Question 116336
I need help with these two matrice problems- solving it by the Gaussian elimination method: 
problem 1.)
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2x + 1y - 3z = 1 
3x - 1y + 4z = 6 
1x + 2y - 1z = 9

Abbreviate this with an augmented matrix:

[2   1  -3  | 1] 
[3  -1   4  | 6] 
[1   2  -1  | 9]

Get a zero where the 3 is by
multiplying the top row by -3,
and the middle row by 2:

[-6   -3   9  | -3] 
[ 6   -2   8  | 12] 
[ 1    2  -1  |  9]

Add the top row to the middle row:

[-6   -3   9  | -3] 
[ 0   -5  17  |  9] 
[ 1    2  -1  |  9]

Get a zero where the 1 is by
multiplying the top row by 1,
and the bottom row by 6:

[-6   -3   9  | -3] 
[ 0   -5  17  |  9] 
[ 6   12  -6  | 54]

Add the top row to the bottom row:

[-6   -3   9  | -3] 
[ 0   -5  17  |  9] 
[ 0    9   3  | 51]

Get a zero where the 9 is on
the bottom row by multiplying 
the middle row by 9,
and the bottom row by 5:

[-6   -3   9  |  -3] 
[ 0  -45 153  |  81] 
[ 0   45  15  | 255]

Add the middle row to the bottom row:

[-6   -3   9  |  -3] 
[ 0  -45 153  |  81] 
[ 0    0 168  | 336]

Rewrite as a new system of equations:

 -6x - 3y +   9z =  -3 
     -45y + 153z =  81 
            168z = 336

Solve the bottom equation for z:

            168z = 336
               z = {{{336/168}}}
               z = 2 

Substitute z = 2 in the middle equation 
and solve for y:

     -45y + 153z = 81
   -45y + 153(2) = 81
      -45y + 306 = 81
            -45y = -225       
               y = {{{(-255)/(-45)}}} 
               y = 5

Substitute y = 5 and z = 2 in the top equation 
and solve for x:

    -6x - 3y + 9z = -3
-6x - 3(5) + 9(2) = -3
    -6x - 15 + 18 = -3
          -6x + 3 = -3
              -6x = -6
                x = {{{(-6)/(-6)}}}
                x = 1

So the solution is (x, y, z) = (1, 5, 2)

Edwin</pre></font>