Matrix

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New answer posted

7 months ago

0 Follower 8 Views

A
alok kumar singh

Contributor-Level 10

2? -2? =112. m=7, n=4. mn=28.

New answer posted

7 months ago

0 Follower 11 Views

A
alok kumar singh

Contributor-Level 10

| B | = b 11 b 12 b 13 b 21 b 22 b 23 b 31 b 32 b 33 = 3 0 a 11 3 1 a 21 3 2 a 31 3 1 a 12 3 2 a 22 3 3 a 32 3 2 a 13 3 3 a 23 3 4 a 33

81 = 3 3 3 3 3 2 | A |

| A | = 1 9

New answer posted

7 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution 

 

New answer posted

7 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

  e ( c o s 2 x + c o s 4 x + c o s 6 x + . . . . . ) l o g e 2

= e c o s 2 x 1 c o s 2 x l o g e 2 = e c o t 2 x l o g e 2 = 2 c o t 2 x            

t2 – 9t + 8 = 0

(t – 8) (t – 1) = 0

t = 2 c o t 2 x = 8 = 2 3        

c o t 2 x = 3 = c o t 2 π 6       

2 s i n x s i n x + 3 c o s x = 2 * 1 2 1 2 + 3 * 3 2 = 1 2

New answer posted

7 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

P = [ 3 1 2 2 0 α 3 5 0 ] a n d Q = [ q i j ] P Q = k l 3           

q 2 3 = k 8 a n d | Q | = k 2 2            

  P Q = k l 3 P 1 = Q k = ( 3 1 2 2 0 α 3 5 0 ) 1

| p | | Q | = ( k l 3 ) 8 . k 2 2 = k 3

k 0 k = 4

α 2 + k 2 = 1 + 1 6 = 1 7 .         

New answer posted

7 months ago

0 Follower 4 Views

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

8 months ago

0 Follower 2 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

8 months ago

0 Follower 2 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

8 months ago

0 Follower 10 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

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