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anonymous
 4 years ago
Let vectora=4i+3jalphak, vector b=2i+alphaj+k and vectorc=5ij+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
anonymous
 4 years ago
Let vectora=4i+3jalphak, vector b=2i+alphaj+k and vectorc=5ij+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

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JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2c 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 nonzero 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 nonzero determinant.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0since alpha is unknown v cn't v shw that it has non zero determinant. isnt t?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for 2 values of alpha the matrix becoes zero. then hw do v shw?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2Yes, 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if 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."

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2It'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 \[ p = \frac{53  2\sqrt{115}}{2(\sqrt{115}22)}, q = \frac{\sqrt{115} 4}{2(\sqrt{115}22}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how do u get this value for alpha? I get alpha as +/sqrt7+2

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2the determinant of the matrix written above is \[ 9\alpha^2  8\alpha + 11 \] That expression is zero, when \( \alpha \) is \[ \alpha = \frac{4 \pm \sqrt{115}}{9} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then hw do v shw that it is nt in the space?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2As 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0extremly sorry 4 troubling u a lt.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanx a lt 4 ur great help.
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