Question

Suppose that X1=-1, x2=2, x3=4, x4=-3 is a solution of a non-homogeneous linear system Ax = b and that the solution set of the homogeneous system Ax=0 is given by the formulas:x1= -3r +4s, x2=r-s, x3=r, x3=s

a) Find the vector form of the general solution of Ax=b

b) Find the vector form of the general solution Ax=0

Answer 1

To find the vector form of the general solution of Ax=b, substitute the given values of x1, x2, x3, and x4 into the formulas for the homogeneous system: x1 = -3r + 4s, x2 = r - s, x3 = r, and x4 = s. To find the vector form of the general solution of Ax=0, set the values of x1, x2, x3, and x4 to zero in the formulas for the homogeneous system.

Step-by-step explanation:

a) To find the vector form of the general solution of Ax=b, we can substitute the given values of x1, x2, x3, and x4 into the formulas for the homogeneous system: x1 = -3r + 4s, x2 = r - s, x3 = r, and x4 = s. So, the vector form of the general solution is:

x = -3r + 4s

y = r - s

z = r

s

b) To find the vector form of the general solution of Ax=0, we set the values of x1, x2, x3, and x4 to zero in the formulas for the homogeneous system. So, the vector form of the general solution is:

x = -3r + 4s

y = r - s

z = r

s

Answer 2

The answers will be as simple as writing the component solutions as components of a vector. For instance,
`\mathbf{Ax}=\mathbf 0` is solved by any vector belonging to the set


`\left\{\mathbf x\in\mathbb R^4~:~x_1=-3r+4s,x_2=r-s,x_3=r,x_4=s,(r,s)\in\mathbb R^2\right\}`

or simply as the vector


`\mathbf x=\begin{bmatrix}-3r+4s\\r-s\\r\\s\end{bmatrix}`

The general solution to
`\mathbf{Ax}=\mathbf b` will be the particular solution plus any vector belonging to the nullspace, so that the general solution would take the form


`\mathbf x=\begin{bmatrix}-3r+4s\\r-s\\r\\s\end{bmatrix}+\begin{bmatrix}-1\\2\\4\\-3\end{bmatrix}`

`\mathbf x=\begin{bmatrix}-1-3r+4s\\2+r-s\\4+r\\-3+s\end{bmatrix}`

where
`r,s` are any real numbers.