Let |(x,2,x)(x^2,x,6)(x,x,6)| = ax^4 + bx^3 + cx^2 + dx + e.

Question

Let \(\begin{vmatrix} x & 2 & x \\[0.3em] x^2 & x & 6 \\[0.3em] x &x &6 \end{vmatrix}\)  = ax⁴ + bx³ + cx² + dx + e. 

Then, 

The value of 5a + 4b + 3c + 2d + e is equal to

A. 0 

B. -16 

C. 16 

D. none of these

Answer 1

We are given that,

\(\begin{vmatrix} x & 2 & x \\[0.3em] x^2 & x & 6 \\[0.3em] x &x &6 \end{vmatrix}\)  = ax⁴ + bx³ + cx² + dx + e.

We need to find the value of 5a + 4b + 3c + 2d + e. 

Determinant of 3 × 3 matrix is given as,

⇒ x⁴ – x³ – 12x² + 12x = ax⁴ + bx³ + cx² + dx + e 

Comparing the left hand side and right hand side of the equation, we get 

a = 1 

b = -1 

c = -12 

d = 12 

e = 0 

Putting these values in 5a + 4b + 3c + 2d + e, we get 

5a + 4b + 3c + 2d + e = 5(1) + 4(-1) + 3(-12) + 2(12) + 0 

⇒ 5a + 4b + 3c + 2d + e = 5 – 4 – 36 + 24 

⇒ 5a + 4b + 3c + 2d + e = 25 – 36 

⇒ 5a + 4b + 3c + 2d + e = -11 

Thus, 

The values of 5a + 4b + 3c + 2d + e is -11.