Question #288318

Q4.

The value of x \in R^4

such that (4, 3, - 3, 2) + 3x = (7, -3, 3, 2) is

(1) x = (4, -4, 3, 0)

(2) x = (4, - 4, 3, 0)

(3) x = (1, -2, 0, 0)

(4) None of the given answers is true.


Q5.

Let W be a subset of R^3 defined as

W = (x, y, z) \in R^3: 2x + y - z - 1 = 0.

Then

(1) W is a subspace of R^3

(2) W is closed under scalar multiplication

(3) W is not a subspace of R^3

(4) None of the given answers is true.


Q6.

The set of differentiable real-valued functions f on the interval (0,3) such that f'(2) = α\alpha is a subspace of R^(0,3). The value of must be

(1) negative

(2) positive

(3) zero

(4) None of the given answers is true.


1
Expert's answer
2022-01-25T10:07:04-0500

Q4. (4,3,3,2)+3x=(7,3,3,2)(4,3,-3,2)+3x=(7,-3,3,2)

    \implies 3x=(7,3,3,2)(4,3,3,2)3x=(7,-3,3,2)-(4,3,-3,2)

Subtract component wisely. This gives

3x=(3,6,6,0)3x=(3,-6,6,0)

Divide both sides by 33 . This gives

x=(1,2,2,0)x=(1,-2,2,0)

(4)None of the given answers is true.


Q5. Let's check if W is a subspace. First we check if it is closed under addition.

[x1y1z1]\begin{bmatrix} x_1 \\ y_1\\ z_1 \end{bmatrix} +[x2y2z2]+\begin{bmatrix} x_2 \\ y_2\\ z_2 \end{bmatrix} =[x1+x2y1+y2z1+z2]=\begin{bmatrix} x_1+x_2 \\ y_1+y_2\\ z_1+z_2 \end{bmatrix}

We will test if the points also lies in the plane. So we take our polynomial 2x+yz1=02x+y-z-1=0 , and substitute xx with x1+x2,yx_1+x_2,y with y1+y2,zy_1+y_2,z with z1+z2z_1+z_2 and get

2(x1+x2)+(y1+y2)(z1+z2)1=02(x_1+x_2)+(y_1+y_2)-(z_1+z_2)-1=0

When we distribute, we get

2x1+2x2+y1+y2z1+z21=02x_1+2x_2+y_1+y_2-z_1+z_2-1=0

We reorganize to get

(2x1+y1z1)+(2x2+y2z2)1=0(2x_1+y_1-z_1)+(2x_2+y_2-z_2)-1=0

(2x1+y1z1)+(2x2+y2z2)=1(2x _1 ​ +y _1 ​ −z _1 ​ )+(2x _2 ​ +y _2 ​ −z _2 ​ )=1

We cannot say it is closed under addition, since it is not equal to 0.

So

(3) W is not a subspace of R3


Q6. Let SS be the set of differentiable real valued function. Then

S=S= {x,f0<x<3,fx,f|0<x<3,f\in R^(0,3) ,f(x)=0f'(x)=0 }

    f(x)=0\implies f'(x) =0 is a subspace of R^(0,3).

Therefore for f(2)=αf'(2)=\alpha to be a subspace, the value of α\alpha must be zero.

(3) zero







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