Question

Verify that vector (1,2,3), (4,1,5) and (-4,6,2) are linearly independent in R3?

Answer 1

SOLUTION

TO VERIFY

The vectors (1,2,3), (4,1,5) and (-4,6,2) are linearly independent in R³

EVALUATION

Here the given vectors are (1,2,3), (4,1,5) and (-4,6,2)

If possible let there exists scalers a , b , c such that

a(1,2,3) + b(4,1,5) + c (-4,6,2) = (0,0,0)

Consequently we get

a + 4b - 4c = 0

2a + b + 6c = 0

3a + 5b + 2c = 0

This is a homogeneous system of equations in a , b , c

Now the coefficient determinant

$= \displaystyle\begin{vmatrix} 1 & 4 & - 4\\ 2 & 1 & 6 \\ 3 & 5& 2 \end{vmatrix}$

$\displaystyle \sf{ =1(2 - 30) - 4(4 - 18) - 4(10 - 3) }$

$\displaystyle \sf{ = - 28 + 56 -28}$

$= 0$

So by Cramer's Rule the system of equations have non zero solution

So there exists non zero values of a , b , c such that

a(1,2,3) + b(4,1,5) + c (-4,6,2) = (0,0,0)

More precisely

- 4 (1,2,3) + 2 (4,1,5) + 1 (-4,6,2) = ( 0,0,0)

So the given set of vectors are dependent

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