Home
Class 12
MATHS
Let A=[(0,2),(0,0)]and (A+1)^100 -100A=[...

Let `A=[(0,2),(0,0)]and (A+1)^100 -100A=[(alpha,beta),(gamma,delta)],` then `alpha+beta+gamma+delta=...`

Text Solution

Verified by Experts

The correct Answer is:
2
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • MATRICES

    AAKASH INSTITUTE ENGLISH|Exercise Assignment (Section - H) Multiple True - False Type Questions|2 Videos

  • MATRICES

    AAKASH INSTITUTE ENGLISH|Exercise Assignment (Section - I) Subjective Type Questions|3 Videos

  • MATRICES

    AAKASH INSTITUTE ENGLISH|Exercise Assignment (Section - F) Matrix - Match Type Question|1 Videos

  • MATHEMATICAL REASONING

    AAKASH INSTITUTE ENGLISH|Exercise Assignment (SECTION-D) (Assertion-Reason Type Questions)|15 Videos

  • PERMUTATIONS AND COMBINATIONS

    AAKASH INSTITUTE ENGLISH|Exercise Assignment Section-J (Aakash Challengers Questions)|7 Videos

Similar Questions

Explore conceptually related problems

If the centroid of triangles formed by the vertices (1, 2, 3), (2, 1, 0) and (3, 1, 4) is (alpha, beta, gamma) then the value of [alpha]+[beta]+[gamma], where [ ] represents the greatest integer function, is _______.

Let A=[(alpha,beta),(gamma,delta)] such that A^3=0, then sum of all the elements of A^2 is

If cos(alpha+beta)*sin(gamma+delta)=cos(alpha-beta)*sin(gamma-delta), prove that cotalphacotbetacotgamma=cotdelta

If cos(alpha+beta)sin(gamma+delta)="cos"(alpha-beta)"sin"(gamma-delta) , prove that cotalphacotbetacotgamma=cotdelta

If the circumcentre of triangle ABC is (alpha,beta,gamma) where A, B, C are ((1, 0, 0) (0, 1, 0) and (0,0,1) respectively then alpha+beta+gamma+=1 (b) alpha=beta gamma=1/3 (d) alpha=1

Prove that |2alpha+beta+gamma+deltaalphabeta+gammadelta alpha+beta+gamma+delta2(alpha+beta)(gamma+delta)alphabeta(gamma+delta)+gammadelta(alpha+beta) alphabeta+gammadeltaalphabeta(gamma+delta)+gammadelta(alpha+beta)2alphabetagammadelta|=0

Prove that |((beta+gamma-alpha-delta)^4,(beta+gamma-alpha-delta)^2,1),((gamma+alpha-beta-delta)^4,(gamma+alpha-beta-delta)^2,1),((alpha+beta-gamma-delta)^4,(alpha+beta-gamma-delta)^2,1)|=-64(alpha-beta)(alpha-gamma)(alpha-delta)(beta-delta)(gamma-delta)

designated by alpha, beta, gamma and delta are in order:

Prove that |2alpha+beta+gamma+deltaalphabeta+gammadeltaalpha+beta+gamma+delta2(alpha+beta)(gamma+delta)alphabeta(gamma+delta)+gammadelta(alpha+beta)alphabeta+gammadeltaalphabeta(gamma+delta)+gammadelta(alpha+beta)2alphabetagammadelta|=0

If alpha,beta,gamma,delta in R satisfy ((alpha+1)^2+(beta+1)^2+(gamma+1)^2+(delta+1)^2)/(alpha+beta+gamma+delta) = 4 If biquadratic equation a_0 x^4 + a_1 x^3 + a_2 x^2 + a_3 x _ a_4 =0 has the roots (alpha+1/beta -1),(beta +1/gamma -1),(gamma+1/delta -1),(delta+1/alpha -1) . then the value of a_2/a_0 is

Home
Profile