Matrices

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New answer posted

a year ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

Δ = 0

sin 3 θ=  1 2

but 0 < θ < π 2

θ= 70°

New answer posted

a year ago

0 Follower 12 Views

V
Vishal Baghel

Contributor-Level 10

A 2 = [ 0 1 0 0 0 1 1 0 0 ] [ 0 1 0 0 0 1 1 0 0 ]

= [ 0 0 1 1 0 0 0 1 0 ]

a R2

= [ 1 0 0 0 0 1 0 1 0 ]

R 2 R 3

= [ 1 0 0 0 1 0 0 0 1 ] =1 

B 0 = A 4 9 + 2 A 9 8 = A + 2 I B n = A d j ( B n 1 ) B 4 = A d j ( A d j ( A d j ( A d j B 0 ) ) )

= | B 0 | ( n 1 ) 4 = | B 0 | 1 6

B 0 = [ 0 1 0 0 0 1 1 0 0 ] + [ 2 0 0 0 2 0 0 0 2 ] = [ 2 1 0 0 2 1 1 0 2 ]

= 2 ( 4 0 ) 1 ( 0 1 ) = 9

B 4 ( 9 ) 1 6 = 3 3 2

New answer posted

a year ago

0 Follower 27 Views

V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

New answer posted

a year ago

0 Follower 13 Views

V
Vishal Baghel

Contributor-Level 10

Here,

A = [ 2 1 1 1 0 1 1 1 0 ]            

we get A2 = A and similarly, for

B = A I = [ 1 1 1 1 1 1 1 1 1 ]            

We get B2 = -B  B3 = B

A n + ( ω B ) n = A + ( ω B ) n f o r n N

For ω n  to be unity n shall be multiple of 3 and for Bn to be B. n shell be 3, 5, 7, ….9

n = { 3 , 9 , 1 5 , . . . . . 9 9 }     

Number of elements = 17.

New answer posted

a year ago

0 Follower 19 Views

V
Vishal Baghel

Contributor-Level 10

A + B = [ β + 1 0 3 α ]

( A + B ) 2 = [ ( β + 1 ) 2 0 3 ( β + 1 ) + 3 α α 2 ]

[ 1 α + 1 2 α + 4 α 2 ] = [ ( β + 1 ) 2 0 3 ( α + β + 1 ) α 2 ]

= 1 = 1

B 2 = [ β 1 1 0 ] [ β 1 1 0 ]

β = 0 , α = 1 = α 2

| α 1 α 2 | = | 1 ( 1 ) | = 2

New answer posted

a year ago

0 Follower 8 Views

V
Vishal Baghel

Contributor-Level 10

[ a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 ] a 1 , b 1 , c 1 t { 1 , 0 , 1 }

a 1 + a 2 + a 3 + . . . . ? 9 t i m e s = 5

Total = 414

New answer posted

a year ago

0 Follower 11 Views

V
Vishal Baghel

Contributor-Level 10

A = [ 0 2 2 0 ]

A 2 = [ 0 2 2 0 ] [ 0 2 2 0 ]

= [ 4 0 0 4 ] = 4 [ 1 0 0 1 ] = 4 l

N 2 = ( 2 2 0 1 ) 2 2 5 l

M N 2 = 4 1 2 5 ( 2 2 0 1 ) 3 l

= [ λ 0 0 0 λ 0 0 0 λ ] , λ k

New answer posted

a year ago

0 Follower 12 Views

V
Vishal Baghel

Contributor-Level 10

A ( 1 1 0 ) = ( 1 1 0 )

A ( 1 0 1 ) = ( 1 0 1 )

A ( 0 0 1 ) = ( 1 1 2 )

A [ 1 1 0 1 0 0 0 1 1 ] = [ 1 1 1 1 0 1 0 1 2 ]

A B = C A = C B 1

| A 2 l | = ( 4 ) ( 1 ) + 3 ( 1 ) + 1 ( 1 )

4 – 3 – 1 = 0

l ( 1 , 0 , 1 ) + m ( 1 , 1 , 0 ) = ( 4 , 3 , 1 ) l m = 4

m = 3 , l = 1

New answer posted

a year ago

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J
Jaya Sharma

Contributor-Level 10

Apart from upper and lower triangular matrices, we have the unit triangular matrix, strictly-triangular matrix and an atomic-triangular matrix.

New answer posted

a year ago

0 Follower 1 View

J
Jaya Sharma

Contributor-Level 10

A few types of matrices are row and column matrices, singleton, null matrix, square, diagonal, scalar, identity, equal, triangular (both upper and lower), singular and non-singular matrices, symmetric, skew-symmetric, hermitian, and orthogonal matrices. NCERT excercise on matrices chapter covers questions related to this topic

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