SalmonPanna2022
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Find the wronskian of the set {1 - x, 1 + x, 1 - 3x}.​

Sagot :

SalmonWorldTopper

Step-by-step explanation:

[tex] \red{\large\underline{\sf{Solution-}}}[/tex]

We know that,

[tex]\rm :\longmapsto\:Wronkian \: of \: the \: set \: {\left \{f_{1}(x),f_{2}(x),f_{3}(x) \right \}} \: is \: [/tex]

[tex]W{\left(f_{1},f_{2},f_{3} \right)}\left(x\right) = \left|\begin{array}{ccc}f_{1}\left(x\right) & f_{2}\left(x\right) & f_{3}\left(x\right)\\f_{1}^{\prime}\left(x\right) & f_{2}^{\prime}\left(x\right) & f_{3}^{\prime}\left(x\right)\\f_{1}^{\prime\prime}\left(x\right) & f_{2}^{\prime\prime}\left(x\right) & f_{3}^{\prime\prime}\left(x\right)\end{array}\right|[/tex]

Here,

[tex]\rm :\longmapsto\:f_1(x) = 1 - x[/tex]

[tex]\rm :\longmapsto\:f_2(x) = 1 + x[/tex]

[tex]\rm :\longmapsto\:f_3(x) = 1 - 3x[/tex]

So, there respective differential coefficients are as follow

[tex]\rm :\longmapsto\:f_1'(x) = 0 - 1 = - 1[/tex]

[tex]\rm :\longmapsto\:f_2'(x) = 0 + 1 = 1[/tex]

[tex]\rm :\longmapsto\:f_3'(x) = 0 - 3 = - 3[/tex]

Also,

[tex]\rm :\longmapsto\:f_1''(x) = 0[/tex]

[tex]\rm :\longmapsto\:f_2''(x) = 0[/tex]

[tex]\rm :\longmapsto\:f_3''(x) = 0[/tex]

So, on substituting the values in

[tex]W{\left(f_{1},f_{2},f_{3} \right)}\left(x\right) = \left|\begin{array}{ccc}f_{1}\left(x\right) & f_{2}\left(x\right) & f_{3}\left(x\right)\\f_{1}^{\prime}\left(x\right) & f_{2}^{\prime}\left(x\right) & f_{3}^{\prime}\left(x\right)\\f_{1}^{\prime\prime}\left(x\right) & f_{2}^{\prime\prime}\left(x\right) & f_{3}^{\prime\prime}\left(x\right)\end{array}\right|.[/tex]

We get

[tex]\rm :\longmapsto\:W= \: \ \: \begin{gathered}\sf \left | \begin{array}{ccc}1 - x&1 + x&1 - 3x\\ - 1&1& - 3\\0&0&0\end{array}\right | \end{gathered} \\ [/tex]

[tex]\rm \:  =  \: 0[/tex]

[ If elements of any row all are 0, its determinant value is zero]

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