# 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

##### Topic

Linear Algebra

##### Difficulty Level

Medium

##### Suggested Book

Introduction to Linear Algebra by Pearson

# Start with hints

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) .(Â 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, characteristicÂ

**values**Â Â andÂ properÂ**values**, or latent roots .) (Eigenvectors are a special set of vectors associated with aÂ linear system of equationsÂ (i.e., aÂ 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 =>Â \(\lambda^2\) – 2\(\lambda\) + 1 = 0 =>\( (\lambda – 1)^2\) = 0 =>Â \(\lambda\) = 1,1 So , what will be the canonical form ?

So , the canonicalÂ matrix will be \begin{pmatrix} 1&0\\0&1 \end{pmatrix} = A’ , i.e \(\le P \epsilon GL_2(\mathbb{C})\) o.t A’ = \(PAP^{-1}\) \((A’)^{10}\) =Â \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((\(A’)^{10}\))Â => (\(1^{10}\)) +Â (\(1^{10}\)) => 2(Ans)

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