Help with Linear Algebra T/F Questions

[stryker105]

Below is some statements for inverse, permutation and transposes. Next to them I will write what I believe to be correct. I know that at least one of my responses is incorrect, can anybody help me? I greatly appreciate it.

Assume that all matrices are n\times n, that P and Q are permutation matrices and that R is a permutation matrix that interchanges two (otherwise unspecified) rows

If A is symmetric and A=LU then L = U^T. F

PQ=QP F

R^-1 = R T

R^15 = R F

The inverse of an invertible symmetric matrix is symmetric. T

A (square) matrix being invertible means the same as it being non-singular. T

A nxn matrix is invertible if and only if elimination, possibly including row interchanges, produces n non-zero pivots. T

The inverse of an invertible matrix is invertible. F

(AB)^-1= (A^-1)(B^-1). F

(AB)^-1= (B^-1)(A^-1). T

if A and B are invertible, then so is AB. T

 

[christoff]

"The inverse of an invertible matrix is symmetric" T
-This is not true. For example, consider the matrix [1,3\\1,2]]. Its inverse is [-2,3\\1,-1]], which is not symmetric.

"The inverse of an invertible matrix is invertible" F
-This actually is true. Let A be invertible, so A^-1=B is its inverse. Then B is certainly invertible, because BA=A^-1*A=I, and AB=A*A^-1=I. That is, A is the inverse of B.

I'm assuming of course that you're working over a field.

 

[HallsofIvy]

christoff, the statement was "The inverse of an invertible symmetric matrix is symmetric" and that is true.

If R is a "permutation matrix that interchanges two (otherwise unspecified) rows" then it is true that $R^{-1}= R$. And because that is true, it follows that $R^2= R*R= R*R^{-1}= I$. From that it follows that R to any even power is I and R to any odd power is R.

 

[christoff]

Ah, my apologies, and thank you for the clarification, HallsofIvy. I guess my eyes skipped over that word.