Question

Is the sector (2,-5,3) in the subspace of R3

s panned by the vectors (1,-3,2), (2,-4,-1),

(1,-5,7).​

Answer 1

Answer:

is the sector(2,_5,3) in the subspace of the planned by the vectors (1,-3,2),(2,-4,-1),(1,-5,1

Answer 2

Answer: The sector (2,-5,3) does not lies in the subspace of R3 spanned by the vectors (1,-3,2), (2,-4,-1) and (1,-5,7)

Step-by-step explanation:

If the sector (2,-5,3) lies in the subspace of R3 spanned by the vectors (1,-3,2), (2,-4,-1) and (1,-5,7) , then  (2,-5,3) can be written in terms of (1,-3,2), (2,-4,-1) and (1,-5,7)

$\left[\begin{array}{ccc}2\\-5\\3\end{array}\right] = x\left[\begin{array}{ccc}1\\-3\\2\end{array}\right] + y\left[\begin{array}{ccc}2\\-4\\-1\end{array}\right] + z\left[\begin{array}{ccc}1\\-5\\7\end{array}\right]$

$\left[\begin{array}{ccc}2\\-5\\3\end{array}\right] = \left[\begin{array}{ccc}(x+2y+z)\\(-3x-4y-5z)\\(2x-y+7z)\end{array}\right]$

$2= x+2y+z$                   eq(1)

$-5 = -3x-4y-5z$          eq(2)

$3 = 2x-y+7z$                 eq(3)

y = 2x + 7z - 3  ( from eq(3) )

Substituting value of y in eq(1) :

2 = x + 2(2x + 7z - 3) + z

2 = x + 4x + 14z - 6 + z

5x + 15z  = 8

x + 3z = $\frac{8}{5}$

Substituting value of y in eq(2) :

-5 = -3x -4(2x + 7z - 3) -5z

-5 =  -3x -8x  -28z + 12 -5z

11x + 33z  = 17

x + 3z  = $\frac{17}{11}$

Value of x + 3z is not unique, which is not possible.

Hence, no possible solutions for x , y and z.

Therefore, (2,-5,3)  does not lies in subspace of the given three vectors.

#SPJ2