Maths Matrices

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New answer posted•a year ago
S
a year ago

A = [ 0 2 k − 1 ] t h e n     A 2 = [ 0 2 k − 1 ] [ 0 2 k − 1 ] = [ 2 k − 2 − k 2 k + 1 ]

A 3 = [ 2 k − 2 − k 2 k + 1 ] [ 0 2 k − 1 ] = [ − 2 k 4 k + 2 2 k 2 + k − 4 k − 1 ]              

Now, A(A3 + 3l) = 2l gives A3 + 3l = 2A-1

⇒ − 2 k + 3 = 1 k ⇒ 2 k 2 − 3 k + 1 = 0    

  k = 1 2 , 1            

& 4 k + 2 = 2 k o r     2 k + 1 − 1 k s o     o r     2 k 2 + k − 1 = 0

=> k = 1 2

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New answer posted•a year ago
S
a year ago

x + y + z = 4

3x + 2y + 5z = 3

9 x + 4 y + ( 2 8 + [ λ ] ) z = [ λ ]           

Δ = | 1 1 1 3 2 5 9 4 2 8 + [ λ ] | = 5 6 + 2 [ λ ] − 2 0 − ( 8 4 + 3 [ λ ] − 4 5 ) + ( − 6 )

= − [ λ ] − 9              

I f [ λ ] + 9 ≠ 0 then unique solution

& if  [ λ ] + 9 = 0     t h e n       Δ 1 = Δ 2 = Δ 3 = 0 so infinite solution will exist

Hence λ ∈ R  

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New answer posted•a year ago
S
a year ago

| A | = | [ x + 1 ] [ x + 2 ] [ x + 3 ] [ x ] [ x + 3 ] [ x + 3 ] [ x ] [ x + 2 ] [ x + 4 ] | = | [ x ] + 1 [ x ] + 2 [ x ] + 3 [ x ] [ x ] + 3 [ x ] + 3 [ x ] [ x ] + 2 [ x ] + 4 |

R 1 → R 1 − R 3 & R 2 → R 2 − R 3

| 1 0 − 1 0 1 − 1 [ x ] [ x ] + 2 [ x ] + 4 | = 1 9 2

⇒ [ x ] = 6 2

⇒ x ∈ [ 6 2 , 6 3 )              

 

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New answer posted•a year ago
S
a year ago

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

A 3 = ( 1 0 0 1 1 1 1 0 0 ) ( 1 0 0 0 1 1 1 0 0 ) = ( 1 0 0 2 1 1 1 0 0 )

A 2 0 2 5 − A 2 0 2 0

= ( 1 0 0 2 0 2 4 1 1 1 0 0 ) − ( 1 0 0 2 0 1 9 1 1 1 0 0 ) = ( 1 0 0 5 0 0 1 0 0 ) = A 6 − A

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New answer posted•a year ago
S
a year ago

AAT = ATA = l

A ( A T B A ) 2 0 2 1 A T              

=  B 2 0 2 1 = ( I + P ) 2 0 2 1 = l + 2 0 2 1 P = [ 1 0 2 0 2 1 i 1 ]

⇒ ( B 2 0 2 1 ) − 1 = [ 1 0 − 2 0 2 1 i 1 ]

where P = [ 0 0 i 0 ]

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

A=[−121−1] EA=[acbd][−121−1]

=[−a+c2a−c−b+d2b−d]

For a = c For −a+c=02a−c=1]→a=1,c=1                E=[1101]

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

b+d=12b−d=−1]→b=0,d=1                      R1→R1→R2[1001]

For −a+c=12a−c=1]→a=0,c=1

For−a+c=−12a−c=2]→a=1,c=0

−b+d=−22b−d=7]→b=5,d=3            [1053]                     [1001]

R2 → 5R1 + 3R2

For For−a+c=−12a−c=2]→a=1,c=1

−b+d=−12b−d=3]→b=2,d=1

(A) R1 → R1 + R2

(B) R2 → R2 + 2R1 [1021]                           [1001]

(C) R2 → 3R2 + 5R1

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New answer posted•a year ago
S
a year ago

Δ = | − k 3 − 1 4 − 1 5 4 − k − 4 1 3 |

= ( − k ) ( 1 2 − k ) + 3 ( 4 k + 4 5 ) − 1 4 ( − 1 5 + 1 6 )

Δ = 0 ⇒ k = ± 1 1             

 For k = -11,

->11x + 3y – 14z = 25

-4x + y + 3z = 4

{ − 1 1 x + 3 y − 1 4 z = 2 5 − 1 5 x + 4 y − 1 1 z = 3 − 4 x + y + 3 z = 4 } → No solution for k = ± 11

               

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