Difference between revisions of "031 Review Part 1, Problem 5"

Revision as of 13:49, 9 October 2017

True or false: If $A$ and $B$ are invertible $n\times n$ matrices, then so is $A+B.$

Solution:
Let $A={\begin{bmatrix}1&1\\0&1\end{bmatrix}}$ and $B={\begin{bmatrix}-1&-1\\0&-1\end{bmatrix}}.$
Notice that
${\text{det}}A={\text{det}}B=1.$
Since ${\text{det}}A,{\text{det}}B\neq 0,$ we know $A$ and $B$ are invertible.
But,
$A+B={\begin{bmatrix}0&0\\0&0\end{bmatrix}},$
which is not invertible.
Hence, this statement is false.

Final Answer:
FALSE

@@ 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=
+    \begin{bmatrix}
+& 1  \\
+& 1
+         \end{bmatrix}</math>&nbsp; and &nbsp;<math style="vertical-align: -20px">B=
+    \begin{bmatrix}
+           -1 & -1 \\
+& -1
+         \end{bmatrix}.</math>&nbsp;
 |-
-|So, we have
+|Notice that
 |-
 |
-&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
+::<math>\text{det} A=\text{det} B=1.</math>
-\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: -6px">\text{det} A,\text{det} B\neq 0,</math>&nbsp; we know &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">B</math>&nbsp; are invertible.
-& \overset{L'H}{=} & \displaystyle{\frac{-4}{10}}\\
+|-
-&&\\
+|But,
-& = & \displaystyle{-\frac{2}{5}}.
+|-
-\end{array}</math>
+|
+::<math style="vertical-align: -31px">A+B=
+    \begin{bmatrix}
+& 0 \\
+& 0
+         \end{bmatrix},</math>&nbsp;
+|-
+|which is not invertible.
+|-
+|Hence, this statement is false.
 |}
@@ Line 21: / Line 38: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; &nbsp; &nbsp; False
+|&nbsp;&nbsp; &nbsp; &nbsp; FALSE
 |}
 [[031_Review_Part_1|'''<u>Return to Sample Exam</u>''']]