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If $W(y_1,y_2,y_3)=\left| \begin{array}{ccc} y_1 & y_2 & y_3 \\ y_1' & y_2' & y_3' \\ y_1'' & y_2'' & y_3'' \end{array}\right|$, how can I show that $W'(y_1,y_2,y_3)=\left| \begin{array}{ccc} y_1 & y_2 & y_3 \\ y_1' & y_2' & y_3' \\ y_1''' & y_2''' & y_3''' \end{array}\right|$? I am not sure how to go about computing the derivative of a determinant. Any input would be greatly appreciated.

nmasanta
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RXY15
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2 Answers2

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$$W^\prime(y_1,y_2,y_3)=\frac{d}{dx}\left| \begin{array}{ccc} y_1 & y_2 & y_3 \\ y_1\prime & y_2\prime & y_3\prime \\ y_1\prime\prime & y_2\prime\prime & y_3\prime\prime \end{array}\right|\\ =\left| \begin{array}{ccc} y_1\prime & y_2\prime & y_3\prime \\ y_1\prime & y_2\prime & y_3\prime \\ y_1\prime\prime & y_2\prime\prime & y_3\prime\prime \end{array}\right|+\left| \begin{array}{ccc} y_1 & y_2 & y_3 \\ y_1\prime\prime & y_2\prime\prime & y_3\prime\prime \\ y_1\prime\prime & y_2\prime\prime & y_3\prime\prime \end{array}\right|+\left| \begin{array}{ccc} y_1 & y_2 & y_3 \\ y_1\prime & y_2\prime & y_3\prime \\ y_1\prime\prime\prime & y_2\prime\prime\prime & y_3\prime\prime\prime \end{array}\right|\\ =\left| \begin{array}{ccc} y_1 & y_2 & y_3 \\ y_1\prime & y_2\prime & y_3\prime \\ y_1\prime\prime\prime & y_2\prime\prime\prime & y_3\prime\prime\prime \end{array}\right|. $$

Chen Wang
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    How does the addition lead to the final answer? I am not sure how $y_1'+y_1+y_1 = y_1$. – RXY15 Mar 14 '14 at 05:15
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    The first two determinants are equal to 0. – Chen Wang Mar 14 '14 at 08:17
  • Spoiler: it is because there is an all zero row! ;) Anyway the source is Elementary Differential Equations and Boundary Value Problems by William E Boyce, Chapter 4.1 Q20. – NetUser5y62 Mar 11 '19 at 02:20
  • Thank you for this answer! I was trying to solve an assignment problem where we have to solve for the relationship of Wronskian and its first derivative, where all n functions in the first row of the Wronskian satisfies a given ODE. I forgot I could differentiate along rows so I wasted so much time trying to do induction, applying cofactor matrix, adjugate matrix, and other useless stuff. – Divide1918 Jun 04 '20 at 14:42
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I'm writing $y_1=x,y_2=y,y_3=z$ Writing out $W(x,y,z)$ gives$$ x''(yz'-zy')-y''(xz'-zx')+z''(xy'-yx') $$ differentiating gives $$ x'''(yz'-zy')-y'''(xz'-zx')+z'''(xy'-yx')+{\color{red}{(x''(yz''-zy'')-y''(xz''-zx'')+z''(xy''-yx''))}} $$The red part cancels out and the first part equal $W'(x,y,z)$ as given in the question.

Kaladin
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