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Home  >>  CBSE XII  >>  Math  >>  Determinants
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Using Cofactors of elements of second row, evaluate $ \Delta = \begin{vmatrix} 5&3&8 \\ 2&0&1 \\ 1&2&3 \end{vmatrix} $

$\begin{array}{1 1} 7 \\ -7 \\ 14 \\ -14 \end{array} $

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Toolbox:
  • Expanding the determinant $\bigtriangleup$ along $R_2$ we have
  • $\bigtriangleup=(-1)^{2+1} a_{21}\begin{vmatrix}a_{12} &a_{13}\\a_{32} & a_{33}\end{vmatrix}+(-1)^{2+2}a_{22}\begin{vmatrix}a_{11} &a_{13}\\a_{31} & a_{33}\end{vmatrix}+(-1)^{2+3}a_{23}\begin{vmatrix}a_{11} &a_{12}\\a_{31} & a_{32}\end{vmatrix}$
  • $\;\;\;=a_{21}A_{21}+a_{22}.A_{22}+a_{23}A_{23}$
Using cofactors of elements of second row,evaluate
$\bigtriangleup=\begin{vmatrix}5 & 3 & 8\\2 & 0 & 1\\1 & 2 & 3\end{vmatrix}$
We know that expanding the determinant $\bigtriangleup$ along $R_2$ we have,
$\bigtriangleup=a_{21}A_{21}+a_{22}A_{22}+a_{23}A_{23}$
Minor of $M_{21}=\begin{vmatrix}3 & 8\\2 & 3\end{vmatrix}$
$\;\;\;=9-16=-7$
Minor of $M_{22}=\begin{vmatrix}5 & 8\\1 & 2\end{vmatrix}=10-8=2$
Minor of $M_{23}=\begin{vmatrix}5 & 3\\1 & 2\end{vmatrix}=10-3=7$
$A_{21}=(-1)^{2+1}\times -7=-(-7)=7$
$A_{22}=(-1)^{2+2}\times 2=2$
$A_{23}=(-1)^{2+3}\times 7=-7$
Hence $a_{21}A_{21}+a_{22}.A_{22}+a_{23}A_{23}$
$\;\;\;=2\times 7+0\times 2+1\times -7$
$\;\;\;=14+0-7$
$\;\;\;=7$
Hence the value of the determinant is 7.
answered Mar 6, 2013 by vijayalakshmi_ramakrishnans
 

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