Linear algebra.?

Question

For what value(s) of t will the set of vectors S = {u1 = (1,2,3) , u2 = (2,-1,4) , u3 = ( 3,t,4)} be linearly dependent ?

Answer 1

Your first step would be to put these vectors in a matrix; each vector represents a column.

[1 2 3]
[2 -1 t]
[3 4 4]

Those vectors will be linearly dependent if the determinant is equal to 0. Solving for the determinant (choosing the first row), we get

1 * det |-1 t | - 2 * det |2 t | + 3 * det |2 -1|
. . . . . . | 4 4 | . . . . . . ..|3 4| . . . . . . . .|3 4 |

1 * ( (-1)(4) - (t)(4) ) - 2 (2*4 - 3t) + 3(8 - (-3))

(-4 - 4t) - 2(8 - 3t) + 3(8 + 3)
(-4 - 4t) - 2(8 - 3t) + 3(11)

-4 - 4t - 16 + 6t + 33

2t + 13

And we want this to equal to 0, so

2t + 13 = 0
2t = -13

t = -13/2

In order to be linearly dependent, t = -13/2

Answer 2

If vectors are linearly dependent, then u3 will be the sum of some multiples of u1 and u2. However, we can break this down further and say that the 1st value of u3 is also a multiple of the 1st values of u1 and u2 and the 2nd value is a multiple of the 2nd values, etc.
So we can say that:
c1*1+c2*2=3
and c1*2+c2*(-1)=t
and finally c1*3+c2*4=4
We can then solve the first and last equations for c1 and c2 ( the constants by which you multiply u1 and u2)
so c1=3-2c2
so 3(3-2c2)+4c2=4. 9-2c2=4
c2=5/2
c1=-2
so c1*2+c2*(-1)=t means that (-2)2+(-1)5/2=t
thus t=-4 - 5/2=-13/2
so t= - 13/2

Answer 3

Take your 3 vectors and put them in matrix form [u1 u2 u3]. Take the determinant. Then, plug in a value of t that makes the determinant non-zero, since linearly independent vectors in matrix form cannon have a determinant equal to zero. Either that use Gaussian elimination to check if the matrix is full rank.