Question

(a) Briefly describe how the resistance R of a metal resistor changes with the temperature T. Sketch a functional form of R(T) and explain why.

(b) Briefly describe how the resistance R of a semiconductor resistor changes with the temperature T. Sketch a functional form of R(T) and explain why.

Answer 1

(A)

To check A is subspace of R3 or not ,let $(x_1,,y_1,z_1),(x_2,y_2,z_2)\epsilon A\&\alpha \mathbb{R}$

$\alpha (x_1,,y_1,z_1)+(x_2,y_2,z_2)=(\alpha x_1+x_2,\alpha y_1+y_2,\alpha z_1+z_2)$since

Now since $(x_1,,y_1,z_1),(x_2,y_2,z_2)\epsilon A$ ,

$x_1<y_1<z_1 \&\\ x_2<y_2<z_2$___(2)

$x_1<y_1<z_1 \\ \Rightarrow \\ \alpha x_1<\alpha y_1<\alpha z_1--(1)$

Now add (1) and (2) ,we get

$\alpha x_1&plus;x_2<\alpha y_1&plus;y_2<\alpha z_1&plus;z_2$

which implies that $\alpha (x_1,,y_1,z_1)&plus;(x_2,y_2,z_2)\epsilon A$

hence A is subspace of R3.

(b)

To check B is subspace of R3 or not ,let $(5,,y_1,z_1),(5,y_2,z_2)\epsilon A\&\alpha \mathbb{R}$

now

$\alpha (5,,y_1,z_1)&plus;(5,y_2,z_2)\\ =(5\alpha &plus;5,\alpha y_1&plus;y_2,\alpha z_1&plus;z_2)\\ =(5(\alpha &plus;1),\alpha y_1&plus;y_2,\alpha z_1&plus;z_2)\\ =(5,\alpha y_1&plus;y_2,\alpha z_1&plus;z_2)$only if $\alpha =0$

hence in genral $\alpha (5,y_1,z_1)&plus;(5,y_2,z_2)\not\epsilon A$

which implies that A is not subspace of R3

Counter example:-Take

$(5,1,2),(5,2,3)\epsilon A\\ then\\ (5,1,2)&plus;(5,2,3)=(10,3,5)\not\epsilon A$,hence A is not subspace of R3.

(c)

To check C is subspace of R3 or not ,let $(2x,6x,9x),(2y,6y,9y)\epsilon A\&\alpha \mathbb{R}$

Now $\alpha (2x,6x,9x)&plus;(2y,6y,9y)=(2(\alpha x&plus;y),6(\alpha x&plus;y),9(\alpha x&plus;y))\epsilon A\Rightarrow$ A is subspace of R3.

(D)

To check D is subspace of R3 or not ,let $(x,y,z),(x',y',z')\epsilon A\&\alpha \mathbb{R}$

$\alpha (x,y,z)&plus;(x',y',z')=(\alpha x&plus;x',\alpha y&plus;y',\alpha z&plus;z')$

Now

$-5(\alpha x&plus;x')&plus;4(\alpha y&plus;y')=\alpha (-5x&plus;4y)&plus;(-5x'&plus;4y')=\alpha *0&plus;0=0\\ \&\\ -3(\alpha x&plus;x')&plus;2(\alpha z&plus;z')=\alpha (-3x&plus;2z)&plus;(-3x'&plus;2z')=\alpha *0&plus;0=0\Rightarrow$

$\alpha (x,y,z)&plus;(x',y',z')\epsilon D\Rightarrow D$ is subspace of R3.

(E)

To check E is subspace of R3 or not ,let $(x,y,z),(x',y',z')\epsilon E\&\alpha \mathbb{R}$

$\alpha (x,y,z)&plus;(x',y',z')=(\alpha x&plus;x',\alpha y&plus;y',\alpha z&plus;z')$

Now

$-5(\alpha x&plus;x')&plus;4(\alpha y&plus;y')&plus;3(\alpha z&plus;z')=(-5x&plus;y&plus;3z)&plus;(-5x'&plus;4y'&plus;3z')=\alpha *0&plus;0=0$

which shows that E is subspace of R3.

(F)

To check F is subspace of R3 or not ,let $(x,y,z),(x',y',z')\epsilon F\&\alpha \mathbb{R}$

$\alpha (x,y,z)&plus;(x',y',z')=(\alpha x&plus;x',\alpha y&plus;y',\alpha z&plus;z')$

Now

$-6(\alpha x&plus;x')-9(\alpha y&plus;y')&plus;7(z&plus;z')=\alpha (-6x-9y&plus;7z)&plus;(-6x'-9y'&plus;7z')=\alpha (-2)&plus;(-2)\neq -2\forall \alpha$

which show sthat In general $\alpha (x,y,z)&plus;(x',y',z')\not \epsilon F\Rightarrow F$ is not subspace of R3.

Counter example

Take

$(-2,0,-2)\epsilon F\because -6(-2)-9(0)&plus;7(-2)=12-14=-2\\ (1/3,0,0)\epsilon F\because -6(1/3)-9(0)&plus;7(0)=-2\\ but\\ (-2,0,-2)&plus;(1/3,0,0)=(-5/3,0,-2)\not\epsilon F\because -6(-5/3)-9(0)&plus;7(-2)=10-14=-4$