If $lim x \to 0 {(\frac{a^{x} + b^{x} + c^{x} + d^{x}}{4})}^{1 / x} = 8,$ then the minimum value of the determinant $∣ ∣ ∣ \begin{matrix} a + i b & c + i d - c + i d & a - i b \end{matrix} ∣ ∣ ∣$ is

Open in App

Solution

We have,
$lim x \to 0 {(\frac{a^{x} + b^{x} + c^{x} + d^{x}}{4})}^{1 / x} = \sqrt[4]{a b c d} = 8 \Rightarrow \sqrt{a b c d} = 64$
Now,
$∣ ∣ ∣ \begin{matrix} a + i b & c + i d - c + i d & a - i b \end{matrix} ∣ ∣ ∣ = (a + i b) (a - i b) - (c + i d) (- c + i d)$
$= (a + i b) (a - i b) + (c + i d) (c - i d)$
$= a^{2} + b^{2} + c^{2} + d^{2}$

We know, $A . M \geq G . M$
$\Rightarrow \frac{a^{2} + b^{2} + c^{2} + d^{2}}{4} \geq \sqrt[4]{a^{2} b^{2} c^{2} d^{2}}$
$\Rightarrow a^{2} + b^{2} + c^{2} + d^{2} \geq 4 \cdot \sqrt{a b c d}$
$\Rightarrow a^{2} + b^{2} + c^{2} + d^{2} \geq 256$

Hence, the minimum value of $∣ ∣ ∣ \begin{matrix} a + i b & c + i d - c + i d & a - i b \end{matrix} ∣ ∣ ∣$ is equal to $256.$

Suggest Corrections

Similar questions

∣ ∣ ∣ \begin{matrix} a + i b & c + i d - c + i d & a - i b \end{matrix} ∣ ∣ ∣

|\begin{matrix} a + i b & c + i d \\ - c + i d & a - i b \end{matrix}| .

A = [\begin{matrix} a + i b & c + i d - c + i d & a - i b \end{matrix}]

a^{2} + b^{2} + c^{2} + d^{2} = 1

A^{- 1}

A = [\begin{matrix} a + i b & c + i d - c + i d & a - i b \end{matrix}]

a^{2} + b^{2} + c^{2} + d^{2} = 1

a, b, c

d

A = ∣ ∣ ∣ \begin{matrix} a + i b & c + i d - c + i d & a - i b \end{matrix} ∣ ∣ ∣

A^{- 1} =

Join BYJU'S Learning Program