document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=17949>Question 17949</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I have been working on trying to find the eigenvalues and eigenvectors of the following matrix:<BR>\n" );
document.write( "3 -1  2<BR>\n" );
document.write( "2  0  2<BR>\n" );
document.write( "5 -1  0\r<BR>\n" );
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document.write( "I have worked it to the point of finding the characteristic polynomial of:<BR>\n" );
document.write( "-Lambda^3 + 3*Lambda^2 + 10*Lambda -8\r<BR>\n" );
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document.write( "This doesn't seem correct to me since, MATLAB gives the eigenvalues of (4,-2,1).<BR>\n" );
document.write( "That is where I am stuck.\r<BR>\n" );
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document.write( "Any help is greatly appreciated. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #8649 by </B><A HREF=/tutors/aboutme.mpl?userid=venugopalramana><B>venugopalramana(3286)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=venugopalramana><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.geocities.com/venugopalramana/FREE_MATHS_ASSISTANCE.html valign=middle><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=8649\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I have been working on trying to find the eigenvalues and eigenvectors of the following matrix:<BR>\n" );
document.write( "3 -1  2<BR>\n" );
document.write( "2  0  2<BR>\n" );
document.write( "5 -1  0\r<BR>\n" );
document.write( "\n" );
document.write( "I have worked it to the point of finding the characteristic polynomial of:<BR>\n" );
document.write( "-Lambda^3 + 3*Lambda^2 + 10*Lambda -8\r<BR>\n" );
document.write( "\n" );
document.write( "This doesn't seem correct to me since, MATLAB gives the eigenvalues of (4,-2,1).<BR>\n" );
document.write( "That is where I am stuck.<BR>\n" );
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document.write( "ok let us check we have..(due to difficulty in depiction ,I am omitting the vertical bars to show determinant..|xx| is shown as xxx)...k is used instead of lambda)<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=matrix%283%2C3%2C%283-k%29%2C-1%2C2%2C2%2C-k%2C2%2C5%2C-1%2C-k%29 ALIGN=MIDDLE  ALT=\"matrix%283%2C3%2C%283-k%29%2C-1%2C2%2C2%2C-k%2C2%2C5%2C-1%2C-k%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >=0<BR>\n" );
document.write( "(3-k)(k^2+2)-2(k+2)+5(2k-2)=0<BR>\n" );
document.write( "3k^2+6-k^3-2k-2k-4+10k-10=0<BR>\n" );
document.write( "-k^3+3k^2+6k-8=0.=f(k)..we find f(1)=0..so k-1 is a factor..dividing with k-1\r<BR>\n" );
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document.write( "1...|...-1....3....6....-8<BR>\n" );
document.write( "....|....0....-1...2....8<BR>\n" );
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document.write( ".........-1...2....8....0<BR>\n" );
document.write( "we get -k^2+2k+8=0<BR>\n" );
document.write( "...-k^2+4k-2k+8=0<BR>\n" );
document.write( "-(k-4)(k+2)=0<BR>\n" );
document.write( "hence k=-2 or 4...or eigen values are 1,-2 and 4..now to find eigen vectors we have to put k=1,-2 and 4 in the matrix eqn (A-kI)(X)=0 and find the solutions <BR>\n" );
document.write( "where as you know A is the given matrix,k is the eigen values we got,I is the unit matrix and X is the vector (x,y,z)<BR>\n" );
document.write( "for k=1,we get (A-kI)as<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=matrix%283%2C3%2C%283-k%29%2C-1%2C2%2C2%2C-k%2C2%2C5%2C-1%2C-k%29 ALIGN=MIDDLE  ALT=\"matrix%283%2C3%2C%283-k%29%2C-1%2C2%2C2%2C-k%2C2%2C5%2C-1%2C-k%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "hence <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=matrix%283%2C3%2C2%2C-1%2C2%2C2%2C-1%2C2%2C5%2C-1%2C-1%29 ALIGN=MIDDLE  ALT=\"matrix%283%2C3%2C2%2C-1%2C2%2C2%2C-1%2C2%2C5%2C-1%2C-1%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >*<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=matrix%283%2C1%2Cx%2Cy%2Cz%29 ALIGN=MIDDLE  ALT=\"matrix%283%2C1%2Cx%2Cy%2Cz%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >=0..or<BR>\n" );
document.write( "2x-y+2z=0....(1)<BR>\n" );
document.write( "2x-y+2z=0....(2)<BR>\n" );
document.write( "5x-y-z=0.....(3)<BR>\n" );
document.write( "(3)-(2) gives3x-3z=0...or x=z..substituting in (3) we get y=4x=4z<BR>\n" );
document.write( "hence the eigen vector is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=matrix%283%2C1%2C1%2C4%2C1%29 ALIGN=MIDDLE  ALT=\"matrix%283%2C1%2C1%2C4%2C1%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >...<BR>\n" );
document.write( "i trust you can find the other 2 eigen vectors for k=-2 and k=4 in a similar way. \r<BR>\n" );
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