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Are you sure its not this:

because when im trying to solve it im getting negative of the expression

i don't like playing with alpha, beta and gammas so im using

alpha = x beta = y gamma = z

Here it is:

you take the right hand side(RHS) of the expression and then expand it

RHS= (y-x)(z-x)(y-z)

= (yz-yx-xz+x²)(y-z)

= y²z-xy²-xyz+x²y-yz²+xyz+xz²-x²z

(now rearrange them and combine like terms)

= y²z-yz²+xz²-x²z-xy²+x²y

= -1(-y²z+yz²-xz²+x²z+xy²-x²y)

= -1[(yz²-y²z)-(xz²-x²z)+(xy²-x²y)]

=

this is what im getting...so check your question again

yes indeed

dim(column space) = 8

i think its more like this

x²+(1/x)²

his ² looks like its only on 1 not on the whole 1/x

anyways...

x²+(1/x)² = 9²

yup, now just multiply and you get:

x²+2x-x-2=4

x²+x-2-4=0

x²+x-6=0

(x-2)(x+3)=0

x=2 and x=-3 are the solutions

yup....in steps it looks like:

√(2x+3) = 3

=> (√(2x+3))² = 3²

=> 2x+3 = 9

=> 2x = 9-3

=> 2x = 6

=> x= (6/2)

=> x=3

that would be if you have to all your steps.....well im sure you understand it now

okay now i don't understand this part:

x + y is mapped to T(x) + T(y),

i know we have to prove: T(x+y) = T(x) + T(y) and then T(cx)=cT(x) but i don't understand how you got there.....can you explain please...

ok now i can read and write urdu and everything but i don't think this is urdu cz when i was trying to read it it sounds like punjabi or something thats why i don't understand it......but i can probably read it to u guys.....well try to:

husan ki toop ka mara kolla tera khukh na rahe

tujh peh gir jaeh katbi tara tera khukh na rahe

(now the next part im not sure im reading it right)

bunkang counsil wale teri har she karki kardein

(the rest im sure is fine)

charh jai tum pe karza bhari tera khukh na rahe

sardi undar neher key baney saari raat khaloote

too na aai laya lara tera khukh na rahe

dil ki choori key ilzam mei police ka chappa parh jai

pharhya jai tabar saara tera khukh na rahe

saari gazal sunna kar bhi sarha dil ka nahin mukya

saadha hai bus ikkoo nara tera khukh na rahe

~ Kalid masood

If anyone can read this and understand it let me know, my mom did indeed taught me how to read an write in urdu and all and she does speak in punjabi with some people she knows but i have no clue.....soo good luck with understanding dat...

**basmah**- Replies: 3

Q). Let *T*: *V* --> *W* be a linear transformation, and let *U* be a subspace of *W*. Prove that the set

*T*-¹ (*U*) = {**v** Є *V*: T(**v**) Є *U*}

is a subspace of *V*?

What is *T*-¹ (*U*) if *U* = {**0**}

I'm normally good with proofs but this one just does not make sense to me ....i'm sooo confused can anyone help me? (also if you can show all your steps, that would be great, its sometimes hard for me to understand when there are alot of steps skipped).

o yaa it kind of make sense now....thanks much!!!

**basmah**- Replies: 2

Let *V* and W be two subspaces of a vector space U. Prove that the set

*V* + *W* = {**u : u = v + w**, where **v** *V* and **w** *W*}

is a subspace of *U*.

*V* = {(x,0) : x is a real number} and *W* = {(0,y) : y is a real number}.

** **Bold** = Vector and "" = "such that"

**basmah**- Replies: 1

There are two proofs that i need help in:

Q#1) Prove that in a given vector space V, the zero vector is unique

and

Q#2) Let *V* and W be two subspaces of a vector space U. Prove that the set

*V* + *W* = {**u : u = v + w**, where **v** *V* and **w** *W*}

is a subspace of *U*.

*V* = {(x,0) : x is a real number} and *W* = {(0,y) : y is a real number}.

** **Bold** = Vector and "" = "such that"

thanks polylog....i mean i really appretiate it....but its too little too late...i did not check late at night since i wanted to get it done....but your answer is the correct one.....if only you could finish on time.....or i checked early in the morning but i did not think anyone would get it, so i just turned it in....all not so very correct....hopefully i'll get some credit on it....but i really want to thank you....there is a new one that is due day after tommorrow that i will post now.....for this one you would have a lot more time....i will try to do it my self and then check my work with what ever responses i get on here.....

let A be nxn

We need to use these facts:

(1) (A^-1)^-1 = A

(2) (k A)^-1 = (1/k) A^-1

(3) det(adj A) = (det(A))^(n-1) where n is the dimension

(4) A^-1 = 1/det(A) adj(A)

begin with (4):

A^-1 = 1/det(A) adj(A)

(A^-1)^-1 = (1/det(A) adj(A))^-1

A = (det A)(adj A)^-1

A = (det A) [ 1/(det (adj A)) adj(adj A) ]

A = (det A)/(det (adj A)) adj(adj A)

A = (det A)/(det A)^(n-1) adj(adj A)

A = (det A)^(2-n) adj(adj A)

adj(adj A) = (det A)^(-(2-n)) A

adj(adj A) = (det A)^(n-2) A

and clearly as a special case for 3x3 matrices:

det(A) = 1 -> adj(adj A) = A

that is the correct answer!!!!....again thx for the effort:D

**basmah**- Replies: 7

I need any kind of help or hints to find the answer to this:

adj (adj A) = ? (adj stands for adjoint and A is any matrix...please express the answer in terms of A and det(A).... det(A) is the determinant of A)

If you are able answer this question, then i would also like you to prove it aswell...if you are not able to prove it, its fine anything would be fine....also it is due tommorrow so as soon as possible please!!!

** the matrix is suppose to be for an n*n matrix

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