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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants1 Mark
MCQ
Let $\begin{vmatrix}\text{x}&2&\text{x}\\\text{x}^2&\text{x}&6\\\text{x}&\text{x}&6\end{vmatrix}=\text{ax}^4+\text{bx}^3+\text{cx}^2+\text{dx}+\text{e}.$ Then, the value of $5a + 4b + 3c + 2d + e$ is equal to:
  • A
    $0$
  • B
    $-16$
  • C
    $16$
  • ✓
    None of these.
✓

Answer

Correct option: D.
None of these.
$\triangle=\begin{vmatrix}\text{x}&2&\text{x}\\\text{x}^2&\text{x}&6\\\text{x}&\text{x}&6\end{vmatrix}$
$=\text{x}\begin{vmatrix}\text{x}&6\\\text{x}&6\end{vmatrix}-\text{x}^2\begin{vmatrix}2&\text{x}\\\text{x}&6\end{vmatrix}+\text{x}\begin{vmatrix}2&\text{x}\\\text{x}&6\end{vmatrix}$
$=0-x^2\left(12-x^2\right)+x\left(12-x^2\right)$
$=x^4-12 x^2+12 x-x^3$
$=a x^4+b x^3+c x^2+d x+e$
$\Rightarrow x^4-12 x^2+12 x-x^3=a x^4+b x^3+c x^2+d x+e$
$\Rightarrow a=1, b=-1, c=-12, d=12, e=0$
Thus,
$5 a+4 b+3 c+2 d+e=5-4-36+24+0=-11$

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