr}(aX + bY) = a \text{Tr}(X) + b \text{Tr}(Y)$ This property is vital for calculating the trace of a linear combination of matrices.

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Step-by-Step Solution

Step 1: Identify Special Properties of Matrix A

The first step in problems involving high powers of a matrix is to check for any special properties that might simplify its powers. Common properties include idempotency ($A^2=A$), nilpotency ($A^k=O$ for some $k$), or periodicity ($A^k=A$ for some $k>1$). Let's calculate $A^2$ for the given matrix $A$: $A = \left( {\begin{matrix} 1 & 0 & 0 \cr 0 & 4 & { - 1} \cr 0 & {12} & { - 3} \cr \end{matrix} } \right)$

Now, we compute $A^2 = A \cdot A$: $A^2 = \left( {\begin{matrix} 1 & 0 & 0 \cr 0 & 4 & { - 1} \cr 0 & {12} & { - 3} \cr \end{matrix} } \right) \left( {\begin{matrix} 1 & 0 & 0 \cr 0 & 4 & { - 1} \cr 0 & {12} & { - 3} \cr \end{matrix} } \right)$

Performing the matrix multiplication:

• $(A^2)_{11} = (1)(1) + (0)(0) + (0)(0) = 1$
• $(A^2)_{12} = (1)(0) + (0)(4) + (0)(12) = 0$
• $(A^2)_{13} = (1)(0) + (0)(-1) + (0)(-3) = 0$
• $(A^2)_{21} = (0)(1) + (4)(0) + (-1)(0) = 0$
• $(A^2)_{22} = (0)(0) + (4)(4) + (-1)(12) = 16 - 12 = 4$
• $(A^2)_{23} = (0)(0) + (4)(-1) + (-1)(-3) = -4 + 3 = -1$
• $(A^2)_{31} = (0)(1) + (12)(0) + (-3)(0) = 0$
• $(A^2)_{32} = (0)(0) + (12)(4) + (-3)(12) = 48 - 36 = 12$
• $(A^2)_{33} = (0)(0) + (12)(-1) + (-3)(-3) = -12 + 9 = -3$

Thus, we find: $A^2 = \left( {\begin{matrix} 1 & 0 & 0 \cr 0 & 4 & { - 1} \cr 0 & {12} & { - 3} \cr \end{matrix} } \right)$ Observation: We notice that $A^2 = A$. This means matrix $A$ is an idempotent matrix.

Why this step is taken: Recognizing that $A$ is idempotent is crucial because it vastly simplifies any power of $A$. If $A^2 = A$, then $A^3 = A^2 \cdot A = A \cdot A = A^2 = A$, and generally, $A^k = A$ for any positive integer $k \ge 1$. This property will allow us to simplify the binomial expansion significantly.

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Step 2: Apply Binomial Theorem to $(A+I)^{11}$

We need to calculate $(A+I)^{11}$. Since $A$ and $I$ (the identity matrix) commute, we can use the binomial theorem for matrices: $(A+I)^{11} = {^{11}C_0}A^{11}I^0 + {^{11}C_1}A^{10}I^1 + \dots + {^{11}C_{10}}A^1I^{10} + {^{11}C_{11}}A^0I^{11}$

Why this step is taken: The binomial theorem provides a systematic way to expand powers of a sum of matrices. Since $A$ and $I$ commute, we can directly apply this theorem to transform the high power of $(A+I)$ into a sum of terms involving powers of $A$ and $I$.

Now, we apply the idempotency property $A^k = A$ for $k \ge 1$. Also, $I^k = I$ for any $k$, and $A^0 = I$ (by definition). Substituting these into the expansion: $(A+I)^{11} = {^{11}C_0}A + {^{11}C_1}A + \dots + {^{11}C_{10}}A + {^{11}C_{11}}I$ Notice that all terms involving $A^k$ for $k \ge 1$ simplify to $A$. The last term, ${^{11}C_{11}}A^0I^{11}$, simplifies to ${^{11}C_{11}}I \cdot I = {^{11}C_{11}}I$.

We can factor out $A$ from the first eleven terms: $(A+I)^{11} = A \left( {^{11}C_0} + {^{11}C_1} + \dots + {^{11}C_{10}} \right) + {^{11}C_{11}}I$

We know that the sum of all binomial coefficients for power $n$ is $\sum_{k=0}^{n} {^nC_k} = 2^n$. For $n=11$, we have $\sum_{k=0}^{11} {^{11}C_k} = 2^{11}$. So, the sum inside the parenthesis is: ${^{11}C_0} + {^{11}C_1} + \dots + {^{11}C_{10}} = \left( \sum_{k=0}^{11} {^{11}C_k} \right) - {^{11}C_{11}} = 2^{11} - 1$ Also, ${^{11}C_{11}} = 1$.

Substituting these values back into the expression for $(A+I)^{11}$: $(A+I)^{11} = A(2^{11}-1) + I$

Why this simplification is important: This step reduces the complex calculation of a high power of a matrix sum to a simple linear combination of matrix $A$ and the identity matrix $I$. This is a massive simplification, making the subsequent trace calculation much easier.

Common Mistake: A common error is to incorrectly handle the $A^0$ term. Remember that $A^0 = I$, and this term is distinct from $A^k = A$ for $k \ge 1$.

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Step 3: Calculate the Trace of $(A+I)^{11}$

The problem asks for the sum of the diagonal elements of $(A+I)^{11}$, which is precisely its trace. We have simplified $(A+I)^{11}$ to $(2^{11}-1)A + I$. Using the linearity property of the trace operator: $\text{Tr}((A+I)^{11}) = \text{Tr}((2^{11}-1)A + I)$ $\text{Tr}((A+I)^{11}) = (2^{11}-1) \text{Tr}(A) + \text{Tr}(I)$

Why this step is taken: The linearity of the trace operator allows us to break down the trace of a sum (or linear combination) into the sum of traces. This means we only need to calculate $\text{Tr}(A)$ and $\text{Tr}(I)$, which are straightforward.

Now, let's calculate the trace of $A$ and $I$: For matrix $A = \left( {\begin{matrix} 1 & 0 & 0 \cr 0 & 4 & { - 1} \cr 0 & {12} & { - 3} \cr \end{matrix} } \right)$: $\text{Tr}(A) = 1 + 4 + (-3) = 5 - 3 = 2$ For the $3 \times 3$ identity matrix $I = \left( {\begin{matrix} 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr \end{matrix} } \right)$: $\text{Tr}(I) = 1 + 1 + 1 = 3$

Finally, substitute these values into the trace expression: We know $2^{11} = 2048$. $\text{Tr}((A+I)^{11}) = (2048-1) \cdot \text{Tr}(A) + \text{Tr}(I)$ $\text{Tr}((A+I)^{11}) = (2047) \cdot 2 + 3$ $\text{Tr}((A+I)^{11}) = 4094 + 3$ $\text{Tr}((A+I)^{11}) = 4097$

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Summary and Key Takeaway

The sum of the diagonal elements of the matrix $(A+I)^{11}$ is $\mathbf{4097}$.

This problem beautifully demonstrates how recognizing special matrix properties can drastically simplify complex calculations. The key steps were:

1. Identify Idempotency: Calculating $A^2$ revealed that $A$ is an idempotent matrix ($A^2=A$), which means $A^k=A$ for $k \ge 1$.
2. Apply Binomial Theorem: The binomial expansion for $(A+I)^{11}$ was used, leveraging the fact that $A$ and $I$ commute.
3. Simplify with Idempotency: The idempotency property allowed us to simplify the binomial expansion into a linear combination: $(A+I)^{11} = (2^{11}-1)A + I$.
4. Use Linearity of Trace: The trace operator's linearity allowed us to compute the trace of the simplified expression by finding $\text{Tr}(A)$ and $\text{Tr}(I)$ separately.

Always look for special properties of matrices (idempotent, nilpotent, periodic, symmetric, skew-symmetric, orthogonal, etc.) when dealing with powers or complex matrix expressions, as they are often the intended shortcut in competitive exams.

The final answer is $\boxed{\text{4097}}$.