# Show that $\begin{vmatrix}x+1 &x+2 &x+a\\x+2 &x+3 &x+b\\x+3 &x+4 &x+c\end{vmatrix}=0$ where $a,b,c$ are in A.P.

## 1 Answer

Toolbox:
• The elementary transformations are
• (i) We can interchange any two rows or two columns.
• (ii) The multiplication of the elements of any row or column by a non-zero number.
• (iii) The addition to the elements of any row or column,the corresponding elements of other row or column multiplied by any non-zero number.
Step 1:
Let $\Delta=\begin{vmatrix}x+1 & x+2 & x+a\\x+2 & x+3 & x+b\\x+3 & x+4 & x+c\end{vmatrix}$
Apply $R_1\rightarrow R_2-R_1$ and $R_2\rightarrow R_3-R_2$
$\Delta=\begin{vmatrix}1 & 1 & b-a\\1 & 1 & c-b\\x+3 & x+4 & x+c\end{vmatrix}$
Apply $C_1\rightarrow C_1-C_2$
$\Delta=\begin{vmatrix}0 & 1 & b-a\\0 & 1 & c-b\\-1 & x+4 & x+c\end{vmatrix}$
Step 2:
Now expanding along $C_1$ we get,
$=0+0-1((b-a)-(c-b))=-b+a+c-b=a+c-2b$
But since a,b,c are in A.P
$2b=a+c$
Therefore $\Delta=2b-2b=0.$
Hence the value is 0.
answered Sep 20, 2013

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