KVPY 2020-21 SB SX Exam Previous Papers

Question : 9 of 120

Marks: +1, -0

Let

{A}=({a}_{{ij}})_{1 \≤ {i}, {j} ≤ 3}

be a

3 \× 3

invertible matrix where each

{a}_{{ij}}

is a real number. Denote the inverse of the matrix

{A}

{A}^{-1}

. If

\∑↙{{j}=1} ↖{3} {a}_{{ij}}=1

for

1 \≤ {i} \≤ 3

, then

Solution:

Since,

A ⋅ A^{-1}=I

⇒[\table a_{1} , a_{2} , a_{3} ; b_{1} , b_{2} , b_{3} ; c_{1} , c_{2} , c_{3} ]

[\table x_{1} , x_{2} , x_{3} ; y_{1} , y_{2} , y_{3} ; z_{1} , z_{2} , z_{3} ]

=[\table 1 , 0 , 0 ; 0 , 1 , 0 ; 0 , 0 , 1 ]

⇒\{\table x_{1} a_{1}+y_{1} a_{2}+z_{1} a_{3}=1 ; x_{2} a_{1}+y_{2} a_{2}+z_{2} a_{3}=0 ; x_{3} a_{1}+y_{3} a_{2}+z_{3} a_{3}=0 .

On adding we get,

(Σ x_{i}) a_{1}+(Σ y_{i}) a_{2}+(Σ z_{1}) a_{3}=1

And it is given that,

a_{1}+a_{2}+a_{3}=1

Therefore,

\;\; Σ x_{i}=1=Σ y_{i}=Σ z_{i}

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