document.write( "Question 601376: | 3 1 -1 2 |
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Algebra.Com's Answer #379733 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "| 3  1 -1  2 |\r\n" );
document.write( "| 0  0 -3  3 |\r\n" );
document.write( "| 9  5  0  1 |\r\n" );
document.write( "|-7  3 -2  0 |\r\n" );
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document.write( "There are already 2 zeros in the 2nd row.\r\n" );
document.write( "So we need to get one more 0 in that row and\r\n" );
document.write( "then we can expand about it.\r\n" );
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document.write( "To get a zero where then -3 is add the 4th column to it:\r\n" );
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document.write( "-1 + 2 =  1\r\n" );
document.write( "-3 + 3 =  0\r\n" );
document.write( " 0 + 1 =  1\r\n" );
document.write( "-2 + 0 = -2\r\n" );
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document.write( "Now the only element in the 2nd row that isn't 0 is the -3,\r\n" );
document.write( "which I've colored red:\r\n" );
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document.write( "| 3  1 -1  1 |\r\n" );
document.write( "| 0  0 -3  0 |\r\n" );
document.write( "| 9  5  0  1 |\r\n" );
document.write( "|-7  3 -2 -2 |\r\n" );
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document.write( "So we expand by the 2nd row.  First we have to decide if the\r\n" );
document.write( "-3 is in a + or a - position in the\r\n" );
document.write( "\"checkerboard\" 4×4 sign-scheme determinant, which is\r\n" );
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document.write( "| +  -  +  - |\r\n" );
document.write( "| -  +  -  + |\r\n" );
document.write( "| +  -  +  - |\r\n" );
document.write( "| -  +  -  + |\r\n" );
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document.write( "So we see the -3 is in a - position, so we will have to multiply\r\n" );
document.write( "the -3 by a -1, and use a +3 multiplier by the 3×3 minor determinant.  \r\n" );
document.write( "To get the minor 3×3 determinant, we erase all the elements in the same row\r\n" );
document.write( "and column with the -3, and we have this:\r\n" );
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document.write( "| 3  1     1 |\r\n" );
document.write( "|      -3    |\r\n" );
document.write( "| 9  5     1 |\r\n" );
document.write( "|-7  3    -2 |\r\n" );
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document.write( "And we put a +3 out front and we have this 3×3 determinant\r\n" );
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document.write( "  | 3  1  1 |\r\n" );
document.write( "3×| 9  5  1 |\r\n" );
document.write( "  |-7  3 -2 |\r\n" );
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document.write( "To get a 0 where the -2 is, multiply the first row by 2 and add\r\n" );
document.write( "that to the 3rd row:\r\n" );
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document.write( "    3  1  1 \r\n" );
document.write( "         ×2\r\n" );
document.write( "    6  2  2\r\n" );
document.write( "   -7  3 -2\r\n" );
document.write( "   -1  5  0\r\n" );
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document.write( "  | 3  1  1 |\r\n" );
document.write( "3×| 9  5  1 |\r\n" );
document.write( "  |-1  5  0 |  \r\n" );
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document.write( " To get a 1 where the 1 is in the second row 3rd column, \r\n" );
document.write( "multiply the first row by -1 and add that to the 2nd row:\r\n" );
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document.write( "    3  1  1 \r\n" );
document.write( "        ×-1\r\n" );
document.write( "   -3 -1 -1\r\n" );
document.write( "    9  5  1\r\n" );
document.write( "    6  4  0\r\n" );
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document.write( "Now the only element in the 3rd column that isn't 0 is the 1,\r\n" );
document.write( "which I've colored red:\r\n" );
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document.write( "  | 3  1  1 |\r\n" );
document.write( "3×| 6  4  0 |\r\n" );
document.write( "  |-1  5  0 |\r\n" );
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document.write( "So we expand by the 3rd column since it has all 0's but one.\r\n" );
document.write( "First we have to decide if the -3 is in a\r\n" );
document.write( "+ or a - position in the \"checkerboard\" 3×3 sign-scheme determinant, \r\n" );
document.write( "which is\r\n" );
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document.write( "| +  -  + |\r\n" );
document.write( "| -  +  - |\r\n" );
document.write( "| +  -  + |\r\n" );
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document.write( "So we see the 1 is in a + position, so we just multiply the red 1\r\n" );
document.write( "by +1, which will not change it and use a +1 multiplier by the minor 2×2\r\n" );
document.write( "determinant.  To get the minor 2×2 determinant, we erase all the elements\r\n" );
document.write( "in the same row and column with the red 1, and we have this:\r\n" );
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document.write( "  |       1 |\r\n" );
document.write( "3×| 6  4    |\r\n" );
document.write( "  |-1  5    |\r\n" );
document.write( "  \r\n" );
document.write( "And we put a +1 out front with the 3 multuiplier that's already out there\r\n" );
document.write( "and we have this 2×2 determinant\r\n" );
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document.write( "3×1×| 6  4 |\r\n" );
document.write( "    |-1  5 |\r\n" );
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document.write( "To get a 0 where the 5 is multiply the 1st column by 5 and add that\r\n" );
document.write( "the the first column:\r\n" );
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document.write( " 6(5) + 4  = 30 + 4 = 34\r\n" );
document.write( "-1(5) + 5  = -5 + 5 =  0\r\n" );
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document.write( "3×1×| 6 34 |\r\n" );
document.write( "    |-1  0 |\r\n" );
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document.write( "So we expand by the 2nd column since it has all 0's but one.\r\n" );
document.write( "First we have to decide if the 34 is in a\r\n" );
document.write( "+ or a - position in the \"checkerboard\" 2×2 sign-scheme determinant, \r\n" );
document.write( "which is\r\n" );
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document.write( "| +  - |\r\n" );
document.write( "| -  + |\r\n" );
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document.write( "So we see the red 34 is in a - position, so we multiply the red 34\r\n" );
document.write( "by -1, and use a -34 multiplier by the minor 1×1\r\n" );
document.write( "determinant.  To get the minor 1×1 determinant, we erase all the \r\n" );
document.write( "elements in the same row and column with the red 34, and we have this:\r\n" );
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document.write( "3×1×|   34 |\r\n" );
document.write( "    |-1    |\r\n" );
document.write( "  \r\n" );
document.write( "And we put a -34 out front with the 3×1 multiplier that's already out there\r\n" );
document.write( "and we have this 1×1 determinant\r\n" );
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document.write( "3×1×(-34)×|-1|\r\n" );
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document.write( "And a 1×1 determinant is equal to its 1 element, so we have\r\n" );
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document.write( "3×1×(-34)×(-1) = 102.\r\n" );
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document.write( "[Notice, we don't have to go all the way to a 1×1 determinant.\r\n" );
document.write( "Some people know a way to expand a 3×3 determinant and stop there\r\n" );
document.write( "and expand by the method they know. Most people stop with a 2×2 \r\n" );
document.write( "determinant and expand it because they know just to subtract the \r\n" );
document.write( "product of the diagonals.  I just wanted to show you that it could \r\n" );
document.write( "be taken all the way down to a 1×1 determinant.  The answer will \r\n" );
document.write( "be 102 regardless of whether you stop along the way and expand or \r\n" );
document.write( "go all the way to a 1×1 determinant]\r\n" );
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document.write( "Edwin
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