## Matrix Properties : A2 + 6A +9I3 = 0, show that A is invertible

Hello, I am having difficulties with this question. A^2+6A+9I3 = 0. A is a 3x3 matrix. I must show that A is invertible. I am tempted to factor but this problem deals with matrices. I know this is wrong but I come to A2+3A+3A+9=0 (does 9I3 = 9, since 9 is a constant?), then A(A+3) + 3(A+3) = 0 but I have no idea what to do after.
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 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus Doesn't $$\left( \begin{array}{ccc} -3 & 0 & 0\\ 0 & -3 & 0 \\ 0 & 0 & 0 \end{array} \right)$$ Constitute a counter-example? Or am I going senile?
 Yeah I knew what I did was wrong. In any case, I have no idea how to show that A is invertible.

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## Matrix Properties : A2 + 6A +9I3 = 0, show that A is invertible

Ok, ignore my above post. I WAS being senile...

What about this approach? the matrix A is similar to a triangular matrix B. This matrix B also satisfied B2+6B+9I3=0. But since B is triangular, it is easy to see that it must have -3's on it's diagonal. Thus the determinant is -9, and hence the matrix is invertible.

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 Quote by Boom101 Hello, I am having difficulties with this question. A is a 3x3 matrix, the first A is squared, and I3 is the 3x3 identity matrix. I must show that A is invertible. I am tempted to factor but this problem deals with matrices. I know this is wrong but I come to A2+3A+3A+9=0
At the very least, write A^2 to indicate the square of a matrix. Better yet would be A2. For the latter form, click the Go Advanced button below the text input area, and then click the X2 button in the menu just above the text input area, and put an exponent of 2 in between the [ sup] and [ /sup] tags.
 Quote by Boom101 (does 9I3 = 9, since 9 is a constant?)
No, 9I3 does not equal 9. 9I3 is a diagonal matrix with 9's down the main diagonal.
 Quote by Boom101 , then A(A+3) + 3(A+3) = 0 but I have no idea what to do after.
A2 + 6A + 9I = 0 can be factored to (A + 3I)2 = 0. One thing that we can say from this is that |A + 3I| = 0.
 Recognitions: Homework Help Science Advisor If A is not invertible, then there is a nonzero vector x such that Ax=0. Is that compatible with A^2+6A+9I=0?
 Recognitions: Gold Member Science Advisor Staff Emeritus If $A^2+ 6A+ 9I= 0$ then $A^2+ 6A= A(A+ 6I)= 9I$ so that $A[(A+ 6I)/9]= I$. Do see how that tells you that A is invertible?
 Thanks Ivy
 I have the same question, I don't see how the matrix is invertible, can anyone please explain?
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus It's inverse is (A+6I)/9
 Thank you!