- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
3.6. $\quad$ Exercises
8. $\quad$ (a) $\quad$ If $F(x) = \begin{vmatrix} f_1(x) & f_2(x) \\ f_1'(x) & f_2'(x) \end{vmatrix},$ prove that $F'(x) = \begin{vmatrix} f_1(x) & f_2(x) \\ f_1''(x) & f_2''(x) \end{vmatrix}.$
(b) $\quad$ State and prove a corresponding result for $3 \times 3$ determinants, assuming the validity of Equation (3.2).
Solution.
(a) $\quad$ If $F(x) = \begin{vmatrix} f_1(x) & f_2(x) \\ f_1'(x) & f_2'(x) \end{vmatrix},$ we have \begin{align*} F(x) &= f_1(x)f'_2(x) - f_2(x)f'_1(x) \\ \\ F'(x) &= [f'_1(x)f'_2(x) + f_1(x)f''_2(x)] - [f'_2(x)f'_1(x) + f_2(x)f''_1(x)] \\ &= f_1(x)f''_2(x) - f_2(x)f''_1(x) \\ &= \begin{vmatrix} f_1(x) & f_2(x) \\ f_1''(x) & f_2''(x) \end{vmatrix}. \quad \blacksquare \end{align*}
(b) $\quad$ We recall that determinants of order three may be computed by Equation (3.2). \begin{align*} (3.2) \qquad \det\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} &= a_{11}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} - a_{12}\begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} + a_{13}\begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix} \end{align*} Thus, if $F(x) = \begin{vmatrix} f_1(x) & f_2(x) & f_3(x) \\ f_1'(x) & f_2'(x) & f_3'(x) \\ f_1''(x) & f_2''(x) & f_3''(x) \\ \end{vmatrix},$ we have \begin{align*} F(x) &= f_1(x)\begin{vmatrix} f_2'(x) & f_3'(x) \\ f_2''(x) & f_3''(x) \end{vmatrix} - f_2(x)\begin{vmatrix} f_1'(x) & f_3'(x) \\ f_1''(x) & f_3''(x) \end{vmatrix} + f_3(x)\begin{vmatrix} f_1'(x) & f_1'(x) \\ f_1''(x) & f_2''(x) \end{vmatrix}. \end{align*} Then, by the product rule for derivatives, $F'(x)$ is given by \begin{align*} F'(x) &= f_1'(x)\begin{vmatrix} f_2'(x) & f_3'(x) \\ f_2''(x) & f_3''(x) \end{vmatrix} + f_1(x)\begin{vmatrix} f_2'(x) & f_3'(x) \\ f_2''(x) & f_3''(x) \end{vmatrix}' \\ &- f_2'(x)\begin{vmatrix} f_1'(x) & f_3'(x) \\ f_1''(x) & f_3''(x) \end{vmatrix} - f_2(x)\begin{vmatrix} f_1'(x) & f_3'(x) \\ f_1''(x) & f_3''(x) \end{vmatrix}' \\ &+ f_3'(x)\begin{vmatrix} f_1'(x) & f_1'(x) \\ f_1''(x) & f_2''(x) \end{vmatrix} + f_3(x)\begin{vmatrix} f_1'(x) & f_1'(x) \\ f_1''(x) & f_2''(x) \end{vmatrix}'. \end{align*} Rearranging terms and applying the result of (a), we have \begin{align*} F'(x) &= f_1'(x)\begin{vmatrix} f_2'(x) & f_3'(x) \\ f_2''(x) & f_3''(x) \end{vmatrix} - f_2'(x)\begin{vmatrix} f_1'(x) & f_3'(x) \\ f_1''(x) & f_3''(x) \end{vmatrix} + f_3'(x)\begin{vmatrix} f_1'(x) & f_1'(x) \\ f_1''(x) & f_2''(x) \end{vmatrix} \\ &+ f_1(x)\begin{vmatrix} f_2'(x) & f_3'(x) \\ f_2''(x) & f_3''(x) \end{vmatrix}' - f_2(x)\begin{vmatrix} f_1'(x) & f_3'(x) \\ f_1''(x) & f_3''(x) \end{vmatrix}' + f_3(x)\begin{vmatrix} f_1'(x) & f_1'(x) \\ f_1''(x) & f_2''(x) \end{vmatrix}' \\ \\ &= f_1'(x)\begin{vmatrix} f_2'(x) & f_3'(x) \\ f_2''(x) & f_3''(x) \end{vmatrix} - f_2'(x)\begin{vmatrix} f_1'(x) & f_3'(x) \\ f_1''(x) & f_3''(x) \end{vmatrix} + f_3'(x)\begin{vmatrix} f_1'(x) & f_1'(x) \\ f_1''(x) & f_2''(x) \end{vmatrix} \\ &+ f_1(x)\begin{vmatrix} f_2'(x) & f_3'(x) \\ f_2'''(x) & f_3'''(x) \end{vmatrix} - f_2(x)\begin{vmatrix} f_1'(x) & f_3(x) \\ f_1'''(x) & f_3'''(x) \end{vmatrix} + f_3(x)\begin{vmatrix} f_1'(x) & f_1'(x) \\ f_1'''(x) & f_2'''(x) \end{vmatrix}. \end{align*} Then, using Equation (3.2) and recalling Axiom 3 of the determinant function, we have \begin{align*} F'(x) &= \begin{vmatrix} f_1'(x) & f_2('x) & f_3'(x) \\ f_1'(x) & f_2'(x) & f_3'(x) \\ f_1''(x) & f_2''(x) & f_3''(x) \\ \end{vmatrix} + \begin{vmatrix} f_1(x) & f_2(x) & f_3(x) \\ f_1'(x) & f_2'(x) & f_3'(x) \\ f_1'''(x) & f_2'''(x) & f_3'''(x) \\ \end{vmatrix} \\ \\ &= \begin{vmatrix} f_1(x) & f_2(x) & f_3(x) \\ f_1'(x) & f_2'(x) & f_3'(x) \\ f_1'''(x) & f_2'''(x) & f_3'''(x) \\ \end{vmatrix}. \quad \blacksquare \end{align*}