Difference between revisions of "031 Review Part 1, Problem 2"
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| − | <span class="exam">True or false: If | + | <span class="exam"> True or false: If a matrix <math style="vertical-align: 0px">A^2</math> is diagonalizable, then the matrix <math style="vertical-align: 0px">A</math> must be diagonalizable as well. |
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Revision as of 12:14, 9 October 2017
True or false: If a matrix is diagonalizable, then the matrix Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A} must be diagonalizable as well.
| Solution: |
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| First, we switch to the limit to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} so that we can use L'Hopital's rule. |
| So, we have |
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Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\lim_{x \rightarrow \infty}\frac{3-2x^2}{5x^2 + x +1}} & \overset{L'H}{=} & \displaystyle{\lim_{x \rightarrow \infty}\frac{-4x}{10x+1}}\\ &&\\ & \overset{L'H}{=} & \displaystyle{\frac{-4}{10}}\\ &&\\ & = & \displaystyle{-\frac{2}{5}}. \end{array}} |
| Final Answer: |
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| False |