Question

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

Answer 1

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

Explanation:

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`