Matrices and Determinants 2 Question 5

5. Let α and β be the roots of the equation x2+x+1=0. Then, for y0 in R, |y+1αβαy+β1β1y+α| is equal to

(a) y(y21)

(b) y(y23)

(c) y31

(d) y3

Show Answer

Answer:

Correct Answer: 5. (d)

Solution:

  1. Given, quadratic equation is x2+x+1=0 having roots α,β.

Then, α+β=1 and αβ=1

Now, given determinant

Δ=|y+1αβαy+β1β1y+α|

On applying R1R1+R2+R3, we get

Δ=|y+1+α+βy+1+α+βy+1+α+βαy+β1β1y+α|=|yyyαy+β1β1y+α|

On applying C2C2C1 and C3C3C1, we get

Δ=|y00αy+βα1αβ1βy+αβ|=y[(y+(βα))(y(βα))(1α)(1β)]

[expanding along R1 ]

=y[y2(βα)2(1αβ+αβ)]

=y[y2β2α2+2αβ1+(α+β)αβ]

=y[y2(α+β)2+2αβ+2αβ1+(α+β)αβ]

=y[y21+311]=y3[α+β=1 and αβ=1]



NCERT Chapter Video Solution

Dual Pane