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If $A$ and $B$ are square matrices of order $2$, then $\det(A+B)=0$ is possible only when:
$(a)$ $\det(A)=0$ or $\det(B)=0$
$(b)$ $\det(A)+\det(B)=0$
$(c)$ $\det(A)=0$ and $\det(B)=0$
$(d)$ $A+B=0$

I was sure that when $A+B=0$, $\det(A+B)=\det(0)=0$ So the answer is $d$.
But I am not able to show that the other three does not meet the condition.

Please offer your assistance?
Thank you :)

chndn
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    This question is broken! Did you translate it into English yourself, perhaps? It's not true that $\det(A+B) = 0$ is possible only when $A+B = 0$. For instance, if $A = 0$ and $B$ is any non-zero matrix with zero determinant, then $\det(A+B) = 0$ but $A+B \ne 0$. (And please don't ask, "Can you prove it?") – TonyK Jun 09 '13 at 07:46
  • I didnt translate it. It is posted as it is. I agree that "It's not true that $det(A+B)=0$ is possible only when $A+B=0$." But what it means in this context is that $det(A+B)$ is true when only one among the four options(conditions) is true. :) – chndn Jun 11 '13 at 14:26
  • and Ill change that prove that. – chndn Jun 11 '13 at 14:36
  • So instead of "is possible only when", you mean "is always true when". Right? They're not the same! – TonyK Jun 11 '13 at 21:48

3 Answers3

5

All the answers aren't true: counterexample $$A = \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix}, B = \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix} $$

3

Counterexamples,

case (a),(b),(c) take $A=I,B=-I$

Then $(A+B)=0\Rightarrow \det(A+B)=0$ but $\det(A)=\det(B)=1$

Abhra Abir Kundu
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1

Hint: consider the counter-example for (a), (b), and (c):

$$A = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}, B = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} $$

Alex Wertheim
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