Differentiation of Determinants

Differentiation of Determinants

To differentiation a determinant, we differentiate one row (or column) at a time, keeping other unchanged. For example, if

\(\Delta \left( x \right)=\left| \begin{matrix}   f(x) & g(x)  \\   u\left( x \right) & v\left( x \right)  \\\end{matrix} \right|\).

Differentiation on both sides,

\(\frac{d}{dx}\left( \Delta x \right)=\left| \begin{matrix}   f'(x) & g'(x)  \\   u(x) & v(x)  \\\end{matrix} \right|+\left| \begin{matrix}   f(x) & g(x)  \\   u'(x) & v'(x)  \\\end{matrix} \right|\).

(or)

\(\frac{d}{dx}\left( \Delta x \right)=\left| \begin{matrix}   f'(x) & g(x)  \\   u'(x) & v(x)  \\\end{matrix} \right|+\left| \begin{matrix}   f(x) & g'(x)  \\   u(x) & v'(x)  \\\end{matrix} \right|\),

Similar results hold for the differentiation of determinants of higher order.

Example: If \(f\left( x \right)=\left| \begin{matrix}   x+{{a}^{2}} & ab & ac  \\   ab & x+{{b}^{2}} & bc  \\   ac & bc & x+{{c}^{2}}  \\\end{matrix} \right|\), then prove that f’(x) = 3x² + 2x (a² + b² + c²) .

Solution: Given that  \(f\left( x \right)=\left| \begin{matrix}   x+{{a}^{2}} & ab & ac  \\   ab & x+{{b}^{2}} & bc  \\   ac & bc & x+{{c}^{2}}  \\\end{matrix} \right|\),

Differentiation with respect x.

\(f’\left( x \right)=\left| \begin{matrix}   (x+{{a}^{2}})’ & (ab)’ & (ac)’  \\   ab & x+{{b}^{2}} & bc  \\   ac & bc & x+{{c}^{2}}  \\\end{matrix} \right|+\left| \begin{matrix}   x+{{a}^{2}} & ab & ac  \\   (ab)’ & (x+{{b}^{2}})’ & (bc)’  \\   ac & bc & x+{{c}^{2}}  \\\end{matrix} \right|+\left| \begin{matrix}   x+{{a}^{2}} & ab & ac  \\   ab & x+{{b}^{2}} & bc  \\   (ac)’ & (bc)’ & (x+{{c}^{2}})’  \\\end{matrix} \right|\),

\(f’\left( x \right)=\left| \begin{matrix}   1 & 0 & 0  \\   ab & x+{{b}^{2}} & bc  \\   ac & bc & x+{{c}^{2}}  \\\end{matrix} \right|+\left| \begin{matrix}   x+{{a}^{2}} & ab & ac  \\   0 & 1 & 0  \\   ac & bc & x+{{c}^{2}}  \\\end{matrix} \right|+\left| \begin{matrix}   x+{{a}^{2}} & ab & ac  \\   ab & x+{{b}^{2}} & bc  \\   0 & 0 & 1  \\\end{matrix} \right|\).

\(=1\left( \left( x+{{b}^{2}} \right)\left( x+{{c}^{2}} \right)-{{\left( bc \right)}^{2}} \right)-0+0-0+1\left( \left( x+{{a}^{2}} \right)\left( x+{{c}^{2}} \right)-{{\left( ac \right)}^{2}} \right)\)\(+0-0+1\left( \left( x+{{a}^{2}} \right){{\left( x+b \right)}^{2}}-{{\left( ab \right)}^{2}} \right)\),

\(=1\left( \left( x+{{b}^{2}} \right)\left( x+{{c}^{2}} \right)-{{b}^{2}}{{c}^{2}} \right)+1\left( \left( x+{{a}^{2}} \right)\left( x+{{c}^{2}} \right)-{{a}^{2}}{{c}^{2}} \right)+1\left( \left( x+{{a}^{2}} \right){{\left( x+b \right)}^{2}}-{{a}^{2}}{{b}^{2}} \right)\),

f’(x) = 3x² + 2x (a² + b² + c²).