If \( \begin{vmatrix}x & 3 & x\\ x^{2} & x & 7\\ x & x & 7\end{vmatrix}=ax^{4}+bx^{3}+cx^{2}+dx+e \), then the value of \( 6a+5b+4c+3d+2e \) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → -20
Expanding the determinant along the first row,
\( \displaystyle x(7x-7x)-3(7x^{2}-7x)+x(x^{3}-x^{2}) \)
\( \displaystyle =-21x^{2}+21x+x^{4}-x^{3} \)
Arranged in powers of x this is \( x^{4}-x^{3}-21x^{2}+21x+0 \). Comparing coefficients with the given form:
\( \displaystyle a=1,\ b=-1,\ c=-21,\ d=21,\ e=0 \)
Hence
\( \displaystyle 6a+5b+4c+3d+2e=6-5-84+63+0 \)
\( \displaystyle =-20 \)