Difference between revisions of "031 Review Part 1, Problem 9"

Revision as of 13:17, 9 October 2017

True or false: If $A$ is an invertible $3\times 3$ matrix, and $B$ and $C$ are $3\times 3$ matrices such that $AB=AC,$ then $B=C.$

Solution:
First, we switch to the limit to $x$ so that we can use L'Hopital's rule.
So, we have
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow \infty }{\frac {3-2x^{2}}{5x^{2}+x+1}}}&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow \infty }{\frac {-4x}{10x+1}}}\\&&\\&{\overset {L'H}{=}}&\displaystyle {\frac {-4}{10}}\\&&\\&=&\displaystyle {-{\frac {2}{5}}}.\end{array}}$

Final Answer:
False

Revision as of 13:13, 9 October 2017 (view source) Kayla Murray (talk \| contribs) (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...")		Revision as of 13:17, 9 October 2017 (view source) Kayla Murray (talk \| contribs) Newer edit →
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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 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>

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