Since the system is inconsistent, there is no solution vector x that satisfies all three equations simultaneously. Therefore, we cannot minimize ||b – Ax|| by finding a solution x. Instead, we need to find the vector x that is the closest possible solution to the inconsistent system.

One way to do this is to use the least squares method. The idea is to find the vector x that minimizes the squared length of the residual vector r = Ax – b, where the residual vector is the vector of differences between the left-hand side and right-hand side of each equation. In other words, we want to minimize ||r||^2.

To find the vector x that minimizes ||r||^2, we can use the normal equations: A^T A x = A^T b. Here, A^T is the transpose of matrix A, and A^T A is a symmetric matrix. The normal equations have the property that if x is a solution, then it is the unique solution that minimizes ||r||^2.

A =
[ 5 3 ]
[ 4 2 ]
[ -2 -1 ]

b =
[ 1 ]
[ 3 ]
[ 3 ]

Therefore, we can calculate:

A^T A =
[ 45 25 ]
[ 25 14 ]

A^T b =
[ 11 ]
[ 6 ]

We can solve the normal equations by computing the inverse of A^T A, which is:

(A^T A)^(-1) =
[ 14/5 -5 ]
[ -5 9 ]

Multiplying both sides by (A^T A)^(-1) and (A^T b), we get:

x = (A^T A)^(-1) A^T b =
[ 4/5 ]
[ -1 ]

Therefore, the vector x that minimizes ||Ax – b|| is x = [ 4/5, -1 ]^T.