031 Review Part 1, Problem 9 - Revision history

Kayla Murray at 19:22, 15 October 2017

2017-10-15T19:22:16Z

Kayla Murray at 21:12, 9 October 2017

2017-10-09T21:12:09Z

Kayla Murray at 20:17, 9 October 2017

2017-10-09T20:17:58Z

Kayla Murray: Created page with "True or false: If all the entries of a $7\times 7$ matrix $A$
2017-10-09T20:13:36Z

Created page with "<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..."

New page
<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>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution:  
|-
|First, we switch to the limit to <math style="vertical-align: 0px">x</math> so that we can use L'Hopital's rule.
|-
|So, we have
|-
|
       <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}}\\
&&\\
& \overset{L'H}{=} & \displaystyle{\frac{-4}{10}}\\
&&\\
& = & \displaystyle{-\frac{2}{5}}.
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|       False
|}
[[031_Review_Part_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 28: / Line 28: @@
 |Hence, the statement is true.
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 34: / Line 35: @@
 |&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>''']]

← Older revision		Revision as of 20:17, 9 October 2017
Line 1:		Line 1:
−	<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 an invertible  <math style="vertical-align: 0px">3\times 3</math>  matrix, and  <math style="vertical-align: 0px">B</math>  and  <math style="vertical-align: 0px">C</math>  are  <math style="vertical-align: 0px">3\times 3</math>  matrices such that  <math style="vertical-align: -4px">AB=AC,</math>  then  <math style="vertical-align: 0px">B=C.</math>

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

@@ Line 1: / Line 1: @@
-<span class="exam">True or false: If &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is an invertible &nbsp;<math style="vertical-align: 0px">3\times 3</math>&nbsp; matrix, and &nbsp;<math style="vertical-align: 0px">B</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">C</math>&nbsp; are &nbsp;<math style="vertical-align: 0px">3\times 3</math>&nbsp; matrices such that &nbsp;<math style="vertical-align: -4px">AB=AC,</math>&nbsp; then &nbsp;<math style="vertical-align: 0px">B=C.</math>
+<span class="exam">True or false: If &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is an invertible &nbsp;<math style="vertical-align: 0px">3\times 3</math>&nbsp; matrix, and &nbsp;<math style="vertical-align: 0px">B</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">C</math>&nbsp; are &nbsp;<math style="vertical-align: 0px">3\times 3</math>&nbsp; matrices such that &nbsp;<math style="vertical-align: -4px">AB=AC,</math>&nbsp;
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !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.
+|Since &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is invertible, &nbsp;<math style="vertical-align: 0px">A^{-1}</math>&nbsp; exists.
 |-
-|So, we have
+|Then, by associativity of matrix multiplication, we have
 |-
+|
 &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}}\\
+\displaystyle{(A^{-1}A)B} & = & \displaystyle{(A^{-1}A)C}\\
 &&\\
-& \overset{L'H}{=} & \displaystyle{\frac{-4}{10}}\\
+\displaystyle{I_3 B} & = & \displaystyle{I_3 C}\\
 &&\\
-& = & \displaystyle{-\frac{2}{5}}.
+\displaystyle{B} & = & \displaystyle{C}
 \end{array}</math>
 |}
@@ Line 21: / Line 32: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; &nbsp; &nbsp; False
+|&nbsp;&nbsp; &nbsp; &nbsp; TRUE
 |}
 [[031_Review_Part_1|'''<u>Return to Sample Exam</u>''']]