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Let f ( x ) = e x − cos x , f(x)=ex−cos⁡x, g ( x ) = e x + cos x , g(x)=ex+cos⁡x, and h ( x ) = cos x . h(x)=cos⁡x. Are f ( x ) , f(x), g ( x ) , g(x), and h ( x ) h(x) linearly independent? If so, show it, if not, find a linear combination that works.

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Answer: No. f(x), g(x), & h(x) are linearly Dependent.

Explanation:

f(x) = eˣ - cos(x) g(x) = eˣ + cos(x) h(x) = cos(x)

Create a 3 x 3 matrix where row 1 is x = 0, row 2 is x = π/2, row 3 = π

then find the determinant.

If det = 0 ⇒ dependent.

If det ≠ 0 ⇒ independent.


det \left[\begin{array}{ccc}f(0)&g(0)&h(0)\\f((\pi)/(2))&g((\pi)/(2))&h((\pi)/(2))\\f(\pi)&g(\pi)&h(\pi)\end{array}\right]


det \left[\begin{array}{ccc}e^0-cos(0)&e^0+cos(0)&cos(0)\\e^{(\pi)/(2)}-cos((\pi)/(2))&e^{(\pi)/(2)}+cos((\pi)/(2))}&cos((\pi)/(2))\\e^\pi-cos(\pi)&e^\pi+cos(0\pi)&cos(0\pi)\end{array}\right] =0

Since determinant = 0, they are linearly dependent.

There are an infinite number of possibilities to make them independent. Any change that makes the determinant ≠ 0

One example would be to make h(x) = sin(x)

User Travis Tidwell
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