Let A = x 1 1 0 ,   x   ε   R and A 4 = a i j . If a 11 = 109 , then a 22 is equal…

Let A=x110, x ε R and A4=aij. If a11=109, then a22 is equal to_____________.

Solution

A2=x110×x110=x2+1xx1

A4=x2+1xx1×x2+1xx1=x4+3x2+1x3+2xx3+2xx2+1

Given a11=109x4+3x2+1=109

x2+12x2-9=0x2=9

a22=x2+1=10

Asked in: JEE Main 2020 (03 Sep Shift 1)

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