Solution - Matrix core operations

$$\det \left(\left[\begin{array}{cc}0& 3\\ 0& -1\end{array}\right]\right)$$

Use elimination to convert A into upper-triangular form; then multiply diagonal entries.

$$\det \left(\left[\begin{array}{cc}0& 3\\ 0& -1\end{array}\right]\right)$$

Every row swap multiplies the determinant by -1. Swap count: 0.

$$\det \left(\left[\begin{array}{cc}0& 3\\ 0& -1\end{array}\right]\right)$$

A zero pivot was reached, so the determinant is 0.

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

This is the upper-triangular matrix used for the determinant product.

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

After triangularization, determinant comes from sign factor and diagonal product.

$$0$$

Determinant is 0, so the matrix is singular (not invertible).

$$0$$

Study tip: determinant zero means not invertible; non-zero means invertible.

$$0$$

Present the final matrix or scalar outcome in canonical form for stable routing and review.