How do you solve the system #x+2y+z=2#, #2x+3y+3z=-3#, and #2x+3y+2z=2#?

We have the equation:

# { (x+2y+z=2), (2x+3y+3z=-3), (2x+3y+2z=2) :} #

In vector matrix form we can write this as:

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

Or as:

# bb(A) bb(ul x) = bb(ul b) #

Where:

# bb(A) = ( (1,2,1), (2,3,3), (2,3,2) ); bb(ul x) = ( (x), (y), (z) ); bb(ul b) = ( (2), (-3), (2) ) #

So then, the solution can be found by inverting the matrix # bb(A)#, to get:

# bb(ul x) = bb(A)^-1 bb(ul b) #

To invert the matrix A, first we compute the matrix of cofactors:

# Cof(bb(A)) = ( (+| (3,3), (3,2) |,-| (2,3), (2,2) |,+| (2,3), (2,3) |), (-| (2,1), (3,2) |,+| (1,1), (2,2) |,-| (1,2), (2,3) |), (+| (2,1), (3,3) |,-| (1,1), (2,3) |,+| (1,2), (2,3) |) ) #

# " " = ( (6-9,-(4-6),6-6), (-(4-3),2-2,-(3-4)), (6-3,-(3-2),3-4) ) #

# " " = ( (-3,2,0), (-1,0,1), (3,-1,-1) ) #

We then compute the adjoint of #A# or #bb(C)^T#

# adj(bb(A)) = ( (-3,-1,3), (2,0,-1), (0,1,-1) ) #

We must also compute the determinant of #bb(A)#:

# det(bb(A)) = (1) | (3,3), (3,2) | - (2) | (2,3), (2,2) | + (1) | (2,3), (2,3) | #
# " " = (6-9) - (2)(4-6) + (6-6) #
# " " = -3+4 #
# " " = 1 #

And so we get the inverse using:

# bb(A)^-1 = 1/det(bb(A)) adj(bb(A)) #
# \ \ \ = 1 * adj(bb(A)) #
# \ \ \ = ( (-3,-1,-3), (2,0,-1), (0,1,-1) ) #

Thus, the solution of the linear system is:

# bb(ul x) = ( (-3,-1,3), (2,0,-1), (0,1,-1) ) ( (2), (-3), (2) ) #

# \ \ \ = ( (-6+3+6), (4+0-2), (0-3-2) ) #
# \ \ \ = ( (3), (2), (-5) ) #

Hence the solution is:

# x=3, y=2, z=-5 #