Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Ok. A determinants question(Link below). What I have done? Well, I tried Row1 = Row 1+Row 2+row 3 but nothing great. Can you guys guide me to get the answer??

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

http://mypat.in/PaperImages/3/425/Q_4_004.gif
can you factor all of the elements
@UnkleRhaukus ,how exactly??

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

\[\Delta=\left|\begin{array}\\1+a^2+a^4&1+ab+a^2b^2&1+ac+a^2c^2\\1+ab+a^2b^2&1+b^2+b^4&1+bc+b^2c^2\\1+ac+a^2c^2&1+bc+b^2c^2&1+c^2+c^4\end{array}\right|\]
yeah maybe not, i really dont want to do it the normal way
Normal way-->never :D It would take ages :P
but it would work in theory, but i would make about ten mistakes
I am sure you play around with determinant properties like Row1 = Row 1 + Row 2+Row3 . I mean some tweaking of that sort
can we just do it the foolish way?
Hmm, tried C1 - C2?
@inkyvoyd Sorry, please, In my exam I have only 3 min(max) to solve this question @apoorvk ,let me give it a try and get back to you
@apoorvk , I did that. But nothing can be taken common out :(
I'm going to try this problem
Have no fear, for inky is here!
I got common stuff out through C1-C2 and C2-C3. but, i am still stuck with a trinomial in each ditch..
@shivam_bhalla I HAVE AN IDEA
aei+bfg+cdh-ceg-bdi-afh =a(ei-fh)+b(fg-dj)+c(dg-eg)
b=d f=h
c=g
aei+bfg+cdh-ceg-bdi-afh aei+bf^2+cbf-ec^2-ah^2
@inkyvoyd , what is a,b,c,d,e,f,g?? Are you expanding the determinant??
I'm wikipediaing, and not making any progress.
LOL. @ "wikipediaing" . :)
wolfram!!!
How to wolfram this ??
@radar sir, any clue ??
factor out the commons!!!
@experimentX , how exactly ?? I mean row wise or column wise. And moreover, what to factor out ??
lol ... factoring out would lead to 3 more matrices ..thats ugly... must be some pattern!!
Oh. Now get what you meant by factoring .:P I tried that too and observed what you said :)
it seems transpose of itself!!
|dw:1336584742255:dw| this is how far i got then..
try making these types to a^4 + a^2 + 1 => a^6 - 1 (ab)^2 + (ab) + 1 => (ab)^3 - 1 ... not sure if it can help ... too tired!!!
This problem really is trollling me very badly :( I will solve this problem before going to sleeep
if you find let met know!!
Oo ...we can drop all ones!!
Oo ... i got zero!! LOL
@shivam_bhalla you still there??
Yes
looks like it got it correct http://www.wolframalpha.com/input/?i=det+{{1%2Ba^2%2Ba^4%2C+1+%2B+%28ab%29+%2B+%28ab%29^2%2C+1+%2B+%28ac%29+%2B+%28ac%29^2}%2C++%09{1+%2B+%28ab%29+%2B+%28ab%29^2%2C+1%2Bb^2%2Bb^4%2C+1+%2B+bc+%2B+%28bc%29^2}%2C++%09{1+%2B+%28ac%29+%2B+%28ac%29^2%2C+1+%2B+bc+%2B+%28bc%29^2%2C+1+%2B+c^2+%2B+c^4}}
LOL @experimentX , that is doing it the long way Atleast I know the answer know. :) Thanks for typing out the huge question on wolframalpha :)
No .. i know the shortcut too!!
actually i verified answer at wolframalpha!!
Ok. Just tell me the method and not the steps because that would literally troll you
take out all ones!! sine they will have same values on row .. they will be zero ..
Sorry, didn't get you @experimentX
|dw:1336588777520:dw|
|dw:1336588827490:dw|
|dw:1336588893052:dw|
i'm not sure how far everyone has got so far but first thoughts: setting a = b we can see that the determinant is zero as R1 = R2 hence (a-b) must be a factor
whenever matrix has same elements on row or column ... determinant is zero.
and by symmetry (b-c) and (a-c) are
@experimentX , I never studied that. I only studied that if two rows or 2 colums are proportional to each other, then the determinant of the matrix is zero @eigenschmeigen , Very logical approach which eliminates option (a) for us :)
;D
proportional ->eg R1 = kR2 so k = 1 is a specific case
Oo ... then i must have mistake!!
no its correct
if R1 = R2 then det(M) = 0
@eigenschmeigen , there is no R1=R2 in @experimentX 's proof
Oh my mistake ... i got the answer in fluke
@experimentX , I wish these flukes come for me too in my exam :P
LOL ... i haven't reviewed matrices in long time!! best of luck if you are having exams right now!!
LOL. I have college entrance exam's right now :P
*exams
is this problem solved or still at large? i can have a proper go now i have a pen
The problem as of now is unsolved @eigenschmeigen
Okay ... i think i got this time!! you can use same technique to reduce matrix!!
|dw:1336589808195:dw|
and then it's simplification!!
apply operation as above!!! |dw:1336589933207:dw|
same as you had in your second place!! -- minus --- one
|dw:1336590227531:dw|
1 Attachment
|dw:1336590458004:dw| It can be proved!!
@experimentX , I will try and let you know in 5-10 minutes @experimentX
okay ... it's just simplification after that. factoring out common factor ... then again you will have 1's and then use same technique ... in the end you will get two linearly dependent columns!! then it will be zero!!
its B
i'm just checking my working
how do i write matrices on here?
@eigenschmeigen , your answer is right. You can use drawing doard :)
ok gimme a sec
*board.
i like the equation writer, i'm quicker that way: \[\left[\begin{matrix}1+a^2 + a^4 & 1+ ab+a^2b^2 & 1 +ac +a^2c^2 \\ 1 + ab +a^2b^2 & 1+b^2 + b^4 & 1 +ac +a^2c^2 \\ 1 +ac +a^2c^2 & 1 +bc+b^2c^2 & 1 + c^2 + c^4\end{matrix}\right]\]
letting a = b we get R1 = R2 letting b = c we get R2 = R3 so the operations to do are R1- R2 = R1' and R2 - R3 = R2'
\[\left[\begin{matrix}a^4 + a^2 - ab - a^2b^2 & a^2b^2 + ab -b^2 -b^4 & a^2c^2 + ac - bc - b^2c^2 \\ a^2b^2 + ab -ac -a^2c^2 & b^4 + b^2 - bc - b^2c^2 & b^2c^2 + bc -c^2 - c^4 \\ 1 +ac +a^2c^2 & 1 +bc+b^2c^2 & 1 + c^2 + c^4\end{matrix}\right]\]
factor out (a-b)(b-c)
\[(a-b)(b-c)\left[\begin{matrix}a^3 + a +ba^2 & b^3 + b + ab^2 & c(ac+bc+1) \\ a(ac+ab+1) & b^3 + b +cb^2 & c^3 + c + bc^2 \\ 1 +ac +a^2c^2 & 1 +bc+b^2c^2 & 1 + c^2 + c^4\end{matrix}\right]\]
letting a = b we now have C1 = C2 so we can see that (a-b) is again a factor. this narrows it down to B as we will have (a-b)^2 in our answer
you can continue to factorise if you want, but this is sufficient to get the right option.
its not said in many textbooks but i find using factor theorem is really useful for factorising determinants, as it tells us an operation that will result in something we can factor
Thank you @eigenschmeigen and @experimentX for both of your solutions :D Thanks for taking your time :)
no problem, love matrices :)
looks like i didn't get that right ... i'm embarrassed. looks like 'linear algebra' is the first thing i should do after i'm done with calc
what else does linear algebra cover?
well ... all forgot!! let's see ... i have a book here. Linear algebra - Serge Lang Vectors, Vector spaces, Matirces, Linear mappings, Linear mappings and matrices, .... Groups, Rings, Convex Sets ... Appendix I thought \( \Delta\) on wolfram was the answer!!

Not the answer you are looking for?

Search for more explanations.

Ask your own question