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**Hint:**Start by finding the value of determinant A and B separately. Use the properties of adjoint matrix $\det (adjM) = {(\det M)^{n - 1}}$, Substitute the values in the given equation and find the value of k . Use the mathematical function of [k], which is the largest or greatest integer function giving output as only the greatest integer.

**Complete step-by-step answer:**

Given, $a_1^2 + a_2^2 + a_3^2 + a_4^2 = 1$

When we observe this question carefully then we come to know that the expression ${\left( {{a_1} - {a_2}} \right)^2} + {\left( {{a_2} - {a_3}} \right)^2} + {\left( {{a_3} - {a_4}} \right)^2} + {\left( {{a_4} - {a_1}} \right)^2}$ is always positive . Since all the terms are squared and the square of the negative term will also be positive therefore the minimum value for this equation is 0. And this can only be true, when all have the same values i.e. ${a_1} = {a_2} = {a_3} = {a_4} = a$

On putting the value of ${a_1},{a_2},{a_3},{a_4}$ as a and in the equation $a_1^2 + a_2^2 + a_3^2 + a_4^2 = 1$ we get the value of

$

{a^2} + {a^2} + {a^2} + {a^2} = 1 \\

\Rightarrow 4{a^2} = 1 \\

\Rightarrow {a^2} = \dfrac{1}{4} \\

\Rightarrow a = \dfrac{1}{2} \\

$

Therefore, we get the values of ${a_1},{a_2},{a_3},{a_4}$ as ${a_1} = {a_2} = {a_3} = {a_4} = a = \dfrac{1}{2}$

So, we found the value of ${a_1},{a_2},{a_3},{a_4}$which is 0.5 and we also came to know that the minimum value of the given equation will be 0, which would lie in the range of (-1.5, 2.5)

**So, the correct answer is “Option B”.**

**Note:**Similar problems can also be asked by using other properties of determinant which might include trace of matrices, cofactor or minor of a matrix, skew or symmetric matrices etc., and students must be well aware of all such important identities and properties. Attention must be given while evaluating the value of determinant as there are chances of making a mistake while considering the factor of ${( - 1)^{i + j}}$.

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