document.write( "Question 464642: Solve using Cramer's Rule.
\n" ); document.write( " x + 3y + 5z = 6
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\n" ); document.write( "9x - 6y + 3z = 3\r
\n" ); document.write( "\n" ); document.write( "I came up with (1,-2,1) but Cramers rule confuses me so I don't know if I did it right
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Algebra.Com's Answer #318297 by MathLover1(20850)\"\" \"About 
You can put this solution on YOUR website!
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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%2B5%2Az=6%2C2%2Ax%2B-4%2Ay%2B6%2Az=14%2C9%2Ax%2B-6%2Ay%2B3%2Az=3%29\"
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\n" ); document.write( " First let \"A=%28matrix%283%2C3%2C1%2C3%2C5%2C2%2C-4%2C6%2C9%2C-6%2C3%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 \"6\", \"14\", and \"3\" 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=288\". 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=-288\". 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=%28-288%29%2F%28288%29=-1\"
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\n" ); document.write( " So the first solution is \"x=-1\"
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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%2C5%2C2%2C-4%2C6%2C9%2C-6%2C3%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=-288\".
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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=%28-288%29%2F%28288%29=-1\"
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\n" ); document.write( " So the second solution is \"y=-1\"
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\n" ); document.write( " Let's reset again by letting \"A=%28matrix%283%2C3%2C1%2C3%2C5%2C2%2C-4%2C6%2C9%2C-6%2C3%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=576\".
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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=%28576%29%2F%28288%29=2\"
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\n" ); document.write( " So the third solution is \"z=2\"
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\n" ); document.write( " Final Answer:
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\n" ); document.write( " So the three solutions are \"x=-1\", \"y=-1\", and \"z=2\" giving the ordered triple (-1, -1, 2)
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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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