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Q : 17     If  a,b,c,  are in A.P, then the determinant
               \dpi{100} \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            (B) 1           (C) x        (D) 2x

Answers (1)

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Given determinant \triangle = \begin{vmatrix} x+2 &x+3 &x+2a \\ x+3 & x+4 & x+2b\\ x+4 & x+5 &x+2c \end{vmatrix} and given that a, b, c are in A.P.

That means , 2b =a+c

\triangle = \begin{vmatrix} x+2 &x+3 &x+2a \\ x+3 & x+4 & x+(a+c)\\ x+4 & x+5 &x+2c \end{vmatrix}

Applying the row transformations,  R_{1} \rightarrow R_{1} -R_{2}  and then R_{3} \rightarrow R_{3} -R_{2} we have;

\triangle = \begin{vmatrix} -1 &-1 &a-c \\ x+3 & x+4 & x+(a+c)\\ 1 & 1 &c-a \end{vmatrix}

Now, applying another row transformation, R_{1} \rightarrow R_{1} + R_{3}, we have

\triangle = \begin{vmatrix} 0 &0 &0 \\ x+3 & x+4 & x+(a+c)\\ 1 & 1 &c-a \end{vmatrix}

Clearly we have the determinant value equal to zero;

Hence the option (A) is correct.

 

 

Posted by

Divya Prakash Singh

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