Maths Matrices

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New answer posted•a year ago
S
a year ago

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 .          

          

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New answer posted•a year ago
S
a year ago

  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

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New answer posted•a year ago
S
a year ago

(I + M)² = (I + M) (I + M) = I + 2M + M² = (I + 2M)
(I + M)³ = (I + 2M) (I + M) = I + 3M + 2M² = (I + 3M)
(I + M)? = I + 50M
det (I + M)? - 50M) = det (I) = 1

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New answer posted•a year ago
S
a year ago

[ 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

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New answer posted•a year ago
S
a year ago

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

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New answer posted•a year ago
S
a year ago

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

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New answer posted•a year ago
S
a year ago

A = [1]
[0 1]
Now A² = [1] [1] = [1 2]
[0 1] [0 1] [0 1]
A³ = A².A = [1 2] [1] = [1 3]
[0 1] [0 1] [0 1]
Similarly A²? ¹¹ = [1 2011]
[0 1]

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New answer posted•a year ago
S
a year ago

A² = [1 2 3; 0 1 2; 0 1]
A³=A².A= [1 3 6; 0 1 3; 0 1]
A²? = [1 20 1+2+3.20; 0 1 20; 0 1] = [1 20 210; 0 1 20; 0 1]
M= [20 210 520; 0 20 210; 0 20]
M (a? )=T? =n (n+1)/2
S? = 1/2 [ n (n+1) (2n+1)/6 + n (n+1)/2 ]
⇒S? =1540
⇒M=2020

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New answer posted•a year ago
S
a year ago

A? =B? (i)
A³B²=A²B³. (ii)
Subtract (i) & (ii)
⇒A³ (A²−B²)=B³ (B²−A²)
⇒ (A²−B²) (A³+B³)=0
A²−B² is invertible matrix
∴A²−B²≠0
⇒A³+B³=0
∴? A³+B³? =0

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New answer posted•a year ago
S
a year ago

x + 2y + z = 2
αx + 3y – z = α
–αx + y + 2z = –α

Δ = | (1, 2, 1), (α, 3, -1), (-α, 1, 2) | = 1 (6+1) – 2 (2α–α) + 1 (α+3α) = 7+2α
α = –7/2

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