Linear independent and combination

[vcb003104]

Homework Statement

let r be an element of R
... 1.... 1 ......r^2.....3 + 2r
u =( 1 )...v = ( 4 )...w = (1 )...b = ( 5 + 12r)
...0.....1......r^2 ...... 2r


(sorry don't know how to type matrices)

1. For which values of r is the set {u, v, w} linearly independent?

2. For which values of r is the vector b a linear combination of u. v, w?

3. For which of these values of r can b be written as a linear combination of u, v and w in more than one way

Homework Equations

(the matrices)

The Attempt at a Solution

So for 1, I reduced the matrix ( u v w ) to become something like this:

_1_1_r^2
(_1_4_1)
0_1_r^2

__1_1_r^2
→ (0_1_r^2)
__0_0_1-4r^2


so for it to be linearly independent 1 - $4r^2$ =/= 0

so r =/= $\pm 1/2$

for part b)

We want something like:

c1(u) + c2(v) + c3(w) = b

I reduced everything and got:


c1 = 3
c2 = 2r - $r^2$ (2/(1 - 2r))
c3 = $\frac{2(1 + 2r)}{(1 + 2r)(1 - 2r)}$

Is it alright to say that for it to be a linear comb. r can't = $\pm1/2$? (Is it correct that I didn't cancel out the 1 + 2r ?)

and I don't really get part C

 

[HallsofIvy]

Homework Statement

let r be an element of R
... 1.... 1 ......r^2.....3 + 2r
u =( 1 )...v = ( 4 )...w = (1 )...b = ( 5 + 12r)
...0.....1......r^2 ...... 2r

Using the "tex" and "/tex" tags (in "[" and "]")
"u= \ begin{pmatrix}1 \\ 1 \\ 0 \ end{pmatrix} v= \ begin{pmatrix}1 \\ 4 \\ 1 \ end{pmatrix} w= \ begin{pmatrix}r^2 \\ 1 \\ r^2\ end{pmatrix} b= \ begin{pmatrix}3+ 2r \\ 5+ 12r \\ 2r\ end{pmatrix}" (without the spaces) will give

$$u= \begin{pmatrix}1 \\ 1 \\ 0 \end{pmatrix} v= \begin{pmatrix}1 \\ 4 \\ 1 \end{pmatrix} w= \begin{pmatrix}r^2 \\ 1 \\ r^2\end{pmatrix} b= \begin{pmatrix}3+ 2r \\ 5+ 12r \\ 2r\end{pmatrix}$$

(sorry don't know how to type matrices)

1. For which values of r is the set {u, v, w} linearly independent?

2. For which values of r is the vector b a linear combination of u. v, w?

3. For which of these values of r can b be written as a linear combination of u, v and w in more than one way

Homework Equations

(the matrices)

The Attempt at a Solution

So for 1, I reduced the matrix ( u v w ) to become something like this:

_1_1_r^2
(_1_4_1)
0_1_r^2

__1_1_r^2
→ (0_1_r^2)
__0_0_1-4r^2


so for it to be linearly independent 1 - $4r^2$ =/= 0

so r =/= $\pm 1/2$

The basic definition of "linear independent" says that four vectors u, v, w, and b, will be linearly independent if and only if the only way we can have pu+ qv+ sw+ tb= 0 is if p= q= s= t= 0. Here, for any numbers, p, q, s, and t
$$pu+ qv+ sw+ tb= \begin{pmatrix}p \\ p \\ 0 \end{pmatrix}+ \begin{pmatrix}q \\ 4q \\ q \end{pmatrix}+ \begin{pmatrix}sr^2 \\ s \\ sr^2\end{pmatrix}+ \begin{pmatrix}t(3+ 2r) \\ t(5+ 12r) \\ t(2r)\end{pmatrix}= \begin{pmatrix}p+ q+ r^2+ 3t+ 2tr \\ p+ 4q+ s+ 5t+12rt \\ q+ sr^2+ 2tr\end{pmatrix}= \begin{pmatrix}0 \\ 0 \\ 0 \end{pmatrix}$$

so we have the equations $p+ q+ sr^2+ 3t+ 2tr= 0$, $p+ 4q+ s+ 4t+ 12rt= 0$, and $q+ sr^2+ 2tr= 0$ which we must solve for p, q, s, and t ("r" is just a parameter). Since there is no "p" in the last equation we can subtract the first equation from the second to eliminate p from those and have $3q+ s(1- r^2)+ t+ 10rt= 0$ and $q+ sr^2+ 2tr= 0$. Multiply that last equation by 3 and subtract from the previous equation to eliminate q and we have $s(1- 2r^2)+ t(1+ 8r)= 0$. If either $1- 2r^2= 0$ or $1+ 8r=0$ that will be true without s and t being equal to 0 so those vectors will NOT be "linearly independent". For any other value of r, they will be which also answers (b).

for part b)

We want something like:

c1(u) + c2(v) + c3(w) = b

I reduced everything and got:


c1 = 3
c2 = 2r - $r^2$ (2/(1 - 2r))
c3 = $\frac{2(1 + 2r)}{(1 + 2r)(1 - 2r)}$

Is it alright to say that for it to be a linear comb. r can't = $\pm1/2$? (Is it correct that I didn't cancel out the 1 + 2r ?)

and I don't really get part C

"Writing b as a linear combination of u, v, and w" means writing
$$\begin{pmatrix}3+ 2r \\ 5+ 12r \\ 2r\end{pmatrix}= x\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}+ y\begin{bmatrix}1 \\ 4 \\ 1 \end{bmatrix}+ z\begin{bmatrix}r^2 \\ 1 \\ r^2\end{bmatrix}= \begin{bmatrix}x+ y+ zr^2 \\ x+ 4y+ z \\ y+ zr^2\end{bmatrix}$$
for some numbers x, y, and z.

That is the same as the equations $3+ 2r= x+ y+ zr^2$, $5+ 12r= x+ 4y+ z$, and $2r= y+ zr^2$.

The question is "for what values of r do those equations have more than one solution for x, y and z?"

 

[vcb003104]

Using the "tex" and "/tex" tags (in "[" and "]")
"u= \ begin{pmatrix}1 \\ 1 \\ 0 \ end{pmatrix} v= \ begin{pmatrix}1 \\ 4 \\ 1 \ end{pmatrix} w= \ begin{pmatrix}r^2 \\ 1 \\ r^2\ end{pmatrix} b= \ begin{pmatrix}3+ 2r \\ 5+ 12r \\ 2r\ end{pmatrix}" (without the spaces) will give

$$u= \begin{pmatrix}1 \\ 1 \\ 0 \end{pmatrix} v= \begin{pmatrix}1 \\ 4 \\ 1 \end{pmatrix} w= \begin{pmatrix}r^2 \\ 1 \\ r^2\end{pmatrix} b= \begin{pmatrix}3+ 2r \\ 5+ 12r \\ 2r\end{pmatrix}$$
The basic definition of "linear independent" says that four vectors u, v, w, and b, will be linearly independent if and only if the only way we can have pu+ qv+ sw+ tb= 0 is if p= q= s= t= 0. Here, for any numbers, p, q, s, and t
$$pu+ qv+ sw+ tb= \begin{pmatrix}p \\ p \\ 0 \end{pmatrix}+ \begin{pmatrix}q \\ 4q \\ q \end{pmatrix}+ \begin{pmatrix}sr^2 \\ s \\ sr^2\end{pmatrix}+ \begin{pmatrix}t(3+ 2r) \\ t(5+ 12r) \\ t(2r)\end{pmatrix}= \begin{pmatrix}p+ q+ r^2+ 3t+ 2tr \\ p+ 4q+ s+ 5t+12rt \\ q+ sr^2+ 2tr\end{pmatrix}= \begin{pmatrix}0 \\ 0 \\ 0 \end{pmatrix}$$

so we have the equations $p+ q+ sr^2+ 3t+ 2tr= 0$, $p+ 4q+ s+ 4t+ 12rt= 0$, and $q+ sr^2+ 2tr= 0$ which we must solve for p, q, s, and t ("r" is just a parameter). Since there is no "p" in the last equation we can subtract the first equation from the second to eliminate p from those and have $3q+ s(1- r^2)+ t+ 10rt= 0$ and $q+ sr^2+ 2tr= 0$. Multiply that last equation by 3 and subtract from the previous equation to eliminate q and we have $s(1- 2r^2)+ t(1+ 8r)= 0$. If either $1- 2r^2= 0$ or $1+ 8r=0$ that will be true without s and t being equal to 0 so those vectors will NOT be "linearly independent". For any other value of r, they will be which also answers (b). "Writing b as a linear combination of u, v, and w" means writing
$$\begin{pmatrix}3+ 2r \\ 5+ 12r \\ 2r\end{pmatrix}= x\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}+ y\begin{bmatrix}1 \\ 4 \\ 1 \end{bmatrix}+ z\begin{bmatrix}r^2 \\ 1 \\ r^2\end{bmatrix}= \begin{bmatrix}x+ y+ zr^2 \\ x+ 4y+ z \\ y+ zr^2\end{bmatrix}$$
for some numbers x, y, and z.

That is the same as the equations $3+ 2r= x+ y+ zr^2$, $5+ 12r= x+ 4y+ z$, and $2r= y+ zr^2$.

The question is "for what values of r do those equations have more than one solution for x, y and z?"

Hi there, but how do I find the ones that have more than one way of writing it?

Do I solve the matrices for $3+ 2r= x+ y+ zr^2$, $5+ 12r= x+ 4y+ z$, and $2r= y+ zr^2$ and make it to have a free parameter so that there are infinite solutions?