Problem on computing a 3x3 determinant by the method of diagonals
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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 method of diagonals for computing the determinant of a 3x3 matrix
The determinant of a $3 \times 3$ matrix can be computing by adding the products of terms on the forward diagonals and subtracting the products of terms on the backward diagonals.
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}
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