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New answer posted

a year ago

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A
alok kumar singh

Contributor-Level 10

M = [ a 1 a 2 a 3 b 1 b 2 c 3 c 1 c 2 c 3 ]    

M T M = [ a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 ] [ a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 ]

T r ( M T M ) = a 1 2 + b 1 2 + c 1 2 + a 2 2 + b 2 2 + c 2 2 + a 3 2 + b 3 2 + c 3 2 = 7           

all   a i , b i , c i { 0 , 1 , 2 } f o r i = 1, 2, 3

Case 1 7 one's and two zeroes which can occur in ways

Case 2 One 2 three 1's five zeroes =

 

total such matrices = 504 + 36 = 540

 

New answer posted

a year ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

A T = A a n d B T = B           

C = A 2 B 2 B 2 A 2           

C T = ( A 2 B 2 ) T ( B 2 A 2 ) T = B 2 A 2 A 2 B 2

 CT = -C. Hence C is skew symmetric metrix

  det (C) = 0

Hence system have infinite solution

 

New answer posted

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

x – 2y = 1, x – y + kz = -2, ky + 4z = 6

 x – 2y + 0. z – 1 = 0

x – y + kz + 2 = 0

0x + ky + 4z – 6 = 0

0x + ky + 4z – 6 = 0

Δ 1 = | 1 2 0 2 1 k 6 k 4 | = ( k + 1 0 ) ( k + 2 )   

For no solution

Δ = 0 , Δ 1 0       

k = 2

 

New answer posted

a year ago

0 Follower 19 Views

A
alok kumar singh

Contributor-Level 10

kx + y + 2z = 1    . (i)

 3x – y – 2z = 2       . (ii)

-2x – 2y – 4z = 3   . (iii)

(ii) * 5 - (i)  (iii) * 3 -> (15 – k) = -6

K = 21

New answer posted

a year ago

0 Follower 41 Views

A
alok kumar singh

Contributor-Level 10

A = [ x y z y z x z x y ] , | A | = 3 x y z ( x 3 + y 3 + z 3 ) = ( x + y + z ) [ ( x + y + z ) 2 3 ( x y + y z + z x ) ]

A2 = l

A. A' = l    (as A = A')

x 2 + y 2 + z 2 = 1 a n d x y + y z + z x = 0         

x 3 + y 3 + z 3 = 3 * 2 + 1 * ( 1 0 ) = 7             

New answer posted

a year ago

0 Follower 21 Views

A
alok kumar singh

Contributor-Level 10

[ 0 t a n θ 2 t a n θ 2 0 ] , I 2 + A = [ 1 t a n θ 2 t a n θ 2 1 ] , I 2 A = [ 1 t a n θ 2 t a n θ 2 1 ]           

  ( I 2 + A ) ( I 2 A ) 1 = [ a b b a ]          

  a 2 + b 2 = | ( I 2 + A ) ( I 2 A ) 1 | = s e c 2 θ 2 * c o s 2 θ 2 = 1          

1 3 ( a 2 + b 2 ) = 1 3 * 1 = 1 3

New answer posted

a year ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

  A = [ a i j ] 3 * 3 = [ a 1 1 a 1 2 a 1 3 a 2 1 a 2 2 a 2 3 a 3 1 a 3 2 a 3 3 ]             

a i 1 + a i 2 + a i 3 = 1 ; i = 1 , 2 , 3            

L e t X = [ 1 1 1 ] t h e m  

given [ a 1 1 a 1 2 a 1 3 a 2 1 a 2 2 a 2 3 a 3 1 a 3 2 a 3 3 ] [ 1 1 1 ] = [ 1 1 1 ]  

->AX = X .(i)

replace x by A x we have

A (AX) = AX

->A2X = AX = X .(ii)

Again replace X by AX

A3X = AX = X.

As  X = [ 1 1 1 ] , Sum of all entries in A3 = sum of entries in X = 1 +1 + 1 = 3

New answer posted

a year ago

0 Follower 1 View

A
Aashi Madavi

Contributor-Level 8

The seat matrix includes reservation quotas for categories such as SC, ST, OBC-NVL, EWS, PwD and many more. Apart from this, the seats are allocated on the basis of NCC, sports, minority, and various other categories. Every category is allocated some percentage of the total seats to be eligible based on the quota or reservation.

New answer posted

a year ago

0 Follower 1 View

A
Aashi Rastogi

Contributor-Level 10

The seat matrix guides seat allocation across various counselling rounds by indicating towards the vacant seats. This process ensures transparency and a fair admission, giving equal opportunity to everyone. In several cases, the vacant seat left by the end of the admission process is occupied on the basis of remaining quotas such as management.

New answer posted

a year ago

0 Follower 1 View

M
Manisha Shukla

Contributor-Level 8

Seat matrix varies as per the type of institution, location, course, and applicable reservation policies. Every institute or university adheres to a different set of policies to conduct seat matrix. These policies vary in facilitating on the basis of quotas, and reservations. Furthermore, the institutes tend to offer Management quota to students based on the provided eligibility criteria.

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