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Using the properties of determinants,\[\begin{vmatrix}a^2+2a &2a+1 & 1\\2a+1 & a+2 & 1\\3 & 3 & 1\end{vmatrix}=(a-1)^3\]

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Toolbox:
  • If each element of a row (or a column)of a determinant is multiplied by a constant k,then |A|=k|A|.
  • If A is a square matrix such that each element of a row (or a column) of A is expressed as a sum of two or more terms,then the determinant of A can be expressed as the sum of the determinants of two or more matrices of the same order.
Let $\Delta=\begin{vmatrix}a^2+2a & 2a+1 & 1\\2a+1 & a+2 & 1\\3 & 3& 1\end{vmatrix}$
 
Apply $R_1\rightarrow R_1-R_2 and R_2\rightarrow R_2-R_3$
 
$\Delta=\begin{vmatrix}a^2-1 & a-1 & 0\\2a-2 & a-1 & 0\\3 & 3& 1\end{vmatrix}=\begin{vmatrix}a^2-1 & a-1 & 0\\2(a-1)& a-1 & 0\\3& 3 & 1\end{vmatrix}$
 
We know ($a^2-1)=(a+1)(a-1)$
 
Take (a-1) as a common factor from $R_1 $ and $R_2$
 
Therefore $\Delta=(a-1)^2\begin{vmatrix}a+1 & 1 & 0\\2 & 1 & 0\\3 & 3& 1\end{vmatrix}$
 
Expand along $R_1$
 
$\quad=(a-1)^2[(a+1)(1)-1(2)+0]$
 
$\quad=(a-1)^2[a+1-2]$
 
$\Delta=(a-1)^3$
 
Hence proved.

 

answered Mar 15, 2013 by sreemathi.v
 

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