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Grade 12Algebra

Let a and b be the roots of equation Ax^2+Bx+C=0. Evaluate the value of determinant | 3 1+a+b 1+a^2 +b^2 || 1+a+b. 1+a^2+b^2 1+a^3+b^3 || 1+a^2+b^2. 1+a^3+b^3. 1+a^4+b^4. |

bhumika gupta

9 Years agoGrade 12

1 Answer

mycroft holmes

9 Years ago

The given determinant can be written as a product:

$\begin{vmatrix} 1 & 1 &1 \\ 1 & a& b\\ 1& a^2& b^2 \end{vmatrix} \ \times \begin{vmatrix} 1 & 1 &1 \\ 1 & a& b\\ 1& a^2 & b^2 \end{vmatrix}$

= $\begin{vmatrix} 1 & 1 &1 \\ 1 & a& b\\ 1& a^2& b^2 \end{vmatrix} ^2$

Using the well known expansion for the Vandermonde Determinant, we can evaluate this as $(a-b)^2 (a-1)^2(b-1)^2$

Since a,b are roots of the quadratic Ax²+Bx+C, we have

$(a-b)^2 = \frac{B^2-4AC}{A^2}$

and $(a-1)^2 (b-1)^2 = [(a-1)(b-1)]^2 = P(1)^2 = (A+B+C)^2$

Hence given expressions equals

$\frac{(B^2-4AC)(A+B+C)^2}{A^2}$

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