document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=54438>Question 54438</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Solve using matrices<BR>\n" );
document.write( "2x-3y=4<BR>\n" );
document.write( "5x+4y=33 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #36688 by </B><A HREF=/tutors/aboutme.mpl?userid=tiffanyc><B>tiffanyc(9)</B></A><img src=\"/images/stars/stars-05.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=tiffanyc><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=36688\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Using matrices is simply using elementary row operations to get 1x=? and 1y=? so the original equation in matrix form becomes the co-efficients of x and y and the constant term:<BR>\n" );
document.write( "2 -3  4<BR>\n" );
document.write( "5 4   33<BR>\n" );
document.write( "Ideally, we want a matrix that is(see below) where ? is the answer for each variable.<BR>\n" );
document.write( "1 0  ?<BR>\n" );
document.write( "0 1  ?\r<BR>\n" );
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document.write( "start by dividing the top row by 2 to get a 1 in the first column of the first row<BR>\n" );
document.write( "1  (-3/2)   2<BR>\n" );
document.write( "5   4       33\r<BR>\n" );
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document.write( "Then get a zero in the first column of the second row by taking row 2 and subtracting 5 times row 1: (note: only row two changes in this step)\r<BR>\n" );
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document.write( "1      (-3/2)             2    =    1  (-3/2)    2<BR>\n" );
document.write( "0   (8/2)-((-3)(5)/2)    33    =    0  (23/2)    23\r<BR>\n" );
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document.write( "Next get a one in the second column of the second row by multiplying the row by (2/23):\r<BR>\n" );
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document.write( "1  (-3/2)   2<BR>\n" );
document.write( "0   1       2\r<BR>\n" );
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document.write( "Now to get a zero in the 2nd column of the first row, take row 1 and add (3/2) times the second row (because the term we're looking to get rid of is negative)<BR>\n" );
document.write( "Note: in the first column (3/2) times 0 is still 0 so 1-0 will still be one.<BR>\n" );
document.write( "Note: multiplying the second row is only for applying it to row 1, after the operation has been completed, the second row goes back to its original form.\r<BR>\n" );
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document.write( "1      (-3/2)      2     =  1    0    5<BR>\n" );
document.write( "0(3/2) 1(3/2)  2(3/2)    =  0    1    2\r<BR>\n" );
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document.write( "Now putting that back into linear algebra form <BR>\n" );
document.write( "1x +0y = 5   x=5<BR>\n" );
document.write( "0x +1y = 2   y=2\r<BR>\n" );
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document.write( "Check this by substituting the x and y values into the original equation:<BR>\n" );
document.write( "2x -3y =4         2(5)-3(2) =4    10-6=4<BR>\n" );
document.write( "5x +4y =33        5(5)+4(2) =33   25+8=33 </SPAN>\n" );
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