Question

The coefficient of $x^2$ in the expansion of the determinant $\begin{vmatrix}x^{2}&x^{3}+1&x^{5}+2\\\ x^{2}+3&x^{3}+x&x^{3}+x^{4}\\\ x+4&x^{3}+x^{5}&2^{3}\end{vmatrix}$ is

• A. -10
• B. -8
• C. -2
• D. -6

Answer 1

Answer: (A)

-10

Answer 2

For the coefficient of $x^{2}$, on expanding along $R_{1}$, we get
$\Delta=x^{2}\left[8 x^{2}+8 x-x^{6}-2 x^{7}-x^{8}\right]-\left(x^{3}+1\right)$
$\left[8 x^{3}+24-x^{4}-x^{5}-4 x^{3}-4 x^{4}\right]+\left(x^{5}+2\right)$
$\left[x^{6}+x^{7}+3 x^{3}+3 x^{4}-x^{3}-x^{2}-4 x^{2}-4 x\right]$
$=8 x^{4}+8 x^{3}-x^{8}-2 x^{9}-x^{10}-8 x^{6}$
$-24 x^{3}+x^{7}+x^{8}+4 x^{6}+4 x^{7}-8 x^{3}-24$
$+x^{4}+x^{5}+4 x^{3}+4 x^{4}+x^{11}+x^{12}$
$+3 x^{8}+3 x^{9}-x^{8}-x^{7}-4 x^{7}-4 x^{6}+2 x^{6}$
$+2 x^{7}+6 x^{3}+6 x^{4}-2 x^{3}-2 x^{2}-8 x^{2}-8 x$
Coefficient of $x^{2}=-2-8=-10$