Using properties of determinates prove ...

CBSE Class 12 Math 2020 Outside Delhi Set 1 Solved Paper

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Section - D

Q. Nos. 33 to 36 carry 6 marks each.

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Question : 33 of 36

Marks: +1, -0

Using properties of determinates prove that:

{|\table a-b , b+c , a ; b-c , c+a , b ; c-a , a+b , c |}=a^3+b^3+c^3-3 a b c

OR
If

A={[\table 1 , 3 , 2 ; 2 , 0 , -1 ; 1 , 2 , 3]}

, then show that

A^3-4 A^2-3 A+11 I=O

. Hence find

A^{-1}

Solution: 👈: Video Solution

{|\table a-b , b+c , a ; b-c , c+a , b ; c-a , a+b , c |}

={|\table -b , b+c+a , a ; -c , c+a+b , b ; -a , a+b+c , c |} [\;\text " Applying "\; C_1 → C_1-C_3 \;\text " and "\; C_2 → C_2+C_3]

=(-1)(a+b+c){|\table b , 1 , a ; c , 1 , b ; a , 1 , c |}

[Taking

(-1)

common from

C_1

and

(a+b+c)

common from

C_2

]

=(-1)(a+b+c){|\table b , 1 , a ; c-b , 0 , b-a ; a-b , 0 , c-a|} [\;\text " Applying "\; R_2 → R_2-R_1 \;\text " and "\; R_3 → R_3-R_1]

=(-1)(a+b+c)[-(c-b)(c-a)+(b-a)(a-b)]

=(-1)(a+b+c) [-c^2+a c+b c-a b+b a-b^2-a^2+a b]

=(-1)(a+b+c) (-a^2-b^2-c^2+a b+b c+a c)

=(a+b+c) (a^2+b^2+c^2-a b-b c-a c)

=a^3+a b^2+a c^2-a^2 b-a b c-a^2 c+b a^2+b^3+b c^2-a b^2-b^2 c-a b c+c a^2+c

b^2+c^3-a c b-b c^2-a c^2

=a^3+b^3+c^3-3 a b c

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