Question

show that the vectors =(6-5, 3) in Rq Three dimensional) cannot be expressed as a linear combination of vectors e1= (1,-3,2) e2=(2,-4,-1) e3=(1,-5,7)​

Answer 1

Answer:

Step-by-step explanation:  

Let A = (6,-5,3)

Let us assume A can be expressed as linear combination of e1 , e2 , e3 .

A = a . e1 + b . e2 + c . e3

⇒ (6 , -5 , 3) = a(1 , -3 , 2) + b(2, -4 , -1) + c(1 , -5 , 7)

⇒ (6 , -5 , 3) = (a+2b+c , -3a-4b-5c , 2a-b+7c)

∴ a+2b+c = 6 →(1)

 -3a-4b-5c = -5 ⇒ 3a+4b+5c = 5 →(2)

  2a-b+7c = 3 →(3)

coefficient matrix is $\left[\begin{array}{ccc}1&2&1\\3&4&5\\2&-1&7\end{array}\right]$

Now determinant of this coefficient matrix is 1 (28+5) -2(21-10) +(-3-8)

= 33-22 -11 = 0

∴ the linear equations (1) , (2) , (3) has no solution

Hence A can not be expressed as linear combination of e1 , e2 , e3