Let|(x^2+3x,x-1,x+3)(x+1,-2x,x-4)(x-3,x+4,3x)| = ax^4 + bx^3 + cx^2 + dx + e be an identify in x,

Question

Let \(\begin{vmatrix} x^2+3x & x-1 &x+3 \\[0.3em] x+1 & -2x &x-4 \\[0.3em] x-3 &x+4 &3x \end{vmatrix}\) = ax⁴ + bx³ + cx² + dx + e 

be an identify in x, were a, b, c, d, e are independent of x. Then the value of e is

A. 4 

B. 0 

C. 1 

D. none of these

Answer 1

B. 0

We are given that,

\(\begin{vmatrix} x^2+3x & x-1 &x+3 \\[0.3em] x+1 & -2x &x-4 \\[0.3em] x-3 &x+4 &3x \end{vmatrix}\) 

= ax⁴ + bx³ + cx² + dx + e 

We need to find the value of e. 

We know that, 

Determinant of 3 × 3 matrix is given as,

⇒ (x² + 3x)[-2x × 3x – (x – 4)(x + 4)] – (x – 1)[(x + 1) × 3x – (x – 4)(x – 3)] + (x + 3)[(x + 1)(x + 4) – (-2x)(x – 3)] 

= ax⁴ + bx³ + cx² + dx + e 

⇒ (x² + 3x)[-6x – (x² – 16)] – (x – 1)[3x(x + 1) – (x² – 3x – 4x + 12)] + (x + 3)[x² + x + 4x + 4 + 2x(x – 3)] 

= ax⁴ + bx³ + cx² + dx + e

⇒ (x² + 3x)[-6x – x² + 16] – (x – 1)[3x² + 3x – x² + 7x – 12] + (x + 3)[x² + 5x + 4 + 2x² – 6x] 

= ax⁴ + bx³ + cx² + dx + e

⇒ -x⁴ – 6x³ + 16x² – 3x³ – 18x² + 48x – (x – 1)[2x² + 10x – 12] + (x + 3)[3x² – x + 4] 

= ax⁴ + bx³ + cx² + dx + e

⇒ -x⁴ – 9x³ – 2x² + 48x – (2x³ – 2x² + 10x² – 10x – 12x + 12) + 3x³ + 9x² – x² – 3x + 4x + 12 

= ax⁴ + bx³ + cx² + dx + e

⇒ -x⁴ – 9x³ – 2x² + 48x – 2x³ + 2x² – 10x² + 10x + 12x – 12 + 3x³ + 9x² – x² – 3x + 4x + 12 

= ax⁴ + bx³ + cx² + dx + e

⇒ -x⁴ – 9x³ – 2x³ + 3x³ – 2x² + 2x² + 9x² – x² + 48x + 10x + 12x – 3x + 4x – 12 + 12 

= ax⁴ + bx³ + cx² + dx + e

⇒ -x⁴ – 8x³ + 8x² + 23x + 0 

= ax⁴ + bx³ + cx² + dx + e

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

e = 0 

Thus, 

e = 0.