Since the system is inconsistent, there is no solution vector x that satisfies all three equations simultaneously. Therefore, we cannot minimize ||Ax – b|| 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 =
[ -1 2 ]
[ -2 5 ]
[ -2 4 ]

b =
[ 5 ]
[ 2 ]
[ 5 ]

Therefore, we can calculate:

A^T A =
[ 9 -20 ]
[ -20 45 ]

A^T b =
[ -19 ]
[ 40 ]

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

(A^T A)^(-1) =
[ 9 4 ]
[ 4 9/5 ]

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

x = (A^T A)^(-1) A^T b =
[ -11 ]
[ -4 ]

Therefore, the vector x that minimizes ||Ax – b|| is x = [ -11, -4 ]^T.