Class 12Mathematics3 and 4 .Determinants and MatricesRank of Matrices , Some special determinants, differentiation and integration of determinants
MediumIf $y = \begin{vmatrix} f(x) & g(x) & h(x) \\ l & m & n \\ a & b & c \end{vmatrix}$,prove that $\frac{dy}{dx} = \begin{vmatrix} f'(x) & g'(x) & h'(x) \\ l & m & n \\ a & b & c \end{vmatrix}$.
(N/A) Given $y = \begin{vmatrix} f(x) & g(x) & h(x) \\ l & m & n \\ a & b & c \end{vmatrix}$.
Expanding the determinant along the first row,we get:
$y = f(x)(mc - nb) - g(x)(lc - na) + h(x)(lb - ma)$.
Differentiating both sides with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}[f(x)(mc - nb)] - \frac{d}{dx}[g(x)(lc - na)] + \frac{d}{dx}[h(x)(lb - ma)]$.
Since $l, m, n, a, b, c$ are constants:
$\frac{dy}{dx} = f'(x)(mc - nb) - g'(x)(lc - na) + h'(x)(lb - ma)$.
This expression is the expansion of the determinant where the first row is differentiated:
$\frac{dy}{dx} = \begin{vmatrix} f'(x) & g'(x) & h'(x) \\ l & m & n \\ a & b & c \end{vmatrix}$.
Hence,the result is proved.
Expanding the determinant along the first row,we get:
$y = f(x)(mc - nb) - g(x)(lc - na) + h(x)(lb - ma)$.
Differentiating both sides with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}[f(x)(mc - nb)] - \frac{d}{dx}[g(x)(lc - na)] + \frac{d}{dx}[h(x)(lb - ma)]$.
Since $l, m, n, a, b, c$ are constants:
$\frac{dy}{dx} = f'(x)(mc - nb) - g'(x)(lc - na) + h'(x)(lb - ma)$.
This expression is the expansion of the determinant where the first row is differentiated:
$\frac{dy}{dx} = \begin{vmatrix} f'(x) & g'(x) & h'(x) \\ l & m & n \\ a & b & c \end{vmatrix}$.
Hence,the result is proved.
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