document.write( "Question 45392This question is from textbook College Algebra
\n" ); document.write( ": Find the inverse of each matrix A if possible. Check that AA^-1=I and A^-1 A=I.\r
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Algebra.Com's Answer #30155 by AnlytcPhil(1810) $\"\"$ $\"About$

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Find the inverse of each matrix A if possible.  \r\n" );
document.write( "Check that AA^-1=I and A^-1 A=I.\r\n" );
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document.write( "[1 -1  2]\r\n" );
document.write( "[1  2  3]\r\n" );
document.write( "[2  1  5]\r\n" );
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document.write( "Augment the matrix with the identity matrix:\r\n" );
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document.write( "[ 1 -1  2 |  1  0  0]\r\n" );
document.write( "[ 1  2  3 |  0  1  0]\r\n" );
document.write( "[ 2  1  5 |  0  0  1]\r\n" );
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document.write( "The idea is to get every element in the\r\n" );
document.write( "left side 0 except the three diagonal elements,\r\n" );
document.write( "which must not be 0.  Then we divide each row\r\n" );
document.write( "through to make the diagonal elements on the\r\n" );
document.write( "left side 1.\r\n" );
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document.write( "Get a 0 where the 1 is in the 2nd row 1st column by\r\n" );
document.write( "multiplying row 1 by -1 and adding it to row 2,\r\n" );
document.write( "restoring row 1\r\n" );
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document.write( "-1[ 1 -1  2 |  1  0  0]\r\n" );
document.write( " 1[ 1  2  3 |  0  1  0]\r\n" );
document.write( "  [ 2  1  5 |  0  0  1]\r\n" );
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document.write( "[ 1 -1  2 |  1  0  0]\r\n" );
document.write( "[ 0  3  1 | -1  1  0]\r\n" );
document.write( "[ 2  1  5 |  0  0  1]\r\n" );
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document.write( "Get a 0 where the 2 is in the 3rd row 1st column by\r\n" );
document.write( "multiplying row 1 by -2 and adding it to row 3,\r\n" );
document.write( "restoring row 1\r\n" );
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document.write( "-2[ 1 -1  2 |  1  0  0]\r\n" );
document.write( "  [ 0  3  1 | -1  1  0]\r\n" );
document.write( " 1[ 2  1  5 |  0  0  1]\r\n" );
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document.write( "[ 1 -1  2 |  1  0  0]\r\n" );
document.write( "[ 0  3  1 | -1  1  0]\r\n" );
document.write( "[ 0  3  1 | -2  0  1]\r\n" );
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document.write( "Get a 0 where the -1 is in the 1st row 2nd column by\r\n" );
document.write( "multiplying row 2 by 1 and adding it to 3 times row 1,\r\n" );
document.write( "restoring row 2\r\n" );
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document.write( " 3[ 1 -1  2 |  1  0  0]\r\n" );
document.write( " 1[ 0  3  1 | -1  1  0]\r\n" );
document.write( "  [ 0  3  1 | -2  0  1]\r\n" );
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document.write( "[ 3  0  7 |  2  1  0]\r\n" );
document.write( "[ 0  3  1 | -1  1  0]\r\n" );
document.write( "[ 0  3  1 | -2  0  1]\r\n" );
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document.write( "Get a 0 where the 3 is in the 3rd row 2nd column by\r\n" );
document.write( "multiplying row 2 by -1 and adding it to 1 times row 3,\r\n" );
document.write( "restoring row 2\r\n" );
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document.write( "  [ 3  0  7 |  2  1  0]\r\n" );
document.write( "-1[ 0  3  1 | -1  1  0]\r\n" );
document.write( " 1[ 0  3  1 | -2  0  1]\r\n" );
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document.write( "[ 3  0  7 |  2  1  0]\r\n" );
document.write( "[ 0  3  1 | -1  1  0]\r\n" );
document.write( "[ 0  0  0 | -2  0  1]\r\n" );
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document.write( "Oh, oh. The element in the 3rd row 2rd column is a 0\r\n" );
document.write( "so there is no way to get 0's where the 7 and 1 are \r\n" );
document.write( "above it.  This matrix has no inverse.\r\n" );
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document.write( "Edwin

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