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If $A =\begin{bmatrix} 2 &i \\ i & 0 \end{bmatrix}$ , the trace of $A^{10}$ is

1. $2$
2. $2(1+i)$
3. $0$
4. $2^{10}$
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$\begin{vmatrix} 2-\lambda & i\\ i& -\lambda \end{vmatrix}$=0

$-(2-\lambda)*\lambda+1=0$

$(\lambda-1)^2=0$

$\lambda=1,1$

Eigen values of A is 1,1

Eigen values of $A^{10}$ will be $1^{10}, 1^{10}$=1,1

Trace of $A^{10}$=Sum of Eigen values=1+1=2
by Active (5k points)
+1 vote
By matrix multiplication , $A^2$ = $\begin{bmatrix} 3 &2i \\ 2i & -1 \end{bmatrix}$ whose trace is 2.

$A^3$ = $\begin{bmatrix} 4 & 3i\\ 3i & -2 \end{bmatrix}$ whose trace is 2.
So we see the trace of the original matrix remains the same irrespective of repeated multiplication with itself. Therefore trace of $A^{10}$ = 2
by Loyal (7.2k points)