SOLUTION: What is w,x,y, and z when using matrices to solve the following system? w-3x-2y+z=-3 2w-7x-y+2z=1 3w-7x-3y+3z=-5 5w+x+4y-2z=18

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1w-3x-2y+1z=-3
2w-7x-1y+2z= 1
3w-7x-3y+3z=-5
5w+1x+4y-2z=18

Erase the letters, replace the ='s by |'s,
put parentheses around:



The idea is to end up with a matrix that looks like this:




where there are various numbers where the X's are:



Multiply the 1st row by -2, and add the second row

 

Replace the 2nd row by the result:



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Multiply the 1st row by -3, and add the 3rd row

 

Replace the 3rd row by the result:



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Multiply the 1st row by -5, and add the 4th row

 

Replace the 3rd row by the result:



Multiply the 2nd row by -1 to get a 1 in the 2nd row, 2nd column:



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Multiply the 2nd row by -2, and add the 3rd row

 

Replace the 3rd row by the result:



----------------------------

Multiply the 2nd row by -16, and add the 4th row

 

Replace the 4th row by the result:



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Multiply the 3rd row by 1/9 to get a 1 in the 3rd row, 3rd column:





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Multiply the 3rd row by -62, and add the 4th row

 

Replace the 4th row by the result:



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Multiply the 4th row by -1/7 to get a 1 in the 4th row, 4th column:



Now we have the matrix in the form, called "triangular form",
we convert it back to a system of equations:



Get rid of the unnecessary or understood 0 terms and 1's



We have the values for y and z, so we substitute y=2
into the 2nd equation:



Finally we substitute x=-1, y=2, z=-3 into the 1st equation:



(w,x,y,z) = (1,-1,2,-3)

Edwin