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.

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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.

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

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