1.

Let |(x,2,x),(x^(2),x,6),(x,x,6)|=Ax^(4)+Bx^(3)+Cx^(2)+Dx+E. Then the value of 5A+4B+3C+2D+E is equal to

Answer»

zero
`-16`
16
`-11`

Solution :LET the given DETERMINANT be equal to `Delta(x)`. Then,
`5A+4B+3C+2D+E=Delta(1)+Delta'(1)`
Now, `Delta(1)=0` as `R_(2) and R_(3)` are IDENTICAL.
`Delta'(x)=|(1,0,1),(x^(2),x,6),(x,x,6)|+|(x,2,x),(2x,1,0),(x,x,6)|+|(x,2,x),(x^(2),x,6),(1,1,0)|`
`Delta'(1)=|(1,2,1),(2,1,0),(1,1,6)|+|(1,2,1),(1,1,6),(1,1,0)|`
`=-17+(12+1-1-6)=-11`


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