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Step-by-Step Solution
Step 1: Analyze the Matrices A and B
Given
A =
\begin{pmatrix}
2 & -1 & -1 \\
1 & 0 & -1 \\
1 & -1 & 0
\end{pmatrix}
and
B = A - I.
First, observe the behavior of powers of these matrices.
Step 2: Compute A²
Notice from the given (and can be verified by direct multiplication) that
A^2 = A.
Once a matrix satisfies
A^2 = A,
it follows by induction that
A^n = A
\text{ for all integers } n \ge 1.
Step 3: Express Matrix B and Compute B²
We have
B = A - I.
Direct computation or using the given information shows
B^2 = -B.
To see why, if
B^2 = -B,
it implies
B^3 = B^2 \cdot B = (-B) \cdot B = -B^2 = -(-B) = B.
Hence for higher powers, you get a repeating pattern:
If n is odd, then
B^n = B.
If n is even, then
B^n = -B.
Step 4: Set up the Required Condition
We want all integers
n \in \{1, 2, \ldots, 100\}
such that
A^n + (\omega B)^n = A + B.
Since
A^n = A
for all
n \ge 1,
the equation becomes
A + (\omega B)^n = A + B
\implies (\omega B)^n = B.
Step 5: Simplify the Condition (\omega B)^n = B
We can write
(\omega B)^n = \omega^n B^n.
So the requirement is
\omega^n B^n = B.
We split this into two parts: one for
B^n
and one for
\omega^n.
Case 1: n is odd
If
n
is odd, then
B^n = B.
Hence
\omega^n B^n = \omega^n B.
We want
\omega^n B = B,
which requires
\omega^n = 1.
Since
\omega
is a (non-real) cube root of unity,
\omega^n = 1
if and only if
3 \mid n.
So in this case,
n
must be both odd and a multiple of 3, namely
3, 9, 15, \ldots
Case 2: n is even
If
n
is even, then
B^n = -B.
So
\omega^n B^n = \omega^n (-B) = -\omega^n B.
For it to equal
B,
we would need
-\omega^n = 1
\implies
\omega^n = -1.
A primitive cube root of unity
\omega
never takes the value
-1
for any positive integer
n.
Hence there are no solutions in the even case.
Step 6: Conclusion on Valid n
Thus, valid
n
are those that are multiples of 3 and odd:
\[
3, 9, 15, 21, \dots, 99.
\]
Each is of the form
3 + 6k
(for
k = 0,1,2,\dots
).
From
3
to
99,
there are exactly 17 such numbers because the sequence
3,9,15,\ldots,99
is an arithmetic progression with first term 3, last term 99, and common difference 6. One can verify via:
\[
\text{Number of terms} = \frac{99 - 3}{6} + 1 = 17.
\]
Final Answer
The number of elements
n \in \{1, 2, \ldots, 100\}
satisfying
A^n + (\omega B)^n = A + B
is
\boxed{17}.
