# For what value of lamda the following vectors will form a basis for E^3 a_1 = (1,5,3) , a_2 = (4,0,lamda) , a_3 = (1,0,0)  ?

Dec 20, 2017

$\lambda \in \mathbb{R} - \left\{0\right\}$.

#### Explanation:

Let the set $B = \left\{{a}_{1} = \left(1 , 5 , 3\right) , {a}_{2} = \left(4 , 0 , \lambda\right) , {a}_{3} = \left(1 , 0 , 0\right)\right\}$

form a Basis for the vector space ${E}^{3}$.

Then an arbitrary vector $v = \left(a , b , c\right) \in {E}^{3}$ can uniquely be

represented as a linear combination of the vectors in $B$.

In other words,

$\exists \text{ unique "l,m,n in RR," s.t., } v = l {a}_{1} + m {a}_{2} + n {a}_{3}$.

Now, v=la_1+ma_2+na_3; l,m,n in RR,

$\Rightarrow \left(a , b , c\right) = l \left(1 , 5 , 3\right) + m \left(4 , 0 , \lambda\right) + n \left(1 , 0 , 0\right) , i . e . ,$,

$\left(a , b , c\right) = \left(l , 5 l , 3 l\right) + \left(4 m , 0 , m \lambda\right) + \left(n , 0 , 0\right) , \mathmr{and} ,$

$\left(a , b , c\right) = \left(l + 4 m + n , 5 l , 3 l + m \lambda\right)$.

By the equality of vectors, then, we have,

$l + 4 m + n = a , 5 l = b , 3 l + m \lambda = c$.

In order that this system of eqns. may have a unique soln.,

we know from Algebra that,

$| \left(1 , 4 , 1\right) , \left(5 , 0 , 0\right) , \left(3 , \lambda , 0\right) | \ne 0$.

$\therefore 1 \left(0\right) - 4 \left(0\right) + 1 \left(5 \lambda - 0\right) \ne 0$.

$\therefore \lambda \ne 0.$

Hence, $\lambda$ can be any non zero real number.