document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1094594>Question 1094594</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Determine a value for x such that the following vectors are linearly independent (1; 1; 2; 1); (2; 1; 2; 3)<BR>\n" );
document.write( "(1; 4; 2; 1); (-1; 3; 5; x) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #709276 by </B><A HREF=/tutors/aboutme.mpl?userid=rothauserc><B>rothauserc(4718)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rothauserc><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/cgi-bin/show-face.mpl?user=rothauserc><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=709276\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the 4 given vectors in R4 are linearly independent if and only if the determinant of the matrix formed by taking the vectors as its columns is non-zero.<BR>\n" );
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document.write( "the matrix A is <BR>\n" );
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document.write( "| 1 2 1 -1 |<BR>\n" );
document.write( "| 1 1 4 3 |<BR>\n" );
document.write( "| 2 2 2 5 |<BR>\n" );
document.write( "| 1 3 1 x |<BR>\n" );
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document.write( "the det of A is expanded into four 3 by 3 determinants<BR>\n" );
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document.write( "1 times<BR>\n" );
document.write( "| 1 4 3 |<BR>\n" );
document.write( "| 2 2 5 | = 1*(2x-5) -4*(2x-15) +3*(2-6) <BR>\n" );
document.write( "| 3 1 x |<BR>\n" );
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document.write( "-2 times<BR>\n" );
document.write( "| 1 4 3 |<BR>\n" );
document.write( "| 2 2 5 | = -2 * {1*(2x-5) -4*(2x-5) +3*(2-2)} <BR>\n" );
document.write( "| 1 1 x |<BR>\n" );
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document.write( "1 times<BR>\n" );
document.write( "| 1 1 3 |<BR>\n" );
document.write( "| 2 2 5 | = 1*(2x-15) -1*(2x-5) +3*(6-2)<BR>\n" );
document.write( "| 1 3 x |<BR>\n" );
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document.write( "1 times<BR>\n" );
document.write( "| 1 1 4 |<BR>\n" );
document.write( "| 2 2 2 | = 1*(2-6) -1*(2-2) +4(6-2) = 12<BR>\n" );
document.write( "| 1 3 1 |<BR>\n" );
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document.write( "the sum is<BR>\n" );
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document.write( "1*(2x-5) -4*(2x-15) -12 -2(2x-5) +8*(2x-5) +(2x-15) -(2x-5) +12 +12 =<BR>\n" );
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document.write( "6*(2x-5) -3*(2x-15) +12 =<BR>\n" );
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document.write( "12x-30 -6x+45 +12 =<BR>\n" );
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document.write( "6x +27<BR>\n" );
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document.write( "6x +27 is not = 0 for all x EXCEPT -27/6 = -9/2 = -4.5<BR>\n" );
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document.write( " </SPAN>\n" );
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