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# Understand the problem

If A = \begin{bmatrix} 2&i\\i&0\end{bmatrix} , the trace of $A^{10}$ is
##### Source of the problem
Sample Questions (MMA):2019
Linear Algebra
Medium
##### Suggested Book
Introduction to Linear Algebra by Pearson

Do you really need a hint? Try it first!

A = \begin{pmatrix} 2&i\\i&0 \end{pmatrix} ; $A^2$ = \begin{pmatrix} 3&2i\\2i&-1 \end{pmatrix} $A^3$ = \begin{pmatrix} 4&3i\\i&-2 \end{pmatrix} Now , we can observe that the trace is always going to be 2 . Now we’re to seal the deal
One way to think about this problem is to go into eigen values as the tr(A) .( Eigenvalues are a special set of scalars associated with a linear system of equations (i.e., a matrix equation) that are sometimes also known as characteristic roots, characteristic values  and  proper values, or latent roots .) (Eigenvectors are a special set of vectors associated with a linear system of equations (i.e., a matrix equation) that are sometimes also known as characteristic vectors, proper vectors, or latent vectors ), remains unaltered in between similar matrices.
| $\lambda$I – A | = 0 => \begin{matrix} \lambda -2&-i\\-i&\lambda \end{matrix} =0 => $\lambda$( $\lambda$ – 2) + 1 = 0 => $\lambda^2$ – 2$\lambda$ + 1 = 0 =>$(\lambda – 1)^2$ = 0 =>  $\lambda$ = 1,1 So , what will be the canonical form ?
So , the canonical  matrix will be \begin{pmatrix} 1&0\\0&1 \end{pmatrix} = A’ , i.e $\le P \epsilon GL_2(\mathbb{C})$ o.t A’ = $PAP^{-1}$ $(A’)^{10}$ =  \begin{pmatrix} 1^{10}&0\\0&1^{10}\end{pmatrix} tr($A^{10}$) = tr(($P^{-1}A’P)^{10}$) => tr($P^{-1}(A’)^{10}P)$ => tr(($A’)^{10}$)  => ($1^{10}$) + ($1^{10}$) => 2(Ans)

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