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Evaluate : $ \begin{vmatrix} x^2-x+1 & x-1 \\ x+1 & x+1 \end{vmatrix} $

1 Answer

Toolbox:
  • For a given determinant A of order 2 $\begin{vmatrix}a_{11}& a_{12}\\a_{21} & a_{22}\end{vmatrix}$
  • To evaluate the value of the given determinants ,let us multiply the elements $a_{11}$ and $a_{22}$ and then subtract $a_{21}\times a_{12}$.
On expanding we get,
$\Delta =(x+1)(x^2-x+1)-(x+1)(x-1)$
But $(x+1)(x^2-x+1)=x^3-1$
$(x+1)(x-1)=x^2-1$
$\therefore \Delta=x^3-1-x^2+1$
$\Delta=x^3-x^2$
$\quad=x^2(x-1)$
answered Nov 6, 2013 by sreemathi.v
 

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