Determinants

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New answer posted

a year ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

This is a  Fill in the blanks as classified in NCERT Exemplar

Sol:

T r u e . L e t                 Δ = | 1 1 1 1 ( 1 + s i n θ ) 1 1 1 1 + c o s θ | C 1 → C 1 − C 2 , C 2 → C 2 − C 3                                         = | 0 0 1 − s i n θ s i n θ 1 0 − c o s θ 1 + c o s θ | E x p a n d i n g     a l o n g     C 3                                         = 1 | − s i n θ s i n θ 0 − c o s θ | = s i n θ c o s θ − 0 = s i n θ c o s θ                                         = 1 2 . 2 s i n θ c o s θ = 1 2 s i n 2 θ                                           = 1 2 * 1 = 1 2                                                                                   [ M a x i m u m     v a l u e     o f     s i n 2 θ = 1 ]

New answer posted

a year ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

This is a  Fill in the blanks as classified in NCERT Exemplar

Sol:

T r u e . G i v e n     t h a t                 Δ = | a p x b q y c r z | = 1 6 L . H . S .         Δ 1 = | p + x a + x a + p q + y b + y b + q r + z c + z c + r | C 1 → C 1 + C 2 + C 3                                 = | 2 p + 2 x + 2 a a + x a + p 2 q + 2 y + 2 b b + y b + q 2 r + 2 z + 2 c c + z c + r |                                 = 2 | p + x + a a + x a + p q + y + b b + y b + q r + z + c c + z c + r |                                   [ T a k i n g     2     c o m m o n     f r o m     C 1 ] C 1 → C 1 − C 2                                 = 2 | p a + x a + p q b + y b + q r c + z c + r | C 3 → C 3 − C 1                                   = 2 | p a + x a q b + y b r c + z c | S p l i t t i n g     u p     C 2                                 = 2 | p a a q b b r c c | + 2 | p x a q y b r z c | = 2 ( 0 ) + 2 | p x a q y b r z c |                               = 2 | p x a q y b r z c | ⇒ 2 | a p x b q y c r z |                               ( C 1 ↔ C 3     a n d     C 2 ↔ C 3 )                               = 2 * 1 6 = 3 2

New answer posted

a year ago

0 Follower 12 Views

V
Vishal Baghel

Contributor-Level 10

This is a  Fill in the blanks as classified in NCERT Exemplar

Sol:

True.Let       Δ=|x+ap+ul+fy+bq+vm+gz+cr+wn+h|Splitting  up  C1                =|xp+ul+fyq+vm+gzr+wn+h|+|ap+ul+fbq+vm+gcr+wn+h|Splitting  up  C2  in  both  determinants                =|xpl+fyqm+gzrn+h|+|xul+fyvm+gzwn+h|+|apl+fbqm+gcrn+h|+|aul+fbvm+gcwn+h|Similarly  by  splitting  C3 in  each determinants ,  we  will  get  8determinants  .

New answer posted

a year ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

This is a  Fill in the blanks as classified in NCERT Exemplar

Sol:

T r u e . L e t     Δ = | s i n A c o s A s i n A + c o s B s i n B c o s A s i n B + c o s B s i n C c o s A s i n C + c o s B | S p l i t t i n g     u p     C 3                                 = | s i n A c o s A s i n A s i n B c o s A s i n B s i n C c o s A s i n C | + | s i n A c o s A c o s B s i n B c o s A c o s B s i n C c o s A c o s B |                                 = 0 + | s i n A c o s A c o s B s i n B c o s A c o s B s i n C c o s A c o s B |                       [ ?     C 1     a n d     C 3     a r e     i d e n t i c a l ]                                 = c o s A c o s B | s i n A 1 1 s i n B 1 1 s i n C 1 1 |                             [ T a k i n g     c o s A     a n d     c o s B     c o m m o n     f r o m     C 2     a n d     C 3     r e s p e c t i v e l y ]                                 = c o s A c o s B ( 0 )                                                                       [ ?     C 2     a n d     C 3     a r e     i d e n t i c a l ]                                 = 0

New answer posted

a year ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

This is a  Fill in the blanks as classified in NCERT Exemplar

Sol:

F a l s e . Since  ? adjA? =? A? n−1  where  n  is  the  order  of  the  square  matrix.

New answer posted

a year ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

This is a  Fill in the blanks as classified in NCERT Exemplar

Sol:

T r u e . L e t                               Δ = | x + 1 x + 2 x + a x + 2 x + 3 x + b x + 3 x + 4 x + c | R 2 → 2 R 2 − ( R 1 + R 3 )                                                           = | x + 1 x + 2 x + a 0 0 2 b − ( a + c ) x + 3 x + 4 x + c | a ,   b ,   c     a r e     i n     A . P . ∴ b − a = c − b           ⇒           2 b = a + c                                                             = | x + 1 x + 2 x + a 0 0 0 x + 3 x + 4 x + c | = 0

New answer posted

a year ago

0 Follower 23 Views

V
Vishal Baghel

Contributor-Level 10

This is a  Fill in the blanks as classified in NCERT Exemplar

Sol:

True.  Since |A|=12If  A  is  a  square  matrix  of  order  nthen              |AdjA|=|A|n−1∴                     |AdjA|=|A|3−1=|A|2= (12)2=144         [n=3]

New answer posted

a year ago

0 Follower 11 Views

V
Vishal Baghel

Contributor-Level 10

This is a  Fill in the blanks as classified in NCERT Exemplar

Sol:

T r u e .         | 3 A B | = 3 3 | A B | = 2 7 | A | | B | = 2 7 * 5 * 3                       [ ? | K A | = K n | A | ]

New answer posted

a year ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

This is a  Fill in the blanks as classified in NCERT Exemplar

Sol:

F a l s e . Since? ? |A? 1|=|A|? 1for? ? a? ? non? singular? ? matrix.

New answer posted

a year ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

This is a  Fill in the blanks as classified in NCERT Exemplar

Sol:

If  a  non−singular  square  matix,  then  for  any  non−zero  scalar  'a',  aA  is  invertible. ∴                             ( a A ) . ( 1 a A − 1 ) = a . 1 a . A . A − 1 = I S o ,     ( a A )     i s     i n v e r s e     o f     ( 1 a A − 1 ) ⇒         ( a A ) − 1 = 1 a A − 1     i s     t r u e .

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