Final answer:
The basis for the solution space of the given linear system is {(-1, 1, 0), (1, -2, 1)}, and the dimension of the solution space is 2.
Step-by-step explanation:
To find a basis for and the dimension of the solution space of the given system of linear equations, we first write down the augmented matrix of the system:
[ -1, 1, 1 | 0 ]
[ 2, -1, 0 | 0 ]
[ 3, -4, -5 | 0 ]
Next, we perform row operations to row-reduce the matrix:
[ 1, -1, -1 | 0 ]
[ 0, 1, 2 | 0 ]
[ 0, 0, 0 | 0 ]
From the row-reduced echelon form, we can see that there are two pivot columns (the first and second), corresponding to the variables x and y. The solution space is then parametrized by the free variable z, so the basis for the solution space is {(-1, 1, 0), (1, -2, 1)}, and the dimension of the solution space is 2.