# If $A = \begin{bmatrix} 1&1&-2\\ 2&1&-3 \\5&4&-9 \end{bmatrix}$, then find $|A|$

$\begin{array}{1 1} 1 \\ -1 \\ 0 \\ None\;of\;the\;above \end{array}$

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:Evaluate: $\mid A\mid=\begin{vmatrix}1 & 1 & -2\\2& 1 &-3\\5 & 4 & -9\end{vmatrix}$

To evaluate the value of the determinant of the matrix A:

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

$Hence \mid A\mid=1[(1\times -9)-(-3\times 4)]-1[(2\times -9)-(-3\times 5]+(-2)[(2\times 4)-(1\times 5)]$

$\qquad=-9+12-[-18+15]-2(8-5)$

$\qquad=3+3-6$

$\qquad=0$