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Choose the correct answer. If a, b, c, are in A.P, then the determinant $\begin{vmatrix} x+2 & x+3 &x+2a \\ x+3 & x+4 &x+2b \\ x+4 & x+5 &x+2c \end{vmatrix}$ is: \[\] $(A)\;0 \hspace{20 mm} (B)\;1 \hspace{20 mm} (C)\; x \hspace{20 mm} (D)\; 2x $

1 Answer

  • If each column of a row (or a column) of a determinant is multiplied by a constant ,then its value gets multiplied by k.
  • By this property we can take out any common factor from any one row or any column.
  • Elementary transformations can be done by
  • 1. Interchanging any two rows or columns.
  • 2. Mutiplication of the elements of any row or column by a non-zero number
  • 3. The addition of any row or column , the corresponding elemnets of any other row or column multiplied by any non zero number.
Let $\Delta=\begin{vmatrix}x+2 & x+3 & x+2a\\x+3 & x+4 & x+2b\\x+4 &x+5 & x+2c\end{vmatrix}$
Apply $R_1\rightarrow R_1+R_2$ and $R_2\rightarrow R_2-R_3$
$\Delta=\begin{vmatrix}-1 & -1 & 2a-2b\\-1 & -1 & 2b-2c\\x+4 &x+5 & x+2c\end{vmatrix}$
$\Delta=\begin{vmatrix}-1 & -1 & 2(a-b)\\-1 & -1 & 2(b-c)\\x+4 &x+5 & x+2c\end{vmatrix}$
Apply $C_1\rightarrow C_1+C_2$
$\Delta=\begin{vmatrix}0 & -1 & 2(a-b)\\0 & -1 & 2(b-c)\\x+4 &x+5 & x+2c\end{vmatrix}$
Now expanding along $C_1$ we get,
$\Delta=0-0+(x+4)[-1\times 2(a-b)+1\times 2(b-c)]$
But given a,b,c are in A.P;hence 2b=a+c.
Therefore $\Delta=(x+4)[-2a+b+a+c-2c]$
$\;\;\;\;\qquad\;\;\;\;\;=(x+4)[-(a+c)+b]$ But (a+c=b)
Therefore $\Delta=(x+4)[-b+b]=0$
Hence the correct answer is A


answered Mar 4, 2013 by sreemathi.v
edited Mar 5, 2013 by vijayalakshmi_ramakrishnans

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