Question

Let vectora=-4i+3j-alphak, vector b=2i+alphaj+k and vectorc=5i-j+alphak, where alpha is a real number. Show that the vector c is not in the space spanned by vectors a and b. Pls help

Answer 1

c is not in the span of a and b, if and only if the three vectors a,b,c are linearly independent. That means that the system of coefficients of the three vectors has non-zero determinant. So one way to show the result of the question is to show that the matrix

-4 3 a
2 a 1
5 -1 a

has non-zero determinant.

Answer 2

since alpha is unknown v cn't v shw that it has non zero determinant. isnt t?

Answer 3

for 2 values of alpha the matrix becoes zero. then hw do v shw?

Answer 4

James pls help.

Answer 5

Yes, you're exactly right. For two values of a, the determinant IS zero and hence the vector c DOES lie in the span of the vectors a and b.

So the question as written is not exactly correct.

Answer 6

if I solve c=pa+qb i get alpha as an imaginary number. so cn v say that "since alpha is given as real number bt v get here alpha as an imaginary number. so c is nt in the spce spand by a nd b."

Answer 7

It's a brutal calculation, but if
\[ \alpha = \frac{1}{9} (-4 \pm \sqrt{115} ) \]
then there are real numbers p and q such that c = pa + qb.

For example, if alpha = (1/9) ( -4 + sqrt(115)), then
\[ p = \frac{53 - 2\sqrt{115}}{2(\sqrt{115}-22)}, q = \frac{\sqrt{115} -4}{2(\sqrt{115}-22}\]

Answer 8

how do u get this value for alpha?
I get alpha as +/-sqrt-7+2

Answer 9

the determinant of the matrix written above is
\[ -9\alpha^2 - 8\alpha + 11 \]
That expression is zero, when \( \alpha \) is
\[ \alpha = \frac{-4 \pm \sqrt{115}}{9} \]

Answer 10

then hw do v shw that it is nt in the space?

Answer 11

As I noted above, the question as written is wrong. For two values of alpha, the vector c DOES lie in span of the vectors a and b.

Answer 12

extremly sorry 4 troubling u a lt.

Answer 13

Thanx a lt 4 ur great help.