document.write( "Question 1042222: or which value of k does the following system of equation have non trivial solution (k-3)x+y=0,x+(k-3)y=0 \n" ); document.write( "

Algebra.Com's Answer #657208 by ikleyn(52781) $\"\"$ $\"About$

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\n" ); document.write( " $\"highlight%28cross%28or%29%29\"$ At which value of k does the following system of equation have non trivial solution (k-3)x+y=0,x+(k-3)y=0 ?
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document.write( "This wording is very usual, standard and typical for systems of linear equations.\r\n" );
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document.write( "The \"trivial solution\", by the definition, is zero solution (x,y) = (0,0).\r\n" );
document.write( "When they are asking about a non-trivial solution, they actually ask: \r\n" );
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document.write( "    \"at which value of \"k\" the system is degenerated ?\"\r\n" );
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document.write( "In other words, \r\n" );
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document.write( "    \"at which value of \"k\" the system has more than one solution ?\"\r\n" );
document.write( "    ( provided that it always has a zero solution (0,0). )\r\n" );
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document.write( "It depends and it is defined entirely by the value of the determinant of the matrix of this system.\r\n" );
document.write( "If the determinant is non-zero, the system has only trivial zero solution.\r\n" );
document.write( "If the determinant is zero, then there are \"non-trivial\" solutions.\r\n" );
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document.write( "The matrix in this case is ,  and its determinant is .\r\n" );
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document.write( "The condition of zeroing the determinant is \r\n" );
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document.write( " = ,  or   = ,  or  k-3 = +/-,  or k-3 = +/-1,\r\n" );
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document.write( "which implies  k = 4  or  k = 2.\r\n" );
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document.write( "At k = 4 the system becomes \r\n" );
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document.write( "x + y = 0,\r\n" );
document.write( "x + y = 0.\r\n" );
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document.write( "Clearly, it is dependent; represents one straight line; and has infinitely many solutions.\r\n" );
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document.write( "At k = 2 the system becomes \r\n" );
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document.write( "-x + y = 0,\r\n" );
document.write( " x - y = 0.\r\n" );
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document.write( "Again, it is dependent; represents one straight line; and has infinitely many solutions.\r\n" );
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