# Evaluate the determinants: $\begin{vmatrix} 0&1&2 \\ -1&0&3 \\-2&3&0 \end{vmatrix}$

$\begin{array}{1 1} 12 \\ -12 \\ 0 \\ \text{none of the above }\end{array}$

## 1 Answer

Toolbox:
• To evaluate a matrix of order $3\times 3$
• $\mid A\mid=\begin{vmatrix}a_{11} & a_{12} & a_{13}\\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33}\end{vmatrix}$
• Therefore $\mid A\mid=a_{11}(a_{22}\times a_{33}-a_{23}\times a_{32})-a_{12}(a_{21}\times a_{32}-a_{23}\times a_{31})+a_{13}(a_{21}\times a_{32}-a_{22}\times a_{31})$
Given:(iii) $Evaluate:\begin{vmatrix}0 & 1 & 2\\-1& 0 &-3\\-2 & 3 & 0\end{vmatrix}$

We know to evaluate the value of the determinant of order $3\times 3$

Therefore $\mid A\mid=a_{11}(a_{22}\times a_{33}-a_{23}\times a_{32})-a_{12}(a_{21}\times a_{32}-a_{23}\times a_{31})+a_{13}(a_{21}\times a_{32}- a_{22}\times a_{31})$

$Hence \mid A\mid=0[(0\times 0-(-3\times 3)]-1[(-1\times 0)-(-3\times -2]+2[(-1\times 3)-(0\times -2)]$

$\qquad=0-1(6)+2(-3)$

$\qquad=-6-6$

$\qquad=-12$
answered Mar 6, 2013
edited Apr 30, 2014

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