document.write( "Question 618812: Please help use Cramer's rule to solve the linear system.\r
\n" ); document.write( "\n" ); document.write( "( 3x-2y-z=3
\n" ); document.write( " x+2y-5z=-31
\n" ); document.write( " -7x+y+z=25 \n" ); document.write( "
Algebra.Com's Answer #389132 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
 
\n" );
document.write( "Hi,
\n" );
document.write( "Please help use Cramer's rule to solve the linear system.
\n" );
document.write( " 3x-2y-z=3
\n" );
document.write( "  x+2y-5z=-31
\n" );
document.write( "  -7x+y+z=25 
\n" );
document.write( " x= -145/31 y = -288/31 z =  48/31)\r
\n" );
document.write( "\n" );
document.write( "\n" );
document.write( "\n" );
document.write( " \n" );
document.write( "\n" );
document.write( "
Solved by pluggable solver: Using Cramer's Rule to Solve Systems with 3 variables

\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  First let \"A=%28matrix%283%2C3%2C3%2C-2%2C-1%2C1%2C2%2C-5%2C-7%2C1%2C1%29%29\". This is the matrix formed by the coefficients of the given system of equations.
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  Take note that the right hand values of the system are \"3\", \"-31\", and \"25\" and they are highlighted here: 
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  These values are important as they will be used to replace the columns of the matrix A.
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  Now let's calculate the the determinant of the matrix A to get \"abs%28A%29=-62\". 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.
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  Notation note: \"abs%28A%29\" denotes the determinant of the matrix A.
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  ---------------------------------------------------------
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\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).
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  Now compute the determinant of \"A%5Bx%5D\" to get \"abs%28A%5Bx%5D%29=290\". 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.
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\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=%28290%29%2F%28-62%29=-145%2F31\"
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  So the first solution is \"x=-145%2F31\"
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  ---------------------------------------------------------
\n" );
document.write( "  
\n" );
document.write( "  
\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%2C3%2C-2%2C-1%2C1%2C2%2C-5%2C-7%2C1%2C1%29%29\" again (this is the coefficient matrix).
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\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).
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  Now compute the determinant of \"A%5By%5D\" to get \"abs%28A%5By%5D%29=576\".
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\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=%28576%29%2F%28-62%29=-288%2F31\"
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  So the second solution is \"y=-288%2F31\"
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  ---------------------------------------------------------
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  Let's reset again by letting \"A=%28matrix%283%2C3%2C3%2C-2%2C-1%2C1%2C2%2C-5%2C-7%2C1%2C1%29%29\" which is the coefficient matrix.
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\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\" 
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  Now compute the determinant of \"A%5Bz%5D\" to get \"abs%28A%5Bz%5D%29=-96\".
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\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=%28-96%29%2F%28-62%29=48%2F31\"
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  So the third solution is \"z=48%2F31\"
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  ====================================================================================
\n" );
document.write( "  
\n" );
document.write( "  Final Answer:
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  So the three solutions are \"x=-145%2F31\", \"y=-288%2F31\", and \"z=48%2F31\" giving the ordered triple (-145/31, -288/31, 48/31)
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\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.
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( "  
\n" );
document.write( " \n" );
document.write( "
\n" );