document.write( "Question 190355: I need HELP using the Gauss-Jordan method to find A^-1 if it exists,\r
\n" ); document.write( "\n" ); document.write( " $\"A=%28matrix%283%2C3%2C1%2C1%2C1%2C2%2C1%2C1%2C2%2C2%2C3%29%29\"$ \n" ); document.write( "

Algebra.Com's Answer #142914 by Edwin McCravy(20060) $\"\"$ $\"About$

You can put this solution on YOUR website!
Edwin's solution:
\n" ); document.write( "

\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "We begin by augmenting that matrix on the right with the\r\n" );
document.write( "identity matrix, and we have\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "We want to use row operations and end up with a matrix \r\n" );
document.write( "that has the identity of the left looking like this:\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "where there are numbers where the letters \"a\" through \"i\"\r\n" );
document.write( "are.  The inverse will then be\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "We start with this:\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "We already have a 1 in the upper left corner.  We need to\r\n" );
document.write( "get a 0 right under it.  So we multiply row 1 \r\n" );
document.write( "by -2 and get \r\n" );
document.write( " and add it to row 2 which is \r\n" );
document.write( " and get \r\n" );
document.write( ".\r\n" );
document.write( "Now we replace row 2 by that and we now have:\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "We need to\r\n" );
document.write( "get a 0 where the 2 is in the lower left corner.  \r\n" );
document.write( "So we multiply row 1 \r\n" );
document.write( "again by -2 and get \r\n" );
document.write( " and add it to row 3 which is \r\n" );
document.write( " and get \r\n" );
document.write( ".\r\n" );
document.write( "Now we replace row 3 by that and we now have:\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "We need to\r\n" );
document.write( "get a 0 where the second 1 is in the top row.  \r\n" );
document.write( "So we just add row 2 \r\n" );
document.write( "to row 1 which is \r\n" );
document.write( " and get  \r\n" );
document.write( "\r\n" );
document.write( "Now we replace row 1 by that and we now have:\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "We need to\r\n" );
document.write( "get a 1 where the -1 is in row column 2, so we just\r\n" );
document.write( "divide that row through by -1, which means we just\r\n" );
document.write( "change the signs of all the elements on row 2:\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Finally we need to\r\n" );
document.write( "get a 0 where 1 is in row 2, column 3.\r\n" );
document.write( " \r\n" );
document.write( "So we multiply row 3 \r\n" );
document.write( "again by -1 and get \r\n" );
document.write( " and add it to row 2 which is \r\n" );
document.write( " and get \r\n" );
document.write( ".\r\n" );
document.write( "Now we replace row 3 by that and we now have: \r\n" );
document.write( "\r\n" );
document.write( " \r\n" );
document.write( "\r\n" );
document.write( "Therefore the inverse of  is the 3x3\r\n" );
document.write( "matrix on the right of the partition bar:\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Edwin

\n" ); document.write( "

\n" );