document.write( "Question 154259: Help evaluate the determinate? Please explain everything so I can comprehend this...please.\r
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\n" ); document.write( "\n" ); document.write( "|1 -3 2 0|
\n" ); document.write( "|-3 -1 0 -2|
\n" ); document.write( "|2 1 3 1|
\n" ); document.write( "|2 0 -2 0 \n" ); document.write( "
Algebra.Com's Answer #113551 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
Help evaluate the determinate? Please explain everything so I can comprehend this...please
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document.write( "| 1  -3   2   0|\r\n" );
document.write( "|-3  -1   0  -2|\r\n" );
document.write( "| 2   1   3   1|\r\n" );
document.write( "| 2   0  -2   0|\r\n" );
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document.write( "Pick a row of column, with the most 0's in it\r\n" );
document.write( "that you can find, and use row and/or column \r\n" );
document.write( "operations to get all but one 0 in that row \r\n" );
document.write( "or column.\r\n" );
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document.write( "The red row below already has two 0's,\r\n" );
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document.write( "| 1  -3   2   0|\r\n" );
document.write( "|-3  -1   0  -2|\r\n" );
document.write( "| 2   1   3   1|\r\n" );
document.write( "| 2   0  -2   0|\r\n" );
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document.write( "and all we have to do to get a 0 where that \r\n" );
document.write( "red -2 is on the bottom row, is to add the \r\n" );
document.write( "numbers in the first column to the corresponding\r\n" );
document.write( "numbers in the third column.  We get this:\r\n" );
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document.write( "| 1  -3   2+1   0|\r\n" );
document.write( "|-3  -1   0-3  -2|\r\n" );
document.write( "| 2   1   3+2   1|\r\n" );
document.write( "| 2   0  -2+2   0|\r\n" );
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document.write( "or\r\n" );
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document.write( "| 1  -3   3   0|\r\n" );
document.write( "|-3  -1  -3  -2|\r\n" );
document.write( "| 2   1   5   1|\r\n" );
document.write( "| 2   0   0   0|\r\n" );
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document.write( "Now all the numbers on the bottom row are 0's\r\n" );
document.write( "all except for one, the 2 in the lower left \r\n" );
document.write( "corner.\r\n" );
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document.write( "So let's cross out all the other elements in the \r\n" );
document.write( "same row and column that that 2 is in:\r\n" );
document.write( "\r\n" );
document.write( "| 1  -3   3   0|\r\n" );
document.write( "|-3  -1  -3  -2|\r\n" );
document.write( "| 2   1   5   1|\r\n" );
document.write( "| 2   0   0   0|  \r\n" );
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document.write( "Now multiply that 2 by the 3x3 determinant formed\r\n" );
document.write( "by the 9 numbers in the upper right:\r\n" );
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document.write( "     |-3   3   0|\r\n" );
document.write( "( )2×|-1  -3  -2|\r\n" );
document.write( "     | 1   5   1|\r\n" );
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document.write( "But we must now check the sign scheme to see whether\r\n" );
document.write( "we keep the positive sign of the 2 or whether we must\r\n" );
document.write( "change its sign to get its opposite, -2. \r\n" );
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document.write( "This is the sign scheme for the 4x4 determinant.\r\n" );
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document.write( "Notice that since the 2 is in the bottom left hand\r\n" );
document.write( "corner and there is a - in the bottom left-corner in\r\n" );
document.write( "the sign scheme, we must change the sign of the 2, to\r\n" );
document.write( "a -2. (If the 2 had been in a position where there is a \r\n" );
document.write( "+ sign in the sign scheme, we would have have just kept\r\n" );
document.write( "the sign and used 2.)\r\n" );
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document.write( "So we have:\r\n" );
document.write( "\r\n" );
document.write( "    |-3   3   0|\r\n" );
document.write( " -2×|-1  -3  -2|\r\n" );
document.write( "    | 1   5   1|\r\n" );
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document.write( "Now we have to expand that determinant.  So we\r\n" );
document.write( "pick a row with the most 0's in it, and use row \r\n" );
document.write( "and/or column operations to get all but one 0 \r\n" );
document.write( "in that row or column.\r\n" );
document.write( "\r\n" );
document.write( "The top (red) row below already has one 0 already,\r\n" );
document.write( "\r\n" );
document.write( "    |-3   3   0|\r\n" );
document.write( " -2×|-1  -3  -2|\r\n" );
document.write( "    | 1   5   1|\r\n" );
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document.write( "and all we have to do to get a 0 where that \r\n" );
document.write( "upper left 3 in on the top row, is to add the \r\n" );
document.write( "numbers in the second column to the corresponding\r\n" );
document.write( "numbers in the first column. \r\n" );
document.write( "\r\n" );
document.write( "    |-3   3   0|\r\n" );
document.write( " -2×|-1  -3  -2|\r\n" );
document.write( "    | 1   5   1|\r\n" );
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document.write( "    |-3+3   3   0|\r\n" );
document.write( " -2×|-1-3  -3  -2|\r\n" );
document.write( "    | 1+5   5   1|\r\n" );
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document.write( "    | 0   3   0|\r\n" );
document.write( " -2×|-4  -3  -2|\r\n" );
document.write( "    | 6   5   1|\r\n" );
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document.write( "So let's cross out all the other elements in the \r\n" );
document.write( "same row and column that that 3 is in:\r\n" );
document.write( "\r\n" );
document.write( "    | 0    3  0|\r\n" );
document.write( " -2×|-4  -3  -2|\r\n" );
document.write( "    | 6   5   1|\r\n" );
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document.write( "Now multiply that 3 by the 2x2 determinant formed\r\n" );
document.write( "by the 4 numbers that haven't been crossed out,\r\n" );
document.write( "so we have:\r\n" );
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document.write( "  -2*( )3*\"abs%28matrix%282%2C2%2C-4%2C-2%2C6%2C1%29%29\"\r\n" );
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document.write( "But we must now check the sign scheme to see whether\r\n" );
document.write( "we keep the positive sign of the 3 or whether we must\r\n" );
document.write( "change its sign to get its opposite, -3. \r\n" );
document.write( "\r\n" );
document.write( "This is the 3x3 sign scheme.\r\n" );
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document.write( "Notice that since the 3 is in the middle of the top row,\r\n" );
document.write( "and there is a + in the middle of the top row in the sign\r\n" );
document.write( "scheme, we change the sign of the 3 to -3. \r\n" );
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document.write( "So we have\r\n" );
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document.write( "  \"-2%2A%28-3%29%2Aabs%28matrix%282%2C2%2C-4%2C-2%2C6%2C1%29%29\"\r\n" );
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document.write( "or\r\n" );
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document.write( "  \"6abs%28matrix%282%2C2%2C-4%2C-2%2C6%2C1%29%29\" =\r\n" );
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document.write( "  \"6%2A%28%28-4%29%281%29-%28-2%29%286%29%29=6%2A%28-4%2B12%29+=+6%2A%288%29+=+48\" \r\n" );
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document.write( "Edwin
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