Question 32

Suppose a, b and c are integers such that a > b > c > 0, and $$A =\begin{bmatrix}a & b & c \\ b & c & a \\c & a & b \end{bmatrix}$$. Then the value of the determinant of A

Let $$M =\begin{bmatrix}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\\\end{bmatrix}$$ :

Determinant of M = $$x_1\left(y_2z_3-y_3z_2\right)-y_1\left(x_2z_3-x_3z_2\right)+z_1\left(x_2y_3-x_3y_2\right)$$

Hence, for the given matrix $$A =\begin{bmatrix}a & b & c \\ b & c & a \\c & a & b \end{bmatrix}$$, the determinant is :

$$a\left(cb-a^2\right)-b\left(b^2-ac\right)+c\left(ba-c^2\right)$$ 

= $$3(abc)-a^3-b^3-c^3$$

Now, it is given in the question that all a,b,c are positive numbers and distinct  :

Hence , we can apply $$AM\ >\ GM$$

$$=\ \dfrac{\ a^3+b^3+c^3}{3}\ >\ abc$$

Therefore the determination will be 3abc - {the value which is greater than 3abc} = {Negative} value.

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