## Problem on computing a 3x3 determinant by the method of diagonals

Use the method of diagonals to compute the determinant $$\text{det } \begin{pmatrix} 0 & 1 & 2 \\ 3 & -1 & 0 \\ 1 & -2 & 1 \end{pmatrix}$$
• ## Solution

Recall that
The forward diagonals are given as
The products of forward diagonals are $0\cdot(-1) \cdot 1$ and $1 \cdot 0 \cdot 1$ and $2 \cdot 3 \cdot (-2)$.
The backward diagonals are given as
The products of backward diagonals are $0\cdot 0 \cdot (-2)$ and $1 \cdot 3 \cdot 1$ and $2 \cdot (-1) \cdot 1$.
Combining these, we see that \begin{align} \text{det } \begin{pmatrix} 0 & 1 & 2 \\ 3 & -1 & 0 \\ 1 & -2 & 1 \end{pmatrix} &= 0\cdot(-1) \cdot 1 + 1 \cdot 0 \cdot 1 + 2 \cdot 3 \cdot (-2) \\ &\phantom{= } - 0\cdot 0 \cdot (-2) - 1 \cdot 3 \cdot 1 - 2 \cdot (-1) \cdot 1 \end{align}
Thus, \begin{align} \text{det } \begin{pmatrix} 0 & 1 & 2 \\ 3 & -1 & 0 \\ 1 & -2 & 1 \end{pmatrix} = 0 + 0 + (-12) - 0 - 3 - (-2) =- 13.\end{align}