[SOLVED] Vectors - Linear independence

**mybrohshi5** · March 28th 2010, 11:58 AM

The vectors
v = $\begin{bmatrix}0\\3\\-6\\ \end{bmatrix}$ $u = \begin{bmatrix}-1\\3\\-22+k\\ \end{bmatrix}$ $w = \begin{bmatrix}-2\\8\\-22\\ \end{bmatrix}$

are linearly independent if and only if k does not equal ______?

Ok so i set up the matrix

$\begin{bmatrix}0&-1&2\\3&3&8\\-6&-22+k&-22\\ \end{bmatrix}$

then i reduced it down to this

$\begin{bmatrix}1&0&\frac{2}{3}\\0&1&2\\0&-16+k&-6\\ \end{bmatrix}$

Not sure what to do now or if i am even headed in the right direction

Thanks for any suggestions.

**vincent** · March 28th 2010, 04:16 PM

you're almost there! u, v and w are linearly independant when the equation $c_1u+c_2v+c_3v=0$ only admits the trivial solution which is when the scalars $c_1= c_2= c_3=0$ . you need to find a k such that your system of equations satisfies this condition. this is the same as saying that your matrix can be reduced to a diagonal matrix!

**Bruno J.** · March 28th 2010, 05:18 PM

Originally Posted by vincent

you're almost there! u, v and w are linearly independant when the equation $c_1u+c_2v+c_3v=0$ only admits the trivial solution which is when the scalars $c_1= c_2= c_3=0$ . you need to find a k such that your system of equations satisfies this condition. this is the same as saying that your matrix can be reduced to a diagonal matrix with nonzero diagonal entries!

A perhaps shorter way is to evaluate the determinant of the matrix whose columns (or rows) are your three vectors, and find the values of $k$ for which this determinant does not vanish. Make sure you understand why this works, however - otherwise it's cheating!

**mybrohshi5** · March 28th 2010, 07:14 PM

Originally Posted by vincent

you're almost there! u, v and w are linearly independant when the equation $c_1u+c_2v+c_3v=0$ only admits the trivial solution which is when the scalars $c_1= c_2= c_3=0$ . you need to find a k such that your system of equations satisfies this condition. this is the same as saying that your matrix can be reduced to a diagonal matrix!

ok so i reduced my matrix to a diagonal with the diagonals having nonzero entries

$\begin{bmatrix} 1&0&\frac{2}{3} \\ 0&1&2 \\ 0&0&2-\frac{6}{16+k} \end{bmatrix}$

I am not sure what to do from here though to solve for k?

**Bruno J.** · March 28th 2010, 07:29 PM

No! First off, you should avoid dividing by an expression containing $k$ because the expression might be zero. In any case, I believe you also made a calculation mistake (the numerator of the fraction should be $16-k$ and not $16+k$ ).

When you reach

$<br /> \begin{bmatrix}1&0&\frac{2}{3}\\0&1&2\\0&-16+k&-6\\ \end{bmatrix}<br />$

you can subtract $-16+k$ times the second row from the third, to get

$<br /> \begin{bmatrix}1&0&\frac{2}{3}\\0&1&2\\0&0&-6-2(-16+k)\\ \end{bmatrix} = \begin{bmatrix}1&0&\frac{2}{3}\\0&1&2\\0&0&26-2k\\ \end{bmatrix}.<br />$

No divisions involved!

Now you want the bottom right corner to be nonzero, i.e. you want $k\neq 13$ .

Check to see if you get the same answer with the determinant method. I didn't check all your calculations so I can't guarantee you that your answer is correct!

**mybrohshi5** · March 28th 2010, 07:35 PM

Thank you. 13 was correct.

I will do it using determinants and see if i get the same answer

Thanks again for all the help

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