Matrices and Determinants 2 Question 23
25. The determinant $\begin{vmatrix}a & b & a \alpha+b \\ b & c & b \alpha+c \\ a \alpha+b & b \alpha+c & 0\end{vmatrix}$ is equal to zero, then
$(1986,2 M)$
(a) $a, b, c$ are in $\mathrm{AP}$
(b) $a, b, c$ are in GP
(c) $a, b, c$ are in $\mathrm{HP}$
(d) $(x-\alpha)$ is a factor of $a x^{2}+2 b x+c$
Numerical Value
Show Answer
Answer:
Correct Answer: 25. (b, d)
Solution:
- Given,
$\begin{vmatrix}a & b & a \alpha+b \\ b & c & b \alpha+c \\ a \alpha+b & b \alpha+c & 0\end{vmatrix}$=0
Applying $C_{3} \rightarrow C_{3}-\left(\alpha C_{1}+C_{2}\right)$
$ \begin{vmatrix} a & b & 0 \\ b & c & 0 \\ a \alpha+b & b \alpha+c & -\left(a \alpha^{2}+2 b \alpha+c\right) \end{vmatrix} $=0
$\Rightarrow -(a \alpha^{2}+2 b \alpha+ct)(a c-b^{2}) =0 $
$\Rightarrow a \alpha^{2}+2 b \alpha+c=0 \text { or } b^{2}= a c $
$\Rightarrow x-\alpha$ is a factor of $a x^{2}+2 b x+c$ or $a, b, c$ are in GP.
