Differentiating and integrating determinants is one of the integral concepts in Mathematics. This lesson will cover the steps on how to differentiate and integrate determinants easily, along with solved examples.
 Contents in Determinants
Differentiation of Determinants
Let
\(\begin{array}{l}\Delta \left( x \right)=\left| \begin{matrix} {{f}_{1}}\left( x \right) & {{g}_{1}}\left( x \right) \\ {{f}_{2}}\left( x \right) & {{g}_{2}}\left( x \right) \\ \end{matrix} \right|,\end{array} \)
where f1(x), f2(x), g1(x), and g2(x) are functions of x. Then,
\(\begin{array}{l}\Delta ‘\left( x \right)=\left| \begin{matrix} {{f}_{1}}’\left( x \right) & {{g}_{1}}’\left( x \right) \\ {{f}_{2}}\left( x \right) & {{g}_{2}}\left( x \right) \\ \end{matrix} \right|+\left| \begin{matrix} {{f}_{1}}\left( x \right) & {{g}_{1}}\left( x \right) \\ {{f}_{2}}’\left( x \right) & {{g}_{2}}’\left( x \right) \\ \end{matrix} \right|\\ Also, \;\;\;\Delta’\left( x \right)=\left| \begin{matrix} {{f}_{1}}’\left( x \right) & {{g}_{1}}\left( x \right) \\ {{f}_{2}}’\left( x \right) & {{g}_{2}}\left( x \right) \\ \end{matrix} \right|+\left| \begin{matrix} {{f}_{1}}\left( x \right) & {{g}_{1}}’\left( x \right) \\ {{f}_{2}}\left( x \right) & {{g}_{2}}’\left( x \right) \\ \end{matrix} \right|\end{array} \)
How to Differentiate a Determinant?
Thus, to differentiate a determinant, we differentiate one row (or column) at a time, keeping others unchanged. If we write
\(\begin{array}{l}\Delta \left( x \right)=\left[ {{C}_{1}}\,\,\,{{C}_{2}} \right],\end{array} \)
where Ci denotes the ith column, then \(\begin{array}{l}\Delta’\left( x \right)=\left[ {{C}_{1}}’\,\,\,{{C}_{2}} \right]+\left[ {{C}_{1}}\,\,\,{{C}_{2}}’ \right],\end{array} \)
where Ci ‘ denotes the column obtained by differentiating functions in the ith column Ci. Also, if \(\begin{array}{l}\Delta \left( x \right)=\left[ \begin{matrix} {{R}_{1}} \\ {{R}_{2}} \\ \end{matrix} \right],\; then \;\Delta’ \left( x \right)=\left[ \begin{matrix} {{R}_{1}}’ \\ {{R}_{2}} \\ \end{matrix} \right]+\left[ \begin{matrix} {{R}_{1}} \\ {{R}_{2}}’ \\ \end{matrix} \right]\end{array} \)
Similarly, we can differentiate determinants of higher order.
Note: Differentiation can also be done column-wise by taking one column at a time.
Integration of Determinants
If f(x), g(x) and h(x) are functions of x and a, b, c, α, β and γ are constants such that
\(\begin{array}{l}\Delta \left( x \right)=\left| \begin{matrix} f\left( x \right) & g\left( x \right) & h\left( x \right) \\ a & b & c \\ \alpha & \beta & \gamma \\ \end{matrix} \right|,\end{array} \)
then the integral of the determinants is given by;
\(\begin{array}{l}\int{\Delta \left( x \right)dx=\left| \begin{matrix} \int{f\left( x \right)dx} & \int{g\left( x \right)dx} & \int{h\left( x \right)dx} \\ a & b & c \\ \alpha & \beta & \gamma \\ \end{matrix} \right|}\end{array} \)
Solved Problems on Differentiation and Integration of Determinants
Example 1: If
\(\begin{array}{l}\Delta \left( x \right)=\left| \begin{matrix} {{\sin }^{2}}x & \log \cos x & \log \tan x \\ {{n}^{2}} & 2n-1 & 2n+1 \\ 1 & -2\log 2 & 0 \\ \end{matrix} \right|,\; then \;evaluate\; \int\limits_{0}^{\pi /2}{\Delta \left( x \right)dx.}\end{array} \)
Solution:
By applying integration on variable elements of determinants, we will solve the given problem.
We have
\(\begin{array}{l}\Delta \left( x \right)=\left| \begin{matrix} {{\sin }^{2}}x & \log \cos x & \log \tan x \\ {{n}^{2}} & 2n-1 & 2n+1 \\ 1 & -2\log 2 & 0 \\ \end{matrix} \right|;\\\int\limits_{0}^{\pi /2}{\Delta \left( x \right)dx=\left| \begin{matrix} \int\limits_{0}^{\pi /2}{{{\sin }^{2}}x\,dx} & \int\limits_{0}^{\pi /2}{\log \,\cos x\,dx} & \int\limits_{0}^{\pi /2}{\log \tan x\,dx} \\ {{n}^{2}} & 2n-1 & 2n+1 \\ 1 & -2\log 2 & 0 \\ \end{matrix} \right|}\end{array} \)
\(\begin{array}{l}=\left| \begin{matrix} \frac{\pi }{4} & -\frac{\pi }{2}\log 2 & 0 \\ {{n}^{2}} & 2n-1 & 2n+1 \\ 1 & -2\log 2 & 0 \\ \end{matrix} \right|\end{array} \)
= (Ï€/2) 2n log 2 +Â (Ï€/2) log 2 –Â (Ï€/2) 2n log 2 –Â (Ï€/2) log 2
= 0
Example 2: If
\(\begin{array}{l}f\left( x \right)=\left| \begin{matrix} {{x}^{n}} & \sin x & \cos x \\ n! & \sin \frac{n\pi }{2} & \cos \frac{n\pi }{2} \\ a & {{a}^{2}} & {{a}^{3}} \\ \end{matrix} \right|,\; then \;show \;that \;\frac{{{d}^{n}}}{d{{x}^{n}}}\left\{ f\left( x \right) \right\}=0 \;at \;\;x=0.\end{array} \)
Solution:
By applying differentiation on variable elements of determinants, we will solve the given problem.
We have,
\(\begin{array}{l}f\left( x \right)=\left| \begin{matrix} {{x}^{n}} & \sin x & \cos x \\ n! & \sin \frac{n\pi }{2} & \cos \frac{n\pi }{2} \\ a & {{a}^{2}} & {{a}^{3}} \\ \end{matrix} \right|; \\\frac{{{d}^{n}}}{d{{x}^{n}}}\left\{ f\left( x \right) \right\}=\left| \begin{matrix} \frac{{{d}^{n}}}{d{{x}^{n}}}\left( {{x}^{n}} \right) & \frac{{{d}^{n}}}{d{{x}^{n}}}\left( \sin x \right) & \frac{{{d}^{n}}}{d{{x}^{n}}}\left( \cos x \right) \\ n! & \sin \frac{n\pi }{2} & \cos \frac{n\pi }{2} \\ a & {{a}^{2}} & {{a}^{3}} \\ \end{matrix} \right|\end{array} \)
\(\begin{array}{l}=\left| \begin{matrix} n! & \sin \left( x+\frac{n\pi }{2} \right) & \cos \left( x+\frac{n\pi }{2} \right) \\ n! & \sin \frac{n\pi }{2} & \cos \frac{n\pi }{2} \\ a & {{a}^{2}} & {{a}^{3}} \\ \end{matrix} \right|\\ {{\left( \frac{{{d}^{n}}}{d{{x}^{n}}}\left\{ f\left( x \right) \right\} \right)}_{x=0}}=\left| \begin{matrix} n! & \sin \frac{n\pi }{2} & \cos \frac{n\pi }{2} \\ n! & \sin \frac{n\pi }{2} & \cos \frac{n\pi }{2} \\ a & {{a}^{2}} & {{a}^{3}} \\ \end{matrix} \right|=0\end{array} \)
Problem-Solving Tactics
Let
\(\begin{array}{l}\Delta \left( x \right)=\left| \begin{matrix} {{f}_{1}}\left( x \right) & {{f}_{2}}\left( x \right) & {{f}_{3}}\left( x \right) \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\ \end{matrix} \right|,\;\;\; then \;\;\; \Delta’ \left( x \right)=\left| \begin{matrix} {{f}_{1}}’\left( x \right) & {{f}_{2}}’\left( x \right) & {{f}_{3}}’\left( x \right) \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\ \end{matrix} \right|,\end{array} \)
and in general
\(\begin{array}{l}{{\Delta }^{n}}\left( x \right)=\left| \begin{matrix} f_{1}^{n}\left( x \right) & f_{2}^{n}\left( x \right) & f_{3}^{n}\left( x \right) \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\ \end{matrix} \right|\end{array} \)
where n is any positive integer and fn(x) denotes the nth derivative of f(x).
Let
\(\begin{array}{l}\Delta \left( x \right)=\left| \begin{matrix} f\left( x \right) & g\left( x \right) & h\left( x \right) \\ a & b & c \\ l & m & n \\ \end{matrix} \right|\end{array} \)
a, b, c, l, m, and n are constants here.
\(\begin{array}{l}\Rightarrow \,\,\int\limits_{a}^{b}{\Delta \left( x \right)dx=\left| \begin{matrix} \int\limits_{a}^{b}{f\left( x \right)dx} & \int\limits_{a}^{b}{g\left( x \right)dx} & \int\limits_{a}^{b}{h\left( x \right)dx} \\ a & b & c \\ l & m & n \\ \end{matrix} \right|}\end{array} \)
If the elements of more than one column or rows are functions of x, then the integration can be done only after the evaluation/expansion of the determinant.
Video Lessons
Integration – Important Questions
Important Theorems of Differentiation for JEE
Frequently Asked Questions
Q1
How is the differentiation of a determinant done?
To differentiate a determinant, we have to differentiate one row or column at a time, keeping others unchanged. Then, add the determinants so obtained.
Q2
How to integrate a determinant?
Consider a determinant with first-row elements as functions of x and other row elements as constants. Then, we have to integrate each element of the first row. If the elements of more than one column or row are functions of x, then we should integrate only after the evaluation/expansion of the determinant.
Q3
Give an application of integration.
We use integration to find the area under the curve of a function that is integrated.
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