document.write( "Question 1129413: Evaluate the determinant, using row or column operations whenever possible to simplify your work.\r
\n" ); document.write( "\n" ); document.write( "[0 0 4 6]
\n" ); document.write( "[2 1 1 3]
\n" ); document.write( "[2 1 2 3]
\n" ); document.write( "[3 0 1 7]
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Algebra.Com's Answer #745986 by t0hierry(194)\"\" \"About 
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\n" ); document.write( "[0 0 4 6]
\n" ); document.write( "[2 1 1 3]
\n" ); document.write( "[2 1 2 3]
\n" ); document.write( "[3 0 1 7]\r
\n" ); document.write( "\n" ); document.write( "We start from the first row and we get:
\n" ); document.write( "4* [2 1 3]
\n" ); document.write( " [2 1 3]
\n" ); document.write( " [3 0 7]
\n" ); document.write( "6* [2 1 1]
\n" ); document.write( " [2 1 2]
\n" ); document.write( " [3 0 1]
\n" ); document.write( "again from last row for the first term
\n" ); document.write( "3*0 + 7*0 = 0
\n" ); document.write( "same for the second term\r
\n" ); document.write( "\n" ); document.write( "3*1 + 0 = 3
\n" ); document.write( "Now for the sign: it's minus\r
\n" ); document.write( "\n" ); document.write( "so I find -3.
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