Consider the similarity transform of a matrix a to A’ given A’ = S-1AS. You need to prove | A n | = ( A’ ) n . Arrange the statements in the correct order so as to obtain to complete proof of this result. See image attached

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Consider the similarity transform of a matrix a to A’ given A’ = S-1AS. You need to prove | A n | = ( A’ ) n . Arrange the statements in the correct order so as to obtain to complete proof of this result. See image attached

Therefore |(A')"| = |S¯¹ A"S|
|(A')"| = |S¹||S||A"|.
|(A')"| = |I||A", where I is the identity
matrix.
Hence |(A')"| = |A"|, as |I| = 1 for any
identity matrix.
(A')¹ = S-¹ AA ... AS, where the A
matrices are repeated n times.
Transcribed Image Text:Therefore |(A')"| = |S¯¹ A"S| |(A')"| = |S¹||S||A"|. |(A')"| = |I||A", where I is the identity matrix. Hence |(A')"| = |A"|, as |I| = 1 for any identity matrix. (A')¹ = S-¹ AA ... AS, where the A matrices are repeated n times.
|(A')"| = |S¯¹ ||A"||S|, where we have
used |AB| = |A||B|.
(A')n = S-¹A¹S
(A')n = [S¹ AS][S¯¹ AS] ... [S¹ AS]
where the terms in square brackets are
repeated n times.
|(A')"| = |S¯¹S||A", where we have
used |AB| = |A||B|.
Transcribed Image Text:|(A')"| = |S¯¹ ||A"||S|, where we have used |AB| = |A||B|. (A')n = S-¹A¹S (A')n = [S¹ AS][S¯¹ AS] ... [S¹ AS] where the terms in square brackets are repeated n times. |(A')"| = |S¯¹S||A", where we have used |AB| = |A||B|.
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