Answer to Question #288318 in Linear Algebra for Sabelo Xulu

Question

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) = $\alpha$ 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.

Answer 1

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

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

Subtract component wisely. This gives

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

Divide both sides by $3$ . This gives

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

(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.

$\begin{bmatrix} x_1 \\ y_1\\ z_1 \end{bmatrix}$ $+\begin{bmatrix} x_2 \\ y_2\\ z_2 \end{bmatrix}$ $=\begin{bmatrix} x_1+x_2 \\ y_1+y_2\\ z_1+z_2 \end{bmatrix}$

We will test if the points also lies in the plane. So we take our polynomial $2x+y-z-1=0$ , and substitute $x$ with $x_1+x_2,y$ with $y_1+y_2,z$ with $z_1+z_2$ and get

$2(x_1+x_2)+(y_1+y_2)-(z_1+z_2)-1=0$

When we distribute, we get

$2x_1+2x_2+y_1+y_2-z_1+z_2-1=0$

We reorganize to get

$(2x_1+y_1-z_1)+(2x_2+y_2-z_2)-1=0$

$(2x _1 ​ +y _1 ​ −z _1 ​ )+(2x _2 ​ +y _2 ​ −z _2 ​ )=1$

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 $S$ be the set of differentiable real valued function. Then

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

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

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

(3) zero