document.write( "Question 239300: I can't get Cramer's Rule to work for Y. It's OK for X and Z.
\n" ); document.write( "the matrix is:\r
\n" ); document.write( "\n" ); document.write( " 1X + 3Y -1Z = 1
\n" ); document.write( "-2X -6Y + 1Z = -3
\n" ); document.write( " 3X + 5Y -2Z = 4\r
\n" ); document.write( "\n" ); document.write( "The Determinate is 2.
\n" ); document.write( "X = 2, Y = 1, Z = 1 using Cramer's Rule, but this doesn't check\r
\n" ); document.write( "\n" ); document.write( "Using RREF, the solution is 2,0,1 which works.\r
\n" ); document.write( "\n" ); document.write( "When I take the determinate of Y in Cramer's rule, it is not 0 which it has to be to make Y=0.\r
\n" ); document.write( "\n" ); document.write( "what am I doing wrong? \r
\n" ); document.write( "\n" ); document.write( "could it be that Cramer's Rule doesn't work some times?
\n" ); document.write( "Thanks
\n" ); document.write( "Photonjohn
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Algebra.Com's Answer #525482 by richwmiller(17219)\"\" \"About 
You can put this solution on YOUR website!
1x+3y-1z=1
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\n" ); document.write( "3x+5y-2z=4\r
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\n" ); document.write( "\n" ); document.write( "D=-4
\n" ); document.write( "Dx=-8
\n" ); document.write( "Dy=0
\n" ); document.write( "Dz=-4\r
\n" ); document.write( "\n" ); document.write( "x=-8/-4= 2
\n" ); document.write( "y=0/-4= 0
\n" ); document.write( "z=-4/-4= 1\r
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Solved by pluggable solver: Using Cramer's Rule to Solve Systems with 3 variables

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\n" ); document.write( " \"system%281%2Ax%2B3%2Ay%2B-1%2Az=1%2C-2%2Ax%2B-6%2Ay%2B1%2Az=-3%2C3%2Ax%2B5%2Ay%2B-2%2Az=4%29\"
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\n" ); document.write( " First let \"A=%28matrix%283%2C3%2C1%2C3%2C-1%2C-2%2C-6%2C1%2C3%2C5%2C-2%29%29\". This is the matrix formed by the coefficients of the given system of equations.
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\n" ); document.write( " Take note that the right hand values of the system are \"1\", \"-3\", and \"4\" and they are highlighted here:
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\n" ); document.write( " These values are important as they will be used to replace the columns of the matrix A.
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\n" ); document.write( " Now let's calculate the the determinant of the matrix A to get \"abs%28A%29=-4\". To save space, I'm not showing the calculations for the determinant. However, if you need help with calculating the determinant of the matrix A, check out this solver.
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\n" ); document.write( " Notation note: \"abs%28A%29\" denotes the determinant of the matrix A.
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\n" ); document.write( " Now replace the first column of A (that corresponds to the variable 'x') with the values that form the right hand side of the system of equations. We will denote this new matrix \"A%5Bx%5D\" (since we're replacing the 'x' column so to speak).
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\n" ); document.write( " Now compute the determinant of \"A%5Bx%5D\" to get \"abs%28A%5Bx%5D%29=-8\". Again, as a space saver, I didn't include the calculations of the determinant. Check out this solver to see how to find this determinant.
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\n" ); document.write( " To find the first solution, simply divide the determinant of \"A%5Bx%5D\" by the determinant of \"A\" to get: \"x=%28abs%28A%5Bx%5D%29%29%2F%28abs%28A%29%29=%28-8%29%2F%28-4%29=2\"
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\n" ); document.write( " So the first solution is \"x=2\"
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\n" ); document.write( " We'll follow the same basic idea to find the other two solutions. Let's reset by letting \"A=%28matrix%283%2C3%2C1%2C3%2C-1%2C-2%2C-6%2C1%2C3%2C5%2C-2%29%29\" again (this is the coefficient matrix).
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\n" ); document.write( " Now replace the second column of A (that corresponds to the variable 'y') with the values that form the right hand side of the system of equations. We will denote this new matrix \"A%5By%5D\" (since we're replacing the 'y' column in a way).
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\n" ); document.write( " Now compute the determinant of \"A%5By%5D\" to get \"abs%28A%5By%5D%29=0\".
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\n" ); document.write( " To find the second solution, divide the determinant of \"A%5By%5D\" by the determinant of \"A\" to get: \"y=%28abs%28A%5By%5D%29%29%2F%28abs%28A%29%29=%280%29%2F%28-4%29=0\"
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\n" ); document.write( " So the second solution is \"y=0\"
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\n" ); document.write( " Let's reset again by letting \"A=%28matrix%283%2C3%2C1%2C3%2C-1%2C-2%2C-6%2C1%2C3%2C5%2C-2%29%29\" which is the coefficient matrix.
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\n" ); document.write( " Replace the third column of A (that corresponds to the variable 'z') with the values that form the right hand side of the system of equations. We will denote this new matrix \"A%5Bz%5D\"
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\n" ); document.write( " Now compute the determinant of \"A%5Bz%5D\" to get \"abs%28A%5Bz%5D%29=-4\".
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\n" ); document.write( " To find the third solution, divide the determinant of \"A%5Bz%5D\" by the determinant of \"A\" to get: \"z=%28abs%28A%5Bz%5D%29%29%2F%28abs%28A%29%29=%28-4%29%2F%28-4%29=1\"
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\n" ); document.write( " So the third solution is \"z=1\"
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\n" ); document.write( " Final Answer:
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\n" ); document.write( " So the three solutions are \"x=2\", \"y=0\", and \"z=1\" giving the ordered triple (2, 0, 1)
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\n" ); document.write( " Note: there is a lot of work that is hidden in finding the determinants. Take a look at this 3x3 Determinant Solver to see how to get each determinant.
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