Class 12th

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New answer posted

a year ago

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V
Vishal Baghel

Contributor-Level 10

(i) |cosθsinθsinθcosθ|

=cosθ*cosθ (sinθ)*sinθ

= cos²θ + sin²θ

=1

(ii) |x2x+1x1x+1x+1|

= (x2x+1) (x+1) (x1) (x+1)

x3 + x2 − x2 − x + x +1 – x2  + 1

= x3x2 + 2

New answer posted

a year ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

|2451|= 2* (−1) – 4* (−5) = −2+20 = 18

New question posted

a year ago

0 Follower 2 Views

New question posted

a year ago

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New answer posted

a year ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

Given, A2= A.

(E) we need to calculate,

(I + A)3- 7A = I3 + A3 + 3IA (I + A) - 7A { (x + y)3 = x3 + y3 + 3xy (a + y)}

New question posted

a year ago

0 Follower 7 Views

New answer posted

a year ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

Given, A is both symmetric and skew-symmetric.

(E) Then, A' = A ____ (1) and A' = -A ____ (2)

So using (2), A' = -A.

A = -A {eqn (I)}

A + A = 0

2A = 0

A = 0.

A is a zero matrix

So, option B is correct.

New answer posted

a year ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

Kindly go through the solution

New answer posted

a year ago

0 Follower 11 Views

V
Vishal Baghel

Contributor-Level 10

We have, AB = BA. (given)

(E) P (n):AB' = B'A.

P (i):AB1 = B1A. Þ AB = BA

so, the result is true for n = 1.

Let the result be true for n = k.

P (k):ABk = BkA

Then,

P (k + 1) : ABk + 1 = A. Bk. B = BkA.B = Bk.BA {? } AB=BA

= Bk + 1.A .

So, ABk + 1 = Bk + 1A.

The result also holds for n = k + 1.

Hence, AB^n = B^n A^n holds for all natural number 'n'.

New answer posted

a year ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

Kindly go through the solution

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