BPSC 46th Exam Prelims 2004 Solved Paper

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Question : 129 of 150

Marks: +1, -0

Let matrices A

=[\table1, -1 ; 2, 2]

, B

=[\table 1, -α ; β, 1]A

(A + B)(A – B)
=

A^2 – B^2

then

α= β = –l
α= β = 0
α– β– 1
α =1, β = –1

Solution:

From the question,

A=[\table 1 , -1 ; 2 , 2], \;\; B=[\table 1 , -2 ; β , 2]

⇒ {A}=[\table 1 , α ; β , 2]

A^{2}=[\table 1 , -1 ; 2 , 2][\table 1 , -1 ; 2 , 2]=[\table -1 , -3 ; 6 , 2]

{B}^{2}=[\table 1 , α ; β , 2][\table 1 , α ; β , 2]=[\table 1+α β , 2 α ; 2 β , α β+1]

A+B=[\table 1 , -1 ; 2 , 2][\table 1 , α ; β , 1]=[\table 2 , α-1 ; 2+β , 3]

{A}- {B}=[\table 1 , -1 ; 2 , 2]-[\table 1 , α ; β , 1]=[\table 0 , -(α-1) ; 2-β , 1]

Now

(A+B) ⋅(A-B)

=[\table (α-1)(2-β) , -2(α+1)+(α-1) ; 3(2-β) , -(2+β)(α+1)+3]

=[\table 2 α-α β-2+β , -2 α-2+α-1 ; 6-3 β , -2 α-2-α β+3]

{A}^{2}- {B}^{2}=[\table -1 , -3 ; 6 , 2][\table 1+α β , 2 α ; 2 β , α β+1]

=[\table -1-1-α β , -3-2 α ; 6-2 β , 2-α β-1]

∵ (A+B)(A-B)=A^{2}-B^{2}

So,

-α-3=-3-2 α

-α+2 α=-3+3

⇒ α=0

Again,

6-3 β=6-2 β

⇒ 0=β ⇒ β=0

∴ α=β=0

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