document.write( "Question 816353: How do you solve a 3x3 Matrix using Cramers rule. I know there is a way to do it by using diagonals but I have forgotten. \n" ); document.write( "
Algebra.Com's Answer #491515 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
 
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document.write( "See example below
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Solved by pluggable solver: Using Cramer's Rule to Solve Systems with 3 variables

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document.write( "  \"system%281%2Ax%2B-1%2Ay%2B2%2Az=6%2C4%2Ax%2B0%2Ay%2B1%2Az=1%2C1%2Ax%2B4%2Ay%2B1%2Az=-15%29\"
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document.write( "  First let \"A=%28matrix%283%2C3%2C1%2C-1%2C2%2C4%2C0%2C1%2C1%2C4%2C1%29%29\". This is the matrix formed by the coefficients of the given system of equations.
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document.write( "  Take note that the right hand values of the system are \"6\", \"1\", and \"-15\" and they are highlighted here: 
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document.write( "  These values are important as they will be used to replace the columns of the matrix A.
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document.write( "  Now let's calculate the the determinant of the matrix A to get \"abs%28A%29=31\". 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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document.write( "  Notation note: \"abs%28A%29\" denotes the determinant of the matrix A.
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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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document.write( "  Now compute the determinant of \"A%5Bx%5D\" to get \"abs%28A%5Bx%5D%29=0\". 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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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=%280%29%2F%2831%29=0\"
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document.write( "  So the first solution is \"x=0\"
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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%2C-1%2C2%2C4%2C0%2C1%2C1%2C4%2C1%29%29\" again (this is the coefficient matrix).
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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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document.write( "  Now compute the determinant of \"A%5By%5D\" to get \"abs%28A%5By%5D%29=-124\".
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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=%28-124%29%2F%2831%29=-4\"
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document.write( "  So the second solution is \"y=-4\"
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document.write( "  Let's reset again by letting \"A=%28matrix%283%2C3%2C1%2C-1%2C2%2C4%2C0%2C1%2C1%2C4%2C1%29%29\" which is the coefficient matrix.
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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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document.write( "  Now compute the determinant of \"A%5Bz%5D\" to get \"abs%28A%5Bz%5D%29=31\".
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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=%2831%29%2F%2831%29=1\"
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document.write( "  So the third solution is \"z=1\"
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document.write( "  ====================================================================================
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document.write( "  Final Answer:
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document.write( "  So the three solutions are \"x=0\", \"y=-4\", and \"z=1\" giving the ordered triple (0, -4, 1)
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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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