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Chapter 8: Problem 66

Evaluate the determinant, in which the entries are functions. Determinants of this type occur when changes of variables are made in calculus. $$\left|\begin{array}{cc} e^{-x} & x e^{-x} \\ -e^{-x} & (1-x) e^{-x} \end{array}\right|$$

Short Answer

Expert verified
The determinant is \(2e^{-2x}-x*e^{-2x}\).

Step by step solution

01

Identify matrix entries

The given matrix is \[\left[\begin{array}{cc} e^{-x} & x e^{-x} \ -e^{-x} & (1-x) e^{-x} \end{array}\right]\]. The entries of the matrix are \(a=e^{-x}\), \(b=xe^{-x}\), \(c=-e^{-x}\), and \(d=(1-x)e^{-x}\).
02

Apply determinant formula

Now apply the determinant formula which is \(ad-bc\). Substitute \(a\), \(b\), \(c\), and \(d\) into the formula. We get determinant = \(a*d - b*c = e^{-x}*(1-x)e^{-x} -xe^{-x}*(-e^{-x})\).
03

Simplifying the expression

Simplify the above expression: determinant = \(e^{-2x} - x*e^{-2x}+ e^{-2x}\). Combine similar terms to get the final answer.

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