document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1057358>Question 1057358</A>: <I><SPAN STYLE=\"word-break: break-all;\"> given a=\r<BR>\n" );
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document.write( " 1 2 3 0  1<BR>\n" );
document.write( " 2 1 3 3 -1<BR>\n" );
document.write( " 1 5 2 5  4\r<BR>\n" );
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document.write( "is b=(1,0,1)T in the span of the columns of a? how do you solve this problem? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #672415 by </B><A HREF=/tutors/aboutme.mpl?userid=solve_for_x><B>solve_for_x(190)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solve_for_x><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=672415\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Use row operations to reduce the matrix \"a\" and determine which columns form a basis<BR>\n" );
document.write( "of the column space.\r<BR>\n" );
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document.write( "The resulting matrix is:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28%0D%0A+matrix%28+3%2C+5%2C+%0D%0A+++1%2C+0%2C+0%2C+4%2C+-1%2C%0D%0A+++0%2C+1%2C+0%2C+1%2C+1%2C%0D%0A+++0%2C+0%2C+1%2C+-2%2C+0%0D%0A+%29%0D%0A%29%0D%0A ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "From this, you can see that the first three columns are linearly independent<BR>\n" );
document.write( "because each column has a single \"1\" in it, and the adjacent columns have <BR>\n" );
document.write( "only 0 entries in the same row (looking only at the first three columns).\r<BR>\n" );
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document.write( "The fourth and fifth columns are linear combinations of the first three columns of a.\r<BR>\n" );
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document.write( "Specifically, the fourth column is a combination of 4 times the first column, <BR>\n" );
document.write( "plus the second column, minus 2 times the third column. (Check this and see <BR>\n" );
document.write( "for yourself.)\r<BR>\n" );
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document.write( "Now, if b = (1, 0, 1)^T is in the span of the columns of a, then there has to be <BR>\n" );
document.write( "some linear combination of the first three columns that is equal to (1, 0, 1)^T.\r<BR>\n" );
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document.write( "That is:\r<BR>\n" );
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document.write( "(1, 2, 1)^T * x + (2, 1, 5)^T * y + (3, 3, 2)^T * z = (1, 0, 1)^T\r<BR>\n" );
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document.write( "This gives three equations:\r<BR>\n" );
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document.write( "x + 2y + 3z = 1\r<BR>\n" );
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document.write( "2x + y + 3z = 0\r<BR>\n" );
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document.write( "x + 5y + 2z = 1\r<BR>\n" );
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document.write( "If there is a solution for this system of equations, then b = (1, 0, 1)^T is in the <BR>\n" );
document.write( "span of the columns of a.\r<BR>\n" );
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document.write( "I'll assume that you can take if from here. :)\r<BR>\n" );
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document.write( " </SPAN>\n" );
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