document.write( "Question 1042312: determine the value of a so that the system
\n" ); document.write( "x+y-z=1
\n" ); document.write( "2x+3y+az=3
\n" ); document.write( "x+ay+3z=2
\n" ); document.write( " has
\n" ); document.write( "1)infinitely many solution
\n" ); document.write( "2)no solution
\n" ); document.write( "3)unique solution
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Algebra.Com's Answer #657363 by ikleyn(52778) $\"\"$ $\"About$

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\n" ); document.write( "determine the value of a so that the system
\n" ); document.write( " x + y - z = 1
\n" ); document.write( "2x + 3y + az = 3
\n" ); document.write( " x + ay + 3z = 2
\n" ); document.write( "has
\n" ); document.write( "1)infinitely many solution
\n" ); document.write( "2)no solution
\n" ); document.write( "3)unique solution
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document.write( " x +  y -  z = 1,   (1)\r\n" );
document.write( "2x + 3y + az = 3,   (2)\r\n" );
document.write( " x + ay + 3z = 2.   (3)\r\n" );
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document.write( "One way is to calculate the determinant of the matrix on the left and equalize it to zero.\r\n" );
document.write( "But calculating the determinant of the 3x3 matrix was, probably, not what you were dreamed on.\r\n" );
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document.write( "So, it is better to reduce the given 3x3 system to a 2x2 system first.\r\n" );
document.write( "How we will do it?  But of course, by eliminating \"x\".\r\n" );
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document.write( "So, multiply the equation (1) by 2 (both sides) and then distract it from the equation (2). You will get\r\n" );
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document.write( "(3y-2y) + (az-(-2z)) = 3-2,   or\r\n" );
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document.write( "y + (a+2)z = 1.            (4)\r\n" );
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document.write( "Next, distract the equation (1) from the equation (3) (both sides). You will get\r\n" );
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document.write( "(ay-y) + (3z - (-z)) = 2-1,   or\r\n" );
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document.write( "(a-1)y + 4z = 1.           (5)\r\n" );
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document.write( "Let us write the equations (4) and (5) as a system, one under another.\r\n" );
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document.write( "     y + (a+2)z = 1.       (4')\r\n" );
document.write( "(a-1)y +     4z = 1.       (5')\r\n" );
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document.write( "This system is equivalent to (1)-(3) in the obvious sense.\r\n" );
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document.write( "Now the determinant of the matrix on the left is \r\n" );
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document.write( "det(A) = 4 - (a+2)*(a-1) = \r\n" );
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document.write( "The theory says:  the system (4'), (5') has  a unique solution               \r\n" );
document.write( "                  if and only if the determinant det(A) is not zero.\r\n" );
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document.write( "                  It has no solution OR infinitely many solution    \r\n" );
document.write( "                  if and and only if the determinant det(A) is zero.\r\n" );
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document.write( "So, let us find the singular values of \"a\" that are the roots of this quadratic equation:\r\n" );
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document.write( " = .\r\n" );
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document.write( "Factor the left side, and you will get\r\n" );
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document.write( "(a+3)*(a-2) = 0\r\n" );
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document.write( "with the roots a = -3  and  a = 2.\r\n" );
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document.write( "At a = -3 the equations (4'), (5') become\r\n" );
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document.write( "     y - z = 1.       (6)\r\n" );
document.write( "   -4y + 4z = 1.      (7)\r\n" );
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document.write( "and evidently have no common solution.\r\n" );
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document.write( "At a = 2 the equations (4'), (5') become\r\n" );
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document.write( "     y + 4z = 1.      (8)\r\n" );
document.write( "     y + 4z = 1.      (9)\r\n" );
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document.write( "and evidently have infinitely many solutions.\r\n" );
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document.write( "Answer.  At a =  2 the original system has infinitely many solutions.\r\n" );
document.write( "         At a = -3 the original system has no solutions.\r\n" );
document.write( "         At all other values of \"a\" the original system has a unique solution.\r\n" );
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