031 Review Part 1, Problem 4 - Revision history

Kayla Murray at 19:16, 15 October 2017

2017-10-15T19:16:48Z

2017-10-09T21:43:26Z

2017-10-09T20:15:25Z

@@ Line 14: / Line 14: @@
 |Therefore, &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is invertible.
 |-
-|Since &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is a diagonal matrix, the eigenvalues of &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; are the entries on the diagonal.
+|Since &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is a triangular matrix, the eigenvalues of &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; are the entries on the diagonal.
 |-
 |Therefore, the only eigenvalue of &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">1.</math>&nbsp; Additionally, there is only one linearly independent eigenvector.
@@ Line 20: / Line 20: @@
 |Hence, &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is not diagonalizable and the statement is false.
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 26: / Line 27: @@
 |&nbsp;&nbsp; &nbsp; &nbsp; FALSE
 |}
-[[031_Review_Part_1|'''<u>Return to Sample Exam</u>''']]
+[[031_Review_Part_1|'''<u>Return to Review Problems</u>''']]

@@ Line 4: / Line 4: @@
 !Solution: &nbsp;
 |-
-|First, we switch to the limit to <math style="vertical-align: 0px">x</math> so that we can use L'Hopital's rule.
+|Let &nbsp;<math style="vertical-align: -20px">A=
 |-
-|So, we have
+|First, notice that &nbsp;<math style="vertical-align: -5px">\text{det }A=1\neq 0.</math>
 |-
+|Therefore, &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is invertible.
-&nbsp; &nbsp; &nbsp; &nbsp;<math>\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}}\\
+|Since &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is a diagonal matrix, the eigenvalues of &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; are the entries on the diagonal.
-&&\\
+|-
-& \overset{L'H}{=} & \displaystyle{\frac{-4}{10}}\\
+|Therefore, the only eigenvalue of &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">1.</math>&nbsp; Additionally, there is only one linearly independent eigenvector.
-&&\\
+|-
-& = & \displaystyle{-\frac{2}{5}}.
+|Hence, &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is not diagonalizable and the statement is false.
 |}
@@ Line 21: / Line 24: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; &nbsp; &nbsp; False
+|&nbsp;&nbsp; &nbsp; &nbsp; FALSE
 |}
 [[031_Review_Part_1|'''<u>Return to Sample Exam</u>''']]

← Older revision		Revision as of 20:15, 9 October 2017
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−	<span class="exam">True or false: If ~~all the entries of a  <math style="vertical-align: 0px">7\times 7</math>  matrix~~  <math style="vertical-align: 0px">A</math>  ~~are  <math style="vertical-align: -4px">7~~,~~</math> ~~ then  <math style="vertical-align: 0px">~~\text{det }~~A</math>  ~~must be  <math style="vertical-align: 0px">7^7~~.~~</math>~~	+	<span class="exam"> True or false: If  <math style="vertical-align: 0px">A</math>  is invertible, then  <math style="vertical-align: 0px">A</math>  is diagonalizable.

	{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"		{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"