(Solved): Use elementary row or column operations to find the determinant. \[ \left|\begin{array}{rrr} 1 & -1 ...

Use elementary row or column operations to find the determinant. \[ \left|\begin{array}{rrr} 1 & -1 & -2 \\ 2 & 1 & 3 \\ -2 & 2 & 4 \end{array}\right| \] [-I1 Points] Use elementary row or column operations to find the determinant. \[ \left|\begin{array}{rrrr} 3 & -6 & 8 & 1 \\ 9 & 4 & 6 & 0 \\ 5 & 9 & -5 & 5 \\ 0 & 6 & 3 & -1 \end{array}\right| \] Find \( \left|A^{T}\right|,\left|A^{2}\right|,\left|A A^{T}\right|,|2 A| \), and \( \left|A^{-1}\right| \) \[ A=\left[\begin{array}{rr} 5 & -6 \\ 4 & 7 \end{array}\right] \] (a) \( \left|A^{T}\right| \) (b) \( \left|A^{2}\right| \) (c) \( \left|A A^{T}\right| \) (d) \( |2 \mathrm{~A}| \) (e) \( \left|A^{-1}\right| \) Find \( |A|,|B|, A B \), and \( |A B| \). Then verify that \( |A||B|=|A B| \). \[ A=\left[\begin{array}{rrrr} 2 & 0 & 1 & 1 \\ 1 & -1 & 0 & 1 \\ 2 & 4 & 1 & 0 \\ 1 & 2 & 4 & 0 \end{array}\right], \quad B=\left[\begin{array}{rrrr} 1 & 0 & -1 & 1 \\ 2 & 1 & 0 & 2 \\ 1 & 1 & -1 & 0 \\ 4 & 2 & 1 & 0 \end{array}\right] \] (a) \( |A| \) (b) \( |B| \) (c) \( A B \) Fina \( 1 A^{-1} \). Begin by findino \( A^{-1} \), and therevaluate its determinant. Verify your result by finding \( |A| \) and then applylng the formula \( \left|A^{-1}\right|=\frac{1}{|A|} \). \[ A=\left[\begin{array}{rrrr} 1 & 0 & 3 & 1 \\ 1 & 0 & -1 & 1 \\ 2 & 0 & -2 & 3 \\ -7 & 1 & -1 & 2 \end{array}\right] \]