As @Foon pointed out, the canonical way to do this is to subtract a column.
However, on a side note, as your problem is overdetermined, you have to use a method such as least squares. By definition, if it's an overdetermined problem, there is no "unique, exact solution". (Otherwise it would be even-determined - A square matrix.)
That aside, here's how you'd go about it:
Let's take your example equation:
|1 0 0 1 0| |n1| |B1|
|1 0 0 0 1| |n2| |B2|
|0 1 0 1 0| X |n3| = |B3|
|0 1 0 0 1| |t1| |B4|
|0 0 1 1 0| |t2| |B5|
|0 0 1 0 1| |B6|
As you noted, this is overdetermined. If we know one of our "model" variables (let's say n1 in this case), it will be even more overdetermined. It's not a problem, but it means we'll need to use least squares, and there isn't a completely unique solution.
So, let's say we know what n1 should be.
In that case, we'd re-state the problem by subtracting n1 multiplied by the first column in the solution matrix from our vector of observations (This is what @Foon suggested):
|0 0 1 0| |n2| |B1 - n1|
|0 0 0 1| |n3| |B2 - n1|
|1 0 1 0| X |t1| = |B3 - 0 |
|1 0 0 1| |t2| |B4 - 0 |
|0 1 1 0| |B5 - 0 |
|0 1 0 1| |B6 - 0 |
Let's use a more concrete example in numpy terms. Let's solve the equation y = Ax^2 + Bx + C. To start with, let's generate our data and "true" model parameters:
import numpy as np
# Randomly generate two of our model variables
a, c = np.random.rand(2)
b = 1
x = np.linspace(0, 2, 6)
y = a * x**2 + b * x + c
noise = np.random.normal(0, 0.1, y.size)
y += noise
First, we'll solve it _without) the knowledge that B = 1. We could use np.polyfit for this, but to lead into the next bit, we'll use a lower-level approach:
# I'm a geophysist, so I tend to use Gm=d instead of Ax=b
G = np.column_stack([x**2, x, np.ones_like(x)])
d = y
m, _, _, _ = np.linalg.lstsq(G, d)
print "Ideally, this would be 1: ", m[1]
As you can see, we'll get something close to, but not quite 1. In this case (I didn't set the seed, so this will vary), the model parameters returned are
[ 0.13392633, 0.97217035, 0.33645734]
While the true parameters are:
[ 0.14592752, 1. , 0.31349185]
Now let's take the fact that we know b exactly into account. We'll make a new G with one less column and subtract that column times b from our observations (d/y):
G = np.column_stack([x**2, np.ones_like(x)])
d = y - b * x
m, _, _, _ = np.linalg.lstsq(G, d)
Now m is [a, c] and we've solved for those two variables using our knowledge of b.
