If begin{vmatrix} x^2+x & x+1 & x-2 | vedclass

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Question

(A) Let the given determinant be $D(x)$. We are given $D(x) = xA + B$.
To find $A$,we differentiate $D(x)$ with respect to $x$ and evaluate at $x=0$.

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(A) Let the given determinant be $D(x)$. We are given $D(x) = xA + B$.
To find $A$,we differentiate $D(x)$ with respect to $x$ and evaluate at $x=0$.
$D'(0) = A$.
Using the property of differentiation of a determinant,$D'(x)$ is the sum of three determinants where each row is differentiated one at a time.
Let $R_1, R_2, R_3$ be the rows of the determinant.
$D'(x) = \begin{vmatrix} 2x+1 & 1 & 1 \ 4x+3 & 3 & 3 \ 2x+2 & 2 & 2 \end{vmatrix} + \begin{vmatrix} x^2+x & x+1 & x-2 \ 4x+3 & 3 & 3 \ x^2+2x+3 & 2x-1 & 2x-1 \end{vmatrix} + \begin{vmatrix} x^2+x & x+1 & x-2 \ 2x^2+3x-1 & 3x & 3x-3 \ 2x+2 & 2 & 2 \end{vmatrix}$.
Evaluating at $x=0$:
$D'(0) = \begin{vmatrix} 1 & 1 & 1 \ 3 & 3 & 3 \ 2 & 2 & 2 \end{vmatrix} + \begin{vmatrix} 0 & 1 & -2 \ 3 & 3 & 3 \ 3 & -1 & -1 \end{vmatrix} + \begin{vmatrix} 0 & 1 & -2 \ -1 & 0 & -3 \ 2 & 2 & 2 \end{vmatrix}$.
The first determinant is $0$ because rows are proportional.
$A = 0 + [0(0 - (-3)) - 1(-3 - 9) - 2(-3 - 0)] + [0(0 - (-6)) - 1(-2 - (-6)) - 2(-2 - 0)]$.
$A = [0 + 12 + 6] + [0 - 4 + 4] = 18 + 0 = 18$.
Wait,re-evaluating the determinant $A$ as a matrix:
$A = \begin{vmatrix} 0 & 1 & -2 \ 3 & 3 & 3 \ 3 & -1 & -1 \end{vmatrix} + \begin{vmatrix} 0 & 1 & -2 \ -1 & 0 & -3 \ 2 & 2 & 2 \end{vmatrix} = 18 + 0 = 18$.
However,checking the options and the structure,the determinant $A$ is defined as a $3 	imes 3$ matrix.
Given the standard form,$|A| = 18$ is not listed. Re-calculating:
$D'(0) = 18$. Thus $|A| = 18$. Given the options,there might be a typo in the question or options. Based on standard evaluation,the result is $18$.

If $\begin{vmatrix} x^2+x & x+1 & x-2 \\ 2x^2+3x-1 & 3x & 3x-3 \\ x^2+2x+3 & 2x-1 & 2x-1 \end{vmatrix} = xA+B$,where $A$ and $B$ are determinants of order $3$ not involving $x$,then $|A|=$

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