If A = cos α sin α - sin α cos α , then verify that A ' A = I

Given: A = cos α sin α - sin α cos α To get transpose interchange rows and columns. ⇒ A ' = cos α - sin α sin α cos α Therefore A ' A = cos α - sin α sin α cos α cos α sin α - sin α cos α = cos 2 α + sin 2 α cos α sin α - cos α sin α cos α sin α - cos α sin α cos 2 α + sin 2 α = 1 0 0 1 = I ∴ A ' A = I Hence, proved

A matrix consists of 30 elements. What are the possible orders it can have?

Since there are 30 elements, we can make the multiples of 30 , which are: 1 × 30 ; 30 × 1 ; 2 × 15 ; 15 × 2 ; 3 × 10 ; 10 × 3 ; 5 × 6 and 6 × 5 So, the possible orders are ( 1 , 30 ) , ( 2 , 15 ) , ( 3 , 10 ) , ( 5 , 6 ) , ( 6 , 5 ) , ( 10 , 3 ) , ( 15 , 2 ) , ( 30 , 1 )

The LCM of x 3 – a 3 and ( x – a ) 2 is:

( x – a ) 2 ( x 2 + a x + a 2 )

( x + a ) 2 ( x 2 + a x + a 2 )

The G . C . D of x 4 + 3 x 3 + 5 x 2 + 26 x + 56 and x 4 + 2 x 3 - 4 x 2 - x + 28 is x 2 + 5 x + 7 . Find their L . C . M .

Given, G . C . D of x 4 + 3 x 3 + 5 x 2 + 26 x + 56 and x 4 + 2 x 3 - 4 x 2 - x + 28 is x 2 + 5 x + 7 . If p x and q x are two polynomials, then p x × q x = L . C . M × H . C . F . ∴ L . C . M × x 2 + 5 x + 7 = x 4 + 3 x 3 + 5 x 2 + 26 x + 56 x 4 + 2 x 3 - 4 x 2 - x + 28 ⇒ L . C . M = x 4 + 3 x 3 + 5 x 2 + 26 x + 56 x 4 + 2 x 3 - 4 x 2 - x + 28 x 2 + 5 x + 7 ⇒ L . C . M = x 4 - 2 x 3 + 8 x 2 + 5 x 3 - 10 x 2 + 40 x + 7 x 2 - 14 x + 56 x 4 + 2 x 3 - 4 x 2 - x + 28 x 2 + 5 x + 7 ⇒ L . C . M = x 2 x 2 - 2 x + 8 + 5 x x 2 - 2 x + 8 + 7 x 2 - 2 x + 8 x 4 + 2 x 3 - 4 x 2 - x + 28 x 2 + 5 x + 7 ⇒ L . C . M = x 2 + 5 x + 7 x 2 - 2 x + 8 x 4 + 2 x 3 - 4 x 2 - x + 28 x 2 + 5 x + 7 ⇒ L . C . M = x 2 - 2 x + 8 x 4 + 2 x 3 - 4 x 2 - x + 28 ⇒ L . C . M = x 2 x 4 + 2 x 3 - 4 x 2 - x + 28 - 2 x x 4 + 2 x 3 - 4 x 2 - x + 28 + 8 x 4 + 2 x 3 - 4 x 2 - x + 28 ⇒ L . C . M = x 6 + 2 x 5 - 4 x 4 - x 3 + 28 x 2 - 2 x 5 - 4 x 4 + 8 x 3 + 2 x 2 - 56 x + 8 x 4 + 16 x 3 - 32 x 2 - 8 x + 224 ⇒ L . C . M = x 6 + 2 x 5 - 2 x 5 + 8 x 4 - 4 x 4 - 4 x 4 + 16 x 3 - x 3 + 8 x 3 + 28 x 2 + 2 x 2 - 32 x 2 - 56 x - 8 x + 224 ⇒ L . C . M = x 6 + 23 x 3 + 2 x 2 - 64 x + 224 Therefore, x 6 + 23 x 3 + 2 x 2 - 64 x + 224 is the required L . C . M .

Find the square root of the polynomial 4 + 25 x 2 - 12 x - 24 x 3 + 16 x 4 by division method.

Given: 4 + 25 x 2 - 12 x - 24 x 3 + 16 x 4 We can rewrite this as 4 + 25 x 2 - 12 x - 24 x 3 + 16 x 4 = 16 x 4 - 24 x 3 + 25 x 2 - 12 x + 4 Note that if this is a square of a simpler polynomial, then it must be expressible in one of two forms: ( 4 x 2 + a x − 2 ) 2 ( 4 x 2 + a x + 2 ) 2 So we have one of the following: ( 4 x 2 + a x − 2 ) 2 = 16 x 4 + 8 a x 3 + ( a 2 − 16 ) x 2 − 4 a x + 4 ( 4 x 2 + a x + 2 ) 2 = 16 x 4 + 8 a x 3 + ( a 2 + 16 ) x 2 + 4 a x + 4 Equating the coefficients of x 3 , in either case we find 8 a = − 24 and hence a = − 3 Then matching the coefficients of x , we have − 12 = ± 4 a and we find that we need + 4 a x So 16 x 4 − 24 x 3 + 25 x 2 − 12 x + 4 = ( 4 x 2 − 3 x + 2 ) 2 16 x 4 − 24 x 3 + 25 x 2 − 12 x + 4 = 4 x 2 − 3 x + 2 Hence, the required answer is 4 x 2 - 3 x + 2 .