document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=161620>Question 161620</A>This question is from textbook <A HREF=http://www.amazon.com/exec/obidos/ASIN/0137003285/algebrahomeworkh>Primer of Linear Algebra</A><BR>\n" );
document.write( ": <I><SPAN STYLE=\"word-break: break-all;\"> Let A represent a matrix.\r<BR>\n" );
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document.write( "Show that if A^3 - 7A^2 + 5In = 0 then A^4 = 49A^2 - 5A - 35In </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #119095 by </B><A HREF=/tutors/aboutme.mpl?userid=kev82><B>kev82(151)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=kev82><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=119095\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I am going to use I instead of In for the the identity. So we are given that A^3 - 7A^2 + 5I = 0 therefore A^3 = 7A^2 - 5I.\r<BR>\n" );
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document.write( "We are interested in A^4 which is A*A^3, but we know what A^3 is, so substitute A^4 = A(7A^2-5I) = 7A^3-5A. WE now have another A^3 we can substitute for, so do that. A^4 = 7(7A^2-5I)-5A = 49A^2-5A-35I. </SPAN>\n" );
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