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Home  >>  CBSE XII  >>  Math  >>  Determinants
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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} $

Can you answer this question?
 
 

1 Answer

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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:
 
$\mid A\mid=a_{11}(a_{22}\times a_{33}-a_{23}\times a_{32})-a_{12}(a_{21}\times a_{32}-a_{23}\times a_{31})+a_{13}(a_{21}\times a_{32}- a_{22}\times a_{31})$
 
$Hence \mid A\mid=1[(1\times -9)-(-3\times 4)]-1[(2\times -9)-(-3\times 5]+(-2)[(2\times 4)-(1\times 5)]$
 
$\qquad=-9+12-[-18+15]-2(8-5)$
 
$\qquad=3+3-6$
 
$\qquad=0$

 

answered Feb 20, 2013 by sreemathi.v
 

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