Question:
If \(A=\begin{bmatrix}1 & 1\\ 0 & i\end{bmatrix}\) and \(A^{42}=\begin{bmatrix}a & b\\ c & d\end{bmatrix}\) then \(a+d\) is equal to
If \(A=\begin{bmatrix}1 & 1\\ 0 & i\end{bmatrix}\) and \(A^{42}=\begin{bmatrix}a & b\\ c & d\end{bmatrix}\) then \(a+d\) is equal to
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Trace of powers = sum of powers of eigenvalues.
Updated On: Apr 30, 2026
- 0
- $i$
- $-i$
- 1
- -1
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The Correct Option is A
Solution and Explanation
Concept:
Upper triangular matrix powers.
Step 1: Observe structure
Eigenvalues: $1$ and $i$
Step 2: Trace relation
\[ \text{trace}(A^{42}) = 1^{42} + i^{42} \]
Step 3: Simplify
\[ i^{42} = i^{2} = -1 \] \[ a+d = 1 + (-1) = 0 \] Final Conclusion:
Option (A)
Step 1: Observe structure
Eigenvalues: $1$ and $i$
Step 2: Trace relation
\[ \text{trace}(A^{42}) = 1^{42} + i^{42} \]
Step 3: Simplify
\[ i^{42} = i^{2} = -1 \] \[ a+d = 1 + (-1) = 0 \] Final Conclusion:
Option (A)
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