Question

If the value of the determinant \left| {\begin{array}{*{20}{c}} a&1&1 \\\ 1&b;&1 \\\ 1&1&c; \end{array}} \right| is positive, then
A. $abc > 1$
B. $abc > - 8$
C. $abc < \- 8$

Answer 1

In the above given question we can see that the determinant is positive. So, we can apply the condition

a&1&1 \\\ 1&b;&1 \\\ 1&1&c; \end{array}} \right| > 0$$ and hence expand the given equation and apply the condition of $$A.M{\text{ }} > {\text{ }}G.M$$ and choose the correct option as per the given information. _**Complete step-by-step answer:**_ Given, that the determinant $$\left| {\begin{array}{*{20}{c}} a&1&1 \\\ 1&b;&1 \\\ 1&1&c; \end{array}} \right|$$ is positive So, expanding the above determinant, we get, $$\left| {\begin{array}{*{20}{c}} a&1&1 \\\ 1&b;&1 \\\ 1&1&c; \end{array}} \right| = a(bc - 1) - (c - 1) + (1 - b)$$ Now, further simplifying the above determinant $$ = a(bc - 1) - (c - 1) + (1 - b) > 0$$ On simplifying we get, $$ = abc - a - c + 1 + 1 - b > 0$$ On taking terms common we get, $$ = abc - (a + b + c) + 2 > 0$$ On rearranging we get $$ = abc + 2 > (a + b + c)$$ Now, applying the condition of $$A.M{\text{ }} > {\text{ }}G.M$$ So, $$a + b + c > 3{(abc)^{\dfrac{1}{3}}}$$ Let $${(abc)^{\dfrac{1}{3}}} = x$$ So it is clear from the above condition that $$abc + 2 > a + b + c > 3{(abc)^{\dfrac{1}{3}}}$$ Now, take the first and last term of the inequality, $$ \Rightarrow {x^3} - 3x + 2 > 0$$ On solving the above inequality,

\Rightarrow {(x - 1)^2}(x + 2) > 0 \\
\Rightarrow x > - 2 \\
\Rightarrow {(abc)^{\dfrac{1}{3}}} > - 2 \\

Here we are not equating $${(x - 1)^2} > 0$$ because the whole square of the number is always positive and thus it cannot make our term ever negative and we are equating the only term $$x + 2 > 0$$ as if it will be positive then the entire term will be positive. And hence on cubing both the sides, $$ = abc > ( - 8)$$ **Hence, option (B) is the correct answer.** **Note:** In mathematics, the inequality of arithmetic and geometric means, or more briefly the AM–GM inequality, states that the arithmetic mean of a list of non-negative real numbers is greater than or equal to the geometric mean of the same list. In linear algebra, the determinant is a scalar value that can be computed from the elements of a square matrix, and which encodes certain properties of the linear transformation described by the matrix. The determinant of a matrix A is denoted $$\det \left( A \right)$$ , det A, or $${\text{|A|}}$$.