1. ## singular matrices 3

if A is singular the A^3+A^2+A is singular

2. Note that $A^{3}+A^{2}+A=A(A^{2}+A+I).$ Does that give you any ideas?

3. Yes, note that

$A(A^2+A+I)$ is a factorization and use the fact that

$\det(XY)=\det(X)\cdot \det(Y)$

too slow

4. how does determinat could tell if its singular?

5. $\det(A)=0$ if and only if $A$ is singular.

6. Originally Posted by transgalactic
how does determinat could tell if its singular?
What is the definition of "singular"?

7. singular means that there could not be inverse

8. You should have learned that det(AB)= det(A)det(B) (as TheEmptySet mentioned). If AB= I then det(A)det(B)= 1 so neither det(A) nor det(B) can be 0.

However, Ackbeet's suggestion gives a more fundamental proof than that. If $A^3+ A^2+ A$ had an inverse, then there would exist some matrix, B, such that $(A^3+ A^2+ A)B= I$ . But that is the same as $A[(A^2+ A+ I)B]= I$ which says that $(A^2+ A+ I)B$ is inverse to A contradicting the fact that A does not have an inverse.