The least squares solution of the system of equations x+y=1, x+2y=3,x+3y=3 and x+4y=5 isA. (0,1.2)B. (0,1.3)C. (0,1.1)D. (1,0)

The least squares solution to a system of equations is found by minimizing the sum of the squares of the residuals. The residuals are the differences between the observed and predicted values.

Here, we have a system of four equations and two variables (x and y). We can write these equations in matrix form as follows:

A = [[1, 1], [1, 2], [1, 3], [1, 4]] b = [1, 3, 3, 5]

The least squares solution is given by the normal equation:

x = (A^T * A)^-1 * A^T * b

where A^T is the transpose of A, and (A^T * A)^-1 is the inverse of (A^T * A).

Let's calculate:

A^T * A = [[4, 10], [10, 30]] A^T * b = [12, 34]

The inverse of A^T * A is [[3, -1], [-1, 0.4]]

So, x = [[3, -1], [-1, 0.4]] * [12, 34] = [0, 1.2]

Therefore, the least squares solution of the system of equations is (0, 1.2), which corresponds to option A.