Understanding Determinant Equations for Linear Algebra | Homework Help

tylerc1991
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On my professor's study guide, he mentions that for the upcoming test we need to be able to solve determinant equations. What does this mean? I have tried to google what a determinant equation is and can't find anything, so am I safe in a assuming that he means that we need to be able to solve a linear system USING determinants (i.e. Cramer's rule)? Or could someone enlighten me as to what a determinant equation is? Thank you!
 
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A "determinant equation" is, of course, an equation that involves a determinant. I presume that the determinant will depend upon some unknown. For example, to find an eigenvalue for, say, the matrix
\begin{bmatrix}1 & 2 & 1 \\ 0 & 1 & 2\\ 3 & 2 & 1\end{bmatrix}
you would have to solve the equation
\left|\begin{array}{ccc}1- \lambda & 2 & 1 \\ 0 & 1- \lambda & 2 \\ 3 & 2 & 1- \lambda\end{array}\right|= 0
a "determinant equation". Evaluating the determinant gives you a cubic equation to solve for \lambda.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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