MCQ
$\left| {\,\begin{array}{*{20}{c}}{a + b}&{b + c}&{c + a}\\{b + c}&{c + a}&{a + b}\\{c + a}&{a + b}&{b + c}\end{array}\,} \right| = K\,\,\left| {\,\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}\,} \right|\,,$ then $K = $
  • A
    $1$
  • $2$
  • C
    $3$
  • D
    $4$

Answer

Correct option: B.
$2$
b
(b) The determinant can be written sum of $2 \times 2 \times 2 = 8$ determinants of which $ 6 $   are reduces to zero because of their two rows are identical. Hence proceed.

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Similar questions

Let $\mathrm{p}$ be an odd prime number and $\mathrm{T}_{\mathrm{p}}$ be the following set of $2 \times 2$ matrices :

$T_p=\left\{A=\left[\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{a}\end{array}\right]: \mathrm{a}, \mathrm{b}, \mathrm{c} \in\{0,1, \ldots ., \mathrm{p}-1\}\right\}$

$1.$ The number of $A$ in $T_p$ such that $A$ is either symmetric or skew-symmetric or both, and $\operatorname{det}(\mathrm{A})$ divisible by $\mathrm{p}$ is

$(A)$ $(\mathrm{p}-1)^2$  $(B)$ $2(\mathrm{p}-1)$

$(C)$ $(\mathrm{p}-1)^2+1$  $(D)$ $2 \mathrm{p}-1$

$2.$ The number of $A$ in $T_p$ such that the trace of $A$ is not divisible by $p$ but det $(A)$ is divisible by $p$ is [Note: The trace of a matrix is the sum of its diagonal entries.]

$(A)$ $(\mathrm{p}-1)\left(\mathrm{p}^2-\mathrm{p}+1\right)$ $(B)$ $\mathrm{p}^3-(\mathrm{p}-1)^2$

$(C)$ $(\mathrm{p}-1)^2$ $(D)$ $(p-1)\left(p^2-2\right)$

$3.$ The number of $A$ in $T_p$ such that det $(A)$ is not divisible by $p$ is

$(A)$ $2 \mathrm{p}^2$ $(B)$ $p^3-5 p$ $(C)$ $p^3-3 p$ $(D)$ $p^3-p^2$

Give the answer question $1,2$ and $3.$

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