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To find Trace of a given Matrix : ISI MMA 2018 Question 13

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" _i="1" _address="0.0.0.1"]If A = \begin{bmatrix} 2&i\\i&0\end{bmatrix} , the trace of \(A^{10}\) is[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" hover_enabled="0" _i="0" _address="0.1.0.0.0"]Sample Questions (MMA):2019[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.0.1" open="off"]Linear Algebra[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]Medium[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="off"]Introduction to Linear Algebra by Pearson[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="1" _address="0.1.0.1"]

Start with hints

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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.2.1"]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[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.2.2"]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.[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.2.3"]| \(\lambda\)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 ?[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" hover_enabled="0" _i="4" _address="0.1.0.2.4"]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)               [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="3" _address="0.1.0.3"]

Watch the video

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Similar Problems

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