Difference between revisions of "031 Review Part 1, Problem 7"

Revision as of 14:33, 9 October 2017

True or false: Let $C=AB$ for $4\times 4$ matrices $A$ and $B.$ If $C$ is invertible, then $A$ is invertible.

Solution:
If $A$ is not invertible, then ${\text{det }}A=0.$
Since $C=AB,$ we have
${\begin{array}{rcl}\displaystyle {{\text{det}}C}&=&\displaystyle {{\text{det}}(AB)}\\&&\\&=&\displaystyle {({\text{det}}A)\cdot ({\text{det}}B)}\\&&\\&=&\displaystyle {0\cdot ({\text{det}}B)}\\&&\\&=&\displaystyle {0.}\end{array}}$
Since ${\text{det }}C=0,$ we know $C$ is not invertible, which is a contradiction.
So, $A$ must be invertible and the statement is true.

Final Answer:
TRUE

@@ 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>
+&&\\
+& = & \displaystyle{0.}
+\end{array}</math>
+|-
+|Since &nbsp;<math style="vertical-align: -5px">\text{det } C =0,</math>&nbsp; we know &nbsp;<math style="vertical-align: 0px">C</math>&nbsp; is not invertible, which is a contradiction.
+|-
+|So, &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; must be invertible and the statement is true.
 |}
@@ 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>''']]