SATHEE: Matrices and Determinants 2 Question 23

Matrices and Determinants 2 Question 23

25. The determinant $| \begin{matrix} a & b & a α + b \\ b & c & b α + c \\ a α + b & b α + c & 0 \end{matrix} |$ is equal to zero, then

$(1986, 2 M)$

(a) $a, b, c$ are in $AP$

(b) $a, b, c$ are in GP

(c) $a, b, c$ are in $HP$

(d) $(x - α)$ is a factor of $a x^{2} + 2 b x + c$

Numerical Value

Show Answer

Answer:

Correct Answer: 25. (b, d)

Solution:

Given,

$| \begin{matrix} a & b & a α + b \\ b & c & b α + c \\ a α + b & b α + c & 0 \end{matrix} |$ =0

Applying $C_{3} \to C_{3} - (α C_{1} + C_{2})$

$| \begin{matrix} a & b & 0 \\ b & c & 0 \\ a α + b & b α + c & - (a α^{2} + 2 b α + c) \end{matrix} |$ =0

$\Rightarrow - (a α^{2} + 2 b α + c t) (a c - b^{2}) = 0$

$\Rightarrow a α^{2} + 2 b α + c = 0 or b^{2} = a c$

$\Rightarrow x - α$ is a factor of $a x^{2} + 2 b x + c$ or $a, b, c$ are in GP.

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