031 Review Part 1, Problem 7 - Revision history

Kayla Murray at 19:20, 15 October 2017

2017-10-15T19:20:50Z

2017-10-09T21:33:25Z

2017-10-09T20:17:06Z

2017-10-09T20:16:53Z

@@ Line 23: / Line 23: @@
 |So, &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; must be invertible and the statement is true.
 |}
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@@ Line 29: / Line 30: @@
 |&nbsp;&nbsp; &nbsp; &nbsp; TRUE
 |}
-[[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.
+|If &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is not invertible, then &nbsp;<math style="vertical-align: 0px">\text{det } A=0.</math>
 |-
-|So, we have
+|Since &nbsp;<math style="vertical-align: -5px">C=AB,</math>&nbsp; we have
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
+::<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}}\\
+\displaystyle{\text{det}C} & = & \displaystyle{\text{det} (AB)}\\
 &&\\
-& \overset{L'H}{=} & \displaystyle{\frac{-4}{10}}\\
+& = & \displaystyle{(\text{det} A) \cdot (\text{det} B)}\\
 &&\\
-& = & \displaystyle{-\frac{2}{5}}.
+& = & \displaystyle{0\cdot (\text{det} B)}\\
-\end{array}</math>
+&&\\
 |}
@@ Line 21: / Line 27: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; &nbsp; &nbsp; False
+|&nbsp;&nbsp; &nbsp; &nbsp; TRUE
 |}
 [[031_Review_Part_1|'''<u>Return to Sample Exam</u>''']]

← Older revision		Revision as of 20:17, 9 October 2017
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−	span class="exam">True or false: Let  <math style="vertical-align: 0px">C=AB</math>  for  <math style="vertical-align: 0px">4\times 4</math>  matrices  <math style="vertical-align: 0px">A</math>  and  <math style="vertical-align: 0px">B.</math>  If  <math style="vertical-align: 0px">C</math>  is invertible, then  <math style="vertical-align: 0px">A</math>  is invertible.	+	<span class="exam">True or false: Let  <math style="vertical-align: 0px">C=AB</math>  for  <math style="vertical-align: 0px">4\times 4</math>  matrices  <math style="vertical-align: 0px">A</math>  and  <math style="vertical-align: 0px">B.</math>  If  <math style="vertical-align: 0px">C</math>  is invertible, then  <math style="vertical-align: 0px">A</math>  is invertible.
−

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← Older revision		Revision as of 20:16, 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: Let  <math style="vertical-align: 0px">C=AB</math>  for  <math style="vertical-align: 0px">4\times 4</math>  matrices  <math style="vertical-align: 0px">A</math>  and  <math style="vertical-align: 0px">B.</math>  If  <math style="vertical-align: 0px">C</math>  is invertible, then  <math style="vertical-align: 0px">A</math>  is invertible.
		+

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