Question

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.

Answer 1

a, b, c and e are true.

d and g are not true

f i'm still debating

Answer 2

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

Answer 3

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.