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The determinant |y^2-x y x^2a b c a ' b ...

The determinant `|y^2-x y x^2a b c a ' b ' c '|` is equal to `|b x+a y c x+b y b^(prime)x+a ' y c^(prime)x+b ' y|` b. `|a x+b y b x+c y a^(prime)x+b ' y b ' x+c ' y|` c. `|b x+c y a x+b y b^(prime)x+c ' y a^(prime)x+b ' y|` d. `|a x+b y b x+c y a^(prime)x+b ' y b^(prime)x+c ' y|`

A

(a)`|{:(bx+ay,cx+by),(b'x+a'y,c'x+b'y):}|`

B

`|{:(a'x+b'y,bx+cy),(ax+by,b'x+c'y):}|`

C

`|{:(bx+cy,ax+by),(b'x+c'y,a'x+b'y):}|`

D

(d)`|{:(ax+by,bx+cy),(a'x+b'y,b'x+c'y):}|`

Text Solution

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The correct Answer is:
D
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The determinant |y^2-x y x^2a b c a ' b ' c '| is equal to a. |b x+a y c x+b y b^(prime)x+a ' y c^(prime)x+b ' y| b. |a x+b y b x+c y a^(prime)x+b ' y b ' x+c ' y| c. |b x+c y a x+b y b^(prime)x+c ' y a^(prime)x+b ' y| d. |a x+b y b x+c y a^(prime)x+b ' y b^(prime)x+c ' y|

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If x\ a n d\ y are two co-primes, then their LCM is (a) x y (b) x+y (c) x/y (d) 1

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If (a x^2+b x+c)y+(a^(prime)x^2+b^(prime)x^2+c^(prime))=0 and x is a rational function of y , then prove that (a c^(prime)-a^(prime)c)^2=(a b^(prime)-a^(prime)b)xx(b c^(prime)-b^(prime)c)dot

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