Question

1. Express the product Ax as a linear combination of the column vectors of A.
Close and open parenthesis 4 1 Close and open Parenthesis 2
-2 3 3

2. Suppose that the solution set of the homogeneous system Ax = 0 is given by the formulas

x1 = − 5r + 6s, x2 = r − 3s, x3 = r, x4 = s

Find a vector form of the general solution of Ax = 0.

3. Suppose that x1 = − 3, x2 = 4, x3 = 1, x4 = − 2 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 = − 5r + 6s, x2 = r − 4s, x3 = r, x4 = s. Find a vector form of the general solution of Ax = b.

Answer 1

1.

$$A=\begin{pmatrix} 2 & -2 \\ 3 & 3 \end{pmatrix}, x=\begin{pmatrix} 4 \\ 1 \end{pmatrix}$$

$$Ax=\begin{pmatrix} 2 & -2 \\ 3 & 3 \end{pmatrix}\begin{pmatrix} 4 \\ 1 \end{pmatrix}=\begin{pmatrix} 2\cdot4+(-2)\cdot1 \\ 3\cdot4+3\cdot1 \end{pmatrix}=$$

$$=4\begin{pmatrix} 2 \\ 3 \end{pmatrix}+1\begin{pmatrix} -2 \\ 3 \end{pmatrix}$$

2.

The solution set of the homogeneous system Ax = 0 is given by the formulas

$$x_1=-5r+6s, x_2=r-3s,x_3=r,x_4=s$$

The solution can be written in matrix form. Then

$$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}=\begin{bmatrix} -5r+6s \\ r-3s \\ r \\ s \end{bmatrix}=r\begin{bmatrix} -5 \\ 1 \\ 1 \\ 0 \end{bmatrix}+s\begin{bmatrix} 6 \\ -3 \\ 0 \\ 1 \end{bmatrix}$$

Then

$$x=r\begin{bmatrix} -5 \\ 1 \\ 1 \\ 0 \end{bmatrix}+s\begin{bmatrix} 6 \\ -3 \\ 0 \\ 1 \end{bmatrix}$$

3.

The solution set of the homogeneous system Ax = 0 is given by the formulas

$$x_1=-5r+6s, x_2=r-4s,x_3=r,x_4=s$$

The solution can be written in matrix form. Then

$$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}=\begin{bmatrix} -5r+6s \\ r-4s \\ r \\ s \end{bmatrix}=r\begin{bmatrix} -5 \\ 1 \\ 1 \\ 0 \end{bmatrix}+s\begin{bmatrix} 6 \\ -4 \\ 0 \\ 1 \end{bmatrix}$$

The solution set of the non homogeneous system Ax = b is given by the formulas

$$x_1=-3, x_2=4,x_3=1,x_4=-2$$

The solution can be written in matrix form. Then

$$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}=\begin{bmatrix} -3 \\ 4 \\ 1 \\ -2 \end{bmatrix}$$

Hence the general solution is

$$x=\begin{bmatrix} -3 \\ 4 \\ 1 \\ -2 \end{bmatrix}+r\begin{bmatrix} -5 \\ 1 \\ 1 \\ 0 \end{bmatrix}+s\begin{bmatrix} 6 \\ -4 \\ 0 \\ 1 \end{bmatrix}$$