Inverse of $$$\left[\begin{array}{ccc}0 & -3 & 0\\-3 & -3 & 1\\0 & 1 & 0\end{array}\right]$$$

Calculate $$$\left[\begin{array}{ccc}0 & -3 & 0\\-3 & -3 & 1\\0 & 1 & 0\end{array}\right]^{-1}$$$ using the Gauss-Jordan elimination.

To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be the inverse matrix.

So, augment the matrix with the identity matrix:

$$$\left[\begin{array}{ccc|ccc}0 & -3 & 0 & 1 & 0 & 0\\-3 & -3 & 1 & 0 & 1 & 0\\0 & 1 & 0 & 0 & 0 & 1\end{array}\right]$$$

Since the element at row $$$1$$$ and column $$$1$$$ (pivot element) equals $$$0$$$, we need to swap the rows.

Find the first nonzero element in column $$$1$$$ under the pivot entry.

The first nonzero element is at row $$$2$$$.

Swap the rows $$$1$$$ and $$$2$$$:

$$$\left[\begin{array}{ccc|ccc}-3 & -3 & 1 & 0 & 1 & 0\\0 & -3 & 0 & 1 & 0 & 0\\0 & 1 & 0 & 0 & 0 & 1\end{array}\right]$$$

Divide row $$$1$$$ by $$$-3$$$: $$$R_{1} = - \frac{R_{1}}{3}$$$.

$$$\left[\begin{array}{ccc|ccc}1 & 1 & - \frac{1}{3} & 0 & - \frac{1}{3} & 0\\0 & -3 & 0 & 1 & 0 & 0\\0 & 1 & 0 & 0 & 0 & 1\end{array}\right]$$$

Divide row $$$2$$$ by $$$-3$$$: $$$R_{2} = - \frac{R_{2}}{3}$$$.

$$$\left[\begin{array}{ccc|ccc}1 & 1 & - \frac{1}{3} & 0 & - \frac{1}{3} & 0\\0 & 1 & 0 & - \frac{1}{3} & 0 & 0\\0 & 1 & 0 & 0 & 0 & 1\end{array}\right]$$$

Subtract row $$$2$$$ from row $$$1$$$: $$$R_{1} = R_{1} - R_{2}$$$.

$$$\left[\begin{array}{ccc|ccc}1 & 0 & - \frac{1}{3} & \frac{1}{3} & - \frac{1}{3} & 0\\0 & 1 & 0 & - \frac{1}{3} & 0 & 0\\0 & 1 & 0 & 0 & 0 & 1\end{array}\right]$$$

Subtract row $$$2$$$ from row $$$3$$$: $$$R_{3} = R_{3} - R_{2}$$$.

$$$\left[\begin{array}{ccc|ccc}1 & 0 & - \frac{1}{3} & \frac{1}{3} & - \frac{1}{3} & 0\\0 & 1 & 0 & - \frac{1}{3} & 0 & 0\\0 & 0 & 0 & \frac{1}{3} & 0 & 1\end{array}\right]$$$

Since row $$$3$$$ consists solely of zeros, the determinant of the matrix equals $$$0$$$.

Thus, the matrix is not invertible.

The matrix is not invertible.