This problem requires us to find the number of integers $n$ in a given range for which a specific matrix equation, $A^n = A$, holds true. The core concept revolves around understanding matrix powers, especially for invertible matrices, and identifying cyclic patterns in these powers.

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1. Analyze the Matrix and Determine Invertibility

First, let's examine the given matrix $A$: $A = \begin{pmatrix} 1 + i & 1 \\ -i & 0 \end{pmatrix}$ To simplify the equation $A^n = A$, we need to determine if $A$ is invertible. An invertible matrix allows us to multiply by its inverse, which can simplify equations involving matrix powers. A matrix is invertible if and only if its determinant is non-zero.

Let's calculate the determinant of $A$: $\det(A) = (1+i)(0) - (1)(-i)$ $\det(A) = 0 - (-i)$ $\det(A) = i$ Since $\det(A) = i \neq 0$, the matrix $A$ is invertible.

Why this step? If $A$ is invertible, we can multiply both sides of $A^n = A$ by $A^{-1}$. This simplifies the equation to $A^{n-1} = I$, where $I$ is the identity matrix. This transformation is crucial as it converts the problem into finding powers of $A$ that result in the identity matrix. If $A$ were not invertible, this simplification would not be valid, and the problem would require a different approach.

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2. Simplify the Equation $A^n = A$

Given the condition $A^n = A$ and knowing that $A$ is invertible: $A^n = A$ Multiply both sides by $A^{-1}$: $A^{-1} A^n = A^{-1} A$ $A^{n-1} = I$ where $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ is the $2 \times 2$ identity matrix.

Why this step? This is a standard algebraic manipulation for matrix equations involving invertible matrices. It transforms the problem into finding $n$ such that $A^{n-1}$ is the identity matrix. This is a common strategy as powers of matrices often exhibit cyclic behavior, eventually returning to the identity matrix. Note that for $n=1$, $A^{1-1} = A^0 = I$, which is true by definition for any invertible matrix. So $n=1$ is a potential solution.

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3. Calculate Powers of A to Find the Cycle

Our goal is to find the smallest positive integer $k$ such that $A^k = I$. This value $k$ is known as the order of the matrix $A$. We will compute successive powers of $A$:

$A = \begin{pmatrix} 1 + i & 1 \\ -i & 0 \end{pmatrix}$

Calculate $A^2$: $A^2 = A \cdot A = \begin{pmatrix} 1 + i & 1 \\ -i & 0 \end{pmatrix} \begin{pmatrix} 1 + i & 1 \\ -i & 0 \end{pmatrix}$ $A^2 = \begin{pmatrix} (1+i)(1+i) + (1)(-i) & (1+i)(1) + (1)(0) \\ (-i)(1+i) + (0)(-i) & (-i)(1) + (0)(0) \end{pmatrix}$ $A^2 = \begin{pmatrix} (1+2i+i^2) - i & 1+i \\ (-i-i^2) & -i \end{pmatrix}$ Since $i^2 = -1$: $A^2 = \begin{pmatrix} (1+2i-1) - i & 1+i \\ (-i-(-1)) & -i \end{pmatrix}$ $A^2 = \begin{pmatrix} i & 1+i \\ 1-i & -i \end{pmatrix}$ This is not $I$.

Calculate $A^3$: $A^3 = A^2 \cdot A = \begin{pmatrix} i & 1+i \\ 1-i & -i \end{pmatrix} \begin{pmatrix} 1 + i & 1 \\ -i & 0 \end{pmatrix}$ $A^3 = \begin{pmatrix} i(1+i) + (1+i)(-i) & i(1) + (1+i)(0) \\ (1-i)(1+i) + (-i)(-i) & (1-i)(1) + (-i)(0) \end{pmatrix}$ $A^3 = \begin{pmatrix} (i+i^2) + (-i-i^2) & i \\ (1-i^2) + i^2 & 1-i \end{pmatrix}$ $A^3 = \begin{pmatrix} (i-1) + (-i+1) & i \\ (1-(-1)) + (-1) & 1-i \end{pmatrix}$ $A^3 = \begin{pmatrix} 0 & i \\ 2-1 & 1-i \end{pmatrix}$ $A^3 = \begin{pmatrix} 0 & i \\ 1 & 1-i \end{pmatrix}$ This is not $I$.

Calculate $A^4$: $A^4 = A^3 \cdot A = \begin{pmatrix} 0 & i \\ 1 & 1-i \end{pmatrix} \begin{pmatrix} 1 + i & 1 \\ -i & 0 \end{pmatrix}$ $A^4 = \begin{pmatrix} 0(1+i) + i(-i) & 0(1) + i(0) \\ 1(1+i) + (1-i)(-i) & 1(1) + (1-i)(0) \end{pmatrix}$ $A^4 = \begin{pmatrix} -i^2 & 0 \\ (1+i) + (-i+i^2) & 1 \end{pmatrix}$ $A^4 = \begin{pmatrix} -(-1) & 0 \\ 1+i-i-1 & 1 \end{pmatrix}$ $A^4 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$ So, we found that $A^4 = I$. The order of matrix $A$ is 4.

Why this step? By finding the smallest positive integer $k$ such that $A^k=I$, we establish the cyclic nature of powers of $A$. Any power of $A$ that equals $I$ must have an exponent that is a multiple of this order $k$. This is a fundamental property of cyclic groups, which matrix powers form.

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4. Determine the Possible Values of $n$

We need to find $n \in \{1, 2, \dots, 100\}$ such that $A^{n-1} = I$. Since $A^4 = I$, it means that $A^k = I$ if and only if $k$ is a multiple of 4. Therefore, the exponent $(n-1)$ must be a multiple of 4. So, we can write $n-1 = 4m$, where $m$ is a non-negative integer. If $m=0$, then $n-1=0 \implies n=1$. In this case, $A^0=I$, which is valid. If $m>0$, then $n-1$ is a positive multiple of 4.

From $n-1 = 4m$, we get $n = 4m + 1$.

Now, we need to find how many such values of $n$ exist in the set $\{1, 2, \dots, 100\}$. We impose the range constraint on $n$: $1 \le n \le 100$ $1 \le 4m + 1 \le 100$ Subtract 1 from all parts of the inequality: $1 - 1 \le 4m \le 100 - 1$ $0 \le 4m \le 99$ Divide by 4: $\frac{0}{4} \le m \le \frac{99}{4}$ $0 \le m \le 24.75$ Since $m$ must be an integer (as $n-1$ must be an integer multiple of 4), the possible integer values for $m$ are: $m \in \{0, 1, 2, \dots, 24\}$

Why this step? This step translates the mathematical condition ($A^{n-1}=I$) into a concrete arithmetic condition on $n$. By finding the range of $m$, we effectively count how many values of $n$ satisfy the criteria within the given domain.

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5. Count the Number of Elements

The number of possible integer values for $m$ is $24 - 0 + 1 = 25$. Each unique value of $m$ corresponds to a unique value of $n$ that satisfies the condition $A^n=A$. For example:

• If $m=0$, $n = 4(0)+1 = 1$. ($A^1=A$)
• If $m=1$, $n = 4(1)+1 = 5$. ($A^5=A$)
• If $m=2$, $n = 4(2)+1 = 9$. ($A^9=A$) ...
• If $m=24$, $n = 4(24)+1 = 96+1 = 97$. ($A^{97}=A$)

Thus, there are 25 such values of $n$.

Relevant Tip: Always double-check matrix multiplications, especially when dealing with complex numbers. A single arithmetic error can lead to a completely different cyclic pattern or order for the matrix. Using the Cayley-Hamilton theorem (as a check, if time permits) can also confirm intermediate powers.

Common Mistake: A common mistake is to forget that $A^0=I$ (for invertible matrices), which means $n=1$ is usually a solution for $A^n=A$. Ensure that $m=0$ is included in the count. Another mistake is to incorrectly deduce the periodicity or order of the matrix.

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Summary:

1. We established that the matrix $A$ is invertible by calculating its determinant.
2. This allowed us to simplify the equation $A^n = A$ to $A^{n-1} = I$.
3. We then calculated successive powers of $A$ and found that $A^4 = I$, meaning the order of the matrix is 4.
4. For $A^{n-1} = I$ to hold, $n-1$ must be a multiple of 4 (i.e., $n-1 = 4m$ for some non-negative integer $m$).
5. Substituting $n = 4m+1$ into the given range $1 \le n \le 100$, we found that $m$ can take any integer value from 0 to 24, inclusive.
6. This gives a total of 25 possible values for $n$.

The number of elements in the set is 25.