Show that the two systems Ax=b RAx=Rb

[orange12]

Hey. Got some troubles with this.

4 1 11
A= 1 3 11
0 2 6
1 1 5


1 -1 2 -2
R= -1 2 -3 2
2 -3 5 -5
-2 2 -5 6

B = 2 -3 6 -5
-2 2 -5 6

1. show that the linear system is Ax=b is only consistent if b € N(B)

2. let b € R4. show that the two systems

Ax=b RAx=Rb

are equivalent

hope somebody can help?

 

[HallsofIvy]

$$A= \begin{bmatrix}4& 1& 11 \\ 1& 3& 11 \\ 0& 2& 6 \\1& 1& 5\end{bmatrix}$$


$$B= \begin{bmatrix}2 & -3 & 6 & -5 \\-2 & 2 & -5 & 6 \end{bmatrix}$$

Row reduce
$$\begin{bmatrix}4& 1& 11 & a\\ 1& 3& 11 & b \\ 0& 2& 6 & c\\1& 1& 5 & d\end{bmatrix}$$
and you will get something like
$$\begin{bmatrix}1& 1& 5 & d\\ 0& 2& 6 & c \\ 0& 0 & 0 & b-d-c\\0& 0 & 0 & a- 4d+ \frac{3}{2}c\end{bmatrix}$$
In order that that be solvable we must have b- d- c= 0 and a- 4d+ (3/2)c= 0. Those lead to a= 4d- (3/2)c and b= c+ d so the space of all such b is spanned by <4, 1, 0 1> and <-3, 2, 0, 1> (by taking d=1, c=0 and then d=0, c= 2).

If you row reduce
$$\begin{bmatrix}2 & -3 & 6 & -5 & 0 \\-2 & 2 & -5 & 6 & 0 \end{bmatrix}$$
you get something like
$$\begin{bmatrix} 2 & - 3 & 6 & -5 \\0 & -1 & 1 & 1\end{bmatrix}$$

The Null space of B is given by <a, b, c, d> with 2a- 3b+ 6c- 5d= 0 and -b+ c+ d= 0.

Show that the solutions to those equations are exactly the space above.

$$R= \begin{bmatrix}1 & -1 & 2 & -2 \\ -1 & 2 & -3 & 2 \\ 2 & -3 & 5 & -5\\ -2 & 2 & -5 & 6\end{bmatrix}$$
That is a square matrix so if you can show it has an inverse, all you have to do is multiply, on the left, on both sides of RAx= Rb by the inverse of R!

 

[orange12]

$$A= \begin{bmatrix}4& 1& 11 \\ 1& 3& 11 \\ 0& 2& 6 \\1& 1& 5\end{bmatrix}$$


$$B= \begin{bmatrix}2 & -3 & 6 & -5 \\-2 & 2 & -5 & 6 \end{bmatrix}$$

Row reduce
$$\begin{bmatrix}4& 1& 11 & a\\ 1& 3& 11 & b \\ 0& 2& 6 & c\\1& 1& 5 & d\end{bmatrix}$$
and you will get something like
$$\begin{bmatrix}1& 1& 5 & d\\ 0& 2& 6 & c \\ 0& 0 & 0 & b-d-c\\0& 0 & 0 & a- 4d+ \frac{3}{2}c\end{bmatrix}$$
In order that that be solvable we must have b- d- c= 0 and a- 4d+ (3/2)c= 0. Those lead to a= 4d- (3/2)c and b= c+ d so the space of all such b is spanned by <4, 1, 0 1> and <-3, 2, 0, 1> (by taking d=1, c=0 and then d=0, c= 2).

If you row reduce
$$\begin{bmatrix}2 & -3 & 6 & -5 & 0 \\-2 & 2 & -5 & 6 & 0 \end{bmatrix}$$
you get something like
$$\begin{bmatrix} 2 & - 3 & 6 & -5 \\0 & -1 & 1 & 1\end{bmatrix}$$

The Null space of B is given by <a, b, c, d> with 2a- 3b+ 6c- 5d= 0 and -b+ c+ d= 0.

Show that the solutions to those equations are exactly the space above.

$$R= \begin{bmatrix}1 & -1 & 2 & -2 \\ -1 & 2 & -3 & 2 \\ 2 & -3 & 5 & -5\\ -2 & 2 & -5 & 6\end{bmatrix}$$
That is a square matrix so if you can show it has an inverse, all you have to do is multiply, on the left, on both sides of RAx= Rb by the inverse of R!

Thx for the help. I can see how a= 4d- (3/2)c and b= c+ d, but why do u choose d=1 c=0 at first, and the d=0 and c=2. And how is The Null space of B is given by <a, b, c, d> with 2a- 3b+ 6c- 5d= 0 and -b+ c+ d= 0. I can't seem to figure this part out, but i understand the answer to part two. Hope u don't mind taking another look, thanks again.

 

[orange12]

If I row reduce to reduced echelon form with the total matrix [B 0 ] I do not get the same as u get. I get it to be:

1 0 3/2 -4 0
0 1 -1 -1 0