Maths Matrices

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New answer posted•a year ago
S
a year ago

| a d j ( a d j ( A ) ) | = | A | 2 2 = | A | 4

∴ | A | 4 = | 1 4 2 8 − 1 4 − 1 4 1 4 2 8 2 5 − 1 4 1 4 |

= ( 1 4 ) 3 | 1 2 − 1 − 1 1 2 2 − 1 1 |

= ( 1 4 ) 3 ( 3 − 2 ( − 5 ) − 1 ( − 1 ) )

| A | 4 = ( 1 4 ) 4 ⇒ | A | = 1 4

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New answer posted•a year ago
S
a year ago

x + 2y + z = 2

α x + 3 y − z = α

− α x + y + 2 z = − α            

Δ = | 1 2 1 α 3 − 1 − α 1 2 | = 1 ( 6 + 1 ) − 2 ( 2 α − α ) + 1 ( α + 3 α ) = 7 + 2 a            

α = − 7 2                

...Read more

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New answer posted•a year ago
S
a year ago

2sin (π22)sin (3π22)sin (5π22)sin (7π22)sin (9π22)

=2sin32π1125sinπ11=116

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New answer posted•a year ago
S
a year ago

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

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New answer posted•a year ago
S
a year ago

A = [ − 1 2 1 − 1 ]

E A = [ a c b d ] [ − 1 2 1 − 1 ]

= [ − a + c 2 a − c − b + d 2 b − d ]

For a = c For − a + c = 0 2 a − c = 1 ] → a = 1 , c = 1                                 E = [ 1 1 0 1 ]  

d = b + 1, d = 1, b = 0

b + d = 1 2 b − d = − 1 ] → b = 0 , d = 1                                             R 1 → R 1 → R 2 [ 1 0 0 1 ]

For  − a + c = 1 2 a − c = 1 ] → a = 0 , c = 1

F o r − a + c = − 1 2 a − c = 2 ] → a = 1 , c = 0

− b + d = − 2 2 b − d = 7 ] → b = 5 , d = 3                         [ 1 0 5 3 ]                                           [ 1 0 0 1 ]

R2 -> 5R1 + 3R2

For F o r − a + c = − 1 2 a − c = 2 ] → a = 1 , c = 1

− b + d = − 1 2 b − d = 3 ] → b = 2 , d = 1

(A) ® R1 ® R1 + R2

(B) ® R2 ® R2 + 2R1 [ 1 0 2 1 ]                                                       [ 1 0 0 1 ]

(C) ® R2 ® 3R2 + 5R1

 

 

 

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New answer posted•a year ago
New answer posted•a year ago
S
a year ago

Let A =  [abcdef9hi]

Now ATA

trace will be a2+b2+c2+d2+e2+f2+92+h2=6

total ways = 9C6.26.1.1.1 = 5376

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New answer posted•a year ago
S
a year ago

A=[100010001]+[0aa00b000]=l+B

B2=[0aa00b000]×[0aa00b000]=[00ab000000]

B3 = 0

∴An=(1+B)n=nC0l+nC1B+nC2B2+nC3B3+....

On comparing we get na = 48, nb = 96 and na + n(n−1)2ab=2160

⇒a=4,n=12  and  b=8

⇒n+a+b=24

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New answer posted•a year ago
S
a year ago

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 =

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New answer posted•a year ago
S
a year ago

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

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