If $$ A=\left|\begin{array}{lll} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{array}\right| $$ then \(\mathrm{A}^{-1}=\ldots\) (a) \(\mathrm{A}\) (b) \(\mathrm{A}^{2}\) (c) \(\mathrm{A}^{3}\) (d) \(\mathrm{A}^{4}\)

First, write down the given matrix A and the 3x3 identity matrix (I). Matrix A: \[ \left[\begin{array}{lll} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{array}\right] \] Identity Matrix (I): \[ \left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \]

Now, we will perform row operations on both A and I simultaneously. The goal is to convert A into the identity matrix (I). 1. Subtract \(\frac{2}{3}\) times row 1 from row 2 (R2 - \(\frac{2}{3}\) * R1): New A: \[ \left[\begin{array}{lll} 3 & -3 & 4 \\ 0 & -1 & 0 \\ 0 & -1 & 1 \end{array}\right] \] New I: \[ \left[\begin{array}{lll} 1 & 0 & 0 \\ -\frac{2}{3} & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \] 2. Add row 2 to row 3 (R3 + R2): New A: \[ \left[\begin{array}{lll} 3 & -3 & 4 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{array}\right] \] New I: \[ \left[\begin{array}{lll} 1 & 0 & 0 \\ -\frac{2}{3} & 1 & 0 \\ -\frac{2}{3} & 1 & 1 \end{array}\right] \] 3. Divide row 2 by -1 (R2 = -R2): New A: \[ \left[\begin{array}{lll} 3 & -3 & 4 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \] New I: \[ \left[\begin{array}{lll} 1 & 0 & 0 \\ \frac{2}{3} & -1 & 0 \\ -\frac{2}{3} & 1 & 1 \end{array}\right] \] 4. Add row 1 to row 2 (R1 + R2): New A: \[ \left[\begin{array}{lll} 3 & 0 & 4 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \] New I: \[ \left[\begin{array}{lll} -\frac{1}{3} & -1 & 0 \\ \frac{2}{3} & -1 & 0 \\ -\frac{2}{3} & 1 & 1 \end{array}\right] \] 5. Divide row 1 by 3 (R1 = \(\frac{1}{3}\) * R1): New A: \[ \left[\begin{array}{lll} 1 & 0 & \frac{4}{3} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \] New I: \[ \left[\begin{array}{lll} -\frac{1}{9} & -\frac{1}{3} & 0 \\ \frac{2}{3} & -1 & 0 \\ -\frac{2}{3} & 1 & 1 \end{array}\right] \] 6. Subtract \(\frac{4}{3}\) times row 3 from row 1 (R1 - \(\frac{4}{3}\) * R3): New A: \[ \left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \] New I: \[ \left[\begin{array}{lll} -\frac{1}{9} & -\frac{1}{3} & \frac{4}{9} \\ \frac{2}{3} & -1 & 0 \\ -\frac{2}{3} & 1 & 1 \end{array}\right] \]

Matrix A has now been converted into the identity matrix (I), and New I is the inverse of the original matrix A. So, A^{-1} = \[ \left[\begin{array}{lll} -\frac{1}{9} & -\frac{1}{3} & \frac{4}{9} \\ \frac{2}{3} & -1 & 0 \\ -\frac{2}{3} & 1 & 1 \end{array}\right] \] This does not match with any of the given options (a), (b), (c), or (d).