Next we find the inverse of the coefficient matrix. One way to do this is to append the identity matrix:
and then use elementary row operations to, in effect, transfer the identity matrix to the front. Adding twice the first row to the second row and adding -3 times the first row to the third row:
Multiply row 2 by 1/7:
Add -2 times the second row to the first row and 8 times the second row to the third row:
Row 3 times 7/27:
At this point I gave up, mostly because the fractions were getting uglier and uglier. Fractions are not necessarily wrong. But most of the time problems are assigned where the fractions don't get too nasty. This led me to think that there was an error in either what you posted or in my work (even though I have checked it several times). Either way I did not think it worth much to either of us for me to continue.
So what I propose for you to do is this:
Double check what you posted. Make sure everything is correct, especially the signs. If there was a mistake, then re-post the corrected problem.
Double check the work I have done above. I think I have explained what I was doing at each step so you should be able to check my work. With a "fresh eye" you might be able to find an error I have not been able to find. If you do find an error in my work...
Correct it.
Re-do the steps from the error to where I left off.
Continue from where I left off:
Add appropriate multiples of the third row to each of the first two rows with the goal of turning the 3rd column in those two rows into zeros (like I used the first row to clear out the 1st column and the second row to clear out the 2nd column.
The first three columns should now be the identity matrix. The last three columns will be the inverse of the coefficient matrix.
Multiply the inverse matrix by the column matrix:
in this order:
inverse matrix * column matrix
The result will be a column matrix of the same size. The elements of this matrix are the solutions for x, y and z (in that order).
I'm sorry I was unable to be of more help with this.
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