## anonymous one year ago Suppose A is an invertible matrix. Which of the following statements are true? Explanations would be greatly appreciated! (a) A can be expressed as a product of elementary matrices. (b) A is row equivalent to the nxn identity matrix. (c) The equation Ax=0 has only the trivial solution. (d) The determinant of A is positive. (e) The transpose of A is also an invertible matrix. (f) A has a unique row echelon form. (g) The equation Ax=b may have two distinct non zero solutions for a non zero vector b.

1. anonymous

a, b, c and e are true. d and g are not true f i'm still debating

2. anonymous

i think f is false. if it were row-reduced echelon form then it would be true. Suppose$A \rightarrow \left[\begin{matrix}1 & 2 \\ 0 & 1\end{matrix}\right] \rightarrow \left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]$ both are in row echelon form (the last is in row-reduced echelon form) thus row echelon form is not unique

3. Empty

Are you familiar with this property of determinants: $det(AB)=det(A)*det(B)$ So if a matrix is invertible, then: $$AA^{-1} = I$$ $det(AA^{-1}) = det(I)$ $det(A)*det(A^{-1}) = 1$ So we can rearrange this, $det(A^{-1}) = \frac{1}{det(A)}$ So if the determinant of A is 0, then it doesn't have an inverse, because you can't divide by zero to find the determinant of of its inverse. Also you get this fancy little relationship between determinants.