# Lin. Dep.

• October 1st 2007, 08:40 PM
Ideasman
Lin. Dep.
1.) Given the set with these two vectors:

\left\{ \left[ \begin {array}{c} a-1\\\noalign{\medskip}a+3
\end {array} \right] , \left[ \begin {array}{c} 1\\\noalign{\medskip}a
\end {array} \right] \right\}

I have to find the value(s) of a that makes it linearly depedent.

WORK:

Well, they are lin. dep. if they are multiples of each other. So I eye-balled it (not sure how else to do it) and saw that a = 3 will make it lin. dep because then you have the two rows are multiples of each other (scalar of 2)

Is that the only value?

2.) Like above..but with 1 more vector..given the set with these 3 vectors:

\left\{ \left[ \begin {array}{c} a+4\\\noalign{\medskip}3
\\\noalign{\medskip}2\end {array} \right] , \left[ \begin {array}{c} a
\\\noalign{\medskip}1\\\noalign{\medskip}0\end {array} \right] ,
\left[ \begin {array}{c} 2\\\noalign{\medskip}1\\\noalign{\medskip}1
\end {array} \right] \right\}

I have to find the value(s) of a that makes it linearly depedent.

WORK:

I can't eye-ball any a that will make these multiples of each other or will make a whole column all 0....I conclude that there is no value a for which this is linearly dependent??
• October 2nd 2007, 06:10 AM
topsquark
Quote:

Originally Posted by Ideasman
1.) Given the set with these two vectors:

\left\{ \left[ \begin {array}{c} a-1\\\noalign{\medskip}a+3
\end {array} \right] , \left[ \begin {array}{c} 1\\\noalign{\medskip}a
\end {array} \right] \right\}

I have to find the value(s) of a that makes it linearly depedent.

As you say, if they are linearly dependent they are multiples of each other. Thus we have the equation:
$\left[ \begin {matrix} a-1 \\ a+3
\end {matrix} \right] = c \left[ \begin {matrix}1 \\ a
\end {matrix} \right]$

where c is some constant.

Thus we have that
$a - 1 = c \cdot 1$
$a + 3 = c \cdot a$

Solving these simultaneously gives:
$a = -1, 3$ (and $c = 2, -2$ respectively.)

-Dan
• October 2nd 2007, 06:17 AM
topsquark
Quote:

Originally Posted by Ideasman
2.) Like above..but with 1 more vector..given the set with these 3 vectors:

\left\{ \left[ \begin {array}{c} a+4\\\noalign{\medskip}3
\\\noalign{\medskip}2\end {array} \right] , \left[ \begin {array}{c} a
\\\noalign{\medskip}1\\\noalign{\medskip}0\end {array} \right] ,
\left[ \begin {array}{c} 2\\\noalign{\medskip}1\\\noalign{\medskip}1
\end {array} \right] \right\}

I have to find the value(s) of a that makes it linearly depedent.

Here we need to define linearly dependent a bit more carefully. If three vectors are linearly dependent we may take a linear combination of any two of them and produce a third. So one example of this is
$\left[ \begin {matrix}a+4 \\ 3
\\ 2 \end {matrix} \right] = c \left[ \begin {matrix} a
\\ 1 \\ 0 \end {matrix} \right] +
d \left[ \begin {matrix} 2\\ 1 \\ 1
\end {matrix} \right]$

where c and d are some constants.

So we have the equations:
$a + 4 = c \cdot a + d \cdot 2$
$3 = c \cdot 1 + d \cdot 1$
$2 = c \cdot 0 + d \cdot 1$

I'll let you solve the system for a, c, and d.

Note: You may set up the 2 other examples, reordering the vectors as you do, and you will get the same value for a, but obviously with different values of c and d for each.

-Dan
• October 2nd 2007, 07:23 PM
Ideasman
Quote:

Originally Posted by topsquark
Here we need to define linearly dependent a bit more carefully. If three vectors are linearly dependent we may take a linear combination of any two of them and produce a third. So one example of this is
$\left[ \begin {matrix}a+4 \\ 3
\\ 2 \end {matrix} \right] = c \left[ \begin {matrix} a
\\ 1 \\ 0 \end {matrix} \right] +
d \left[ \begin {matrix} 2\\ 1 \\ 1
\end {matrix} \right]$

where c and d are some constants.

So we have the equations:
$a + 4 = c \cdot a + d \cdot 2$
$3 = c \cdot 1 + d \cdot 1$
$2 = c \cdot 0 + d \cdot 1$

I'll let you solve the system for a, c, and d.

Note: You may set up the 2 other examples, reordering the vectors as you do, and you will get the same value for a, but obviously with different values of c and d for each.

-Dan

Ohhh!! Wow, completed missed the other solution for #1.

For #2, I can't find any a that makes it depedent. Whatever I try makes it independent :(

I keep getting:

\left[ \begin {array}{ccc} 1&0&8\\\noalign{\medskip}0&1&10
\\\noalign{\medskip}0&0&45\end {array} \right]

Is there no a that makes this lin dep? I'm going crazy.