MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Logarithmic Differentiation

Expressing ...

Question

Expressing $A$ as the sum of Hermitian and a Skew-Hermitian matrix, where,
$A = ⎡ ⎢ ⎣ \begin{matrix} 2 + 3 i & 2 & 5 - 3 - i & 7 & 3 - i 3 - 2 i & i & 2 + i \end{matrix} ⎤ ⎥ ⎦$
be $A = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 2 & - \frac{1}{2} + \frac{i}{2} & m + k i - \frac{1}{2} - \frac{i}{2} & 7 & \frac{3}{2} - 1 4 - i & \frac{3}{2} + 1 & 2 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ + ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 3 i & \frac{5}{2} - \frac{i}{2} & n + p i - \frac{5}{2} - \frac{i}{2} & 0 & \frac{3}{2} - 1 - i & - \frac{3}{2} & i \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$ .Find $m + k + n + p$ ?

Open in App

Solution

$A = ⎡ ⎢ ⎣ \begin{matrix} 2 + 3 i & 2 & 5 - 3 - i & 7 & 3 - i 3 - 2 i & i & 2 + i \end{matrix} ⎤ ⎥ ⎦$
$∴ A^{'} = ⎡ ⎢ ⎣ \begin{matrix} 2 + 3 i & - 3 - i & 3 - 2 i 2 & 7 & i 5 & 3 - i & 2 + i \end{matrix} ⎤ ⎥ ⎦$
$¯ ¯¯¯¯¯¯¯¯ ¯ (A^{'}) = ⎡ ⎢ ⎣ \begin{matrix} 2 - 3 i & - 3 + i & 3 + 2 i 2 & 7 & - i 5 & 3 + i & 2 - i \end{matrix} ⎤ ⎥ ⎦$
Or $A^{⊖} = ⎡ ⎢ ⎣ \begin{matrix} 2 - 3 i & - 3 + i & 3 + 2 i 2 & 7 & - i 5 & 3 + i & 2 - i \end{matrix} ⎤ ⎥ ⎦$
$∴ A + A^{⊖} = ⎡ ⎢ ⎣ \begin{matrix} 2 + 3 i & 2 & 5 - 3 - i & 7 & 3 - i 3 - 2 i & i & 2 + i \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} 2 - 3 i & - 3 + i & 3 + 2 i 2 & 7 & - i 5 & 3 + i & 2 - i \end{matrix} ⎤ ⎥ ⎦$
$= ⎡ ⎢ ⎣ \begin{matrix} 4 & - 1 + i & 8 + 2 i - 1 - i & 14 & 3 - 2 i 8 - 2 i & 3 + 2 i & 4 \end{matrix} ⎤ ⎥ ⎦ . . . (1)$
And $A - A^{⊖} = ⎡ ⎢ ⎣ \begin{matrix} 2 + 3 i & 2 & 5 - 3 - i & 7 & 3 - i 3 - 2 i & i & 2 + i \end{matrix} ⎤ ⎥ ⎦ - ⎡ ⎢ ⎣ \begin{matrix} 2 - 3 i & - 3 + i & 3 + 2 i 2 & 7 & - i 5 & 3 + i & 2 - i \end{matrix} ⎤ ⎥ ⎦$
$= ⎡ ⎢ ⎣ \begin{matrix} 6 i & 5 - i & 2 - 2 i - 5 - i & 0 & 3 - 2 - 2 i & - 3 & 2 i \end{matrix} ⎤ ⎥ ⎦ . . . (2)$
Adding $(1)$ and $(2)$ , we get
$2 A = ⎡ ⎢ ⎣ \begin{matrix} 4 & - 1 + i & 8 + 2 i - 1 - i & 14 & 3 - 2 i 8 - 2 i & 3 + 2 i & 4 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} 6 i & 5 - i & 2 - 2 i - 5 - i & 0 & 3 - 2 - 2 i & - 3 & 2 i \end{matrix} ⎤ ⎥ ⎦$
Hence,
$A = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 2 & - \frac{1}{2} + \frac{i}{2} & 8 + 2 i - \frac{1}{2} - \frac{i}{2} & 7 & \frac{3}{2} - 1 4 - i & \frac{3}{2} + 1 & 2 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ + ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 3 i & \frac{5}{2} - \frac{i}{2} & 1 - i - \frac{5}{2} - \frac{i}{2} & 0 & \frac{3}{2} - 1 - i & - \frac{3}{2} & i \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$
$\Rightarrow m = 8, k = 2, n = 1$ and $p = - 1$
$∴ m + k + n + p = 10$

Suggest Corrections

thumbs-up

0

Similar questions

Q. Express the following complex numbers in the standard form a + i b:
(i)

(1 + i) (1 + 2 i)

\frac{3 + 2 i}{- 2 + i}

\frac{1}{(2 + i)^{2}}

\frac{1 - i}{1 + i}

\frac{(2 + i)^{3}}{2 + 3 i}

\frac{(1 + i) (1 + \sqrt{3} i)}{1 - i}

\frac{2 + 3 i}{4 + 5 i}

\frac{(1 - i)^{3}}{1 - i^{3}}

(1 + 2 i)^{- 3}

\frac{3 - 4 i}{(4 - 2 i) (1 + i)}

(\frac{1}{1 - 4 i} - \frac{2}{1 + i}) (\frac{1 - 4 i}{5 + i})

\frac{5 + \sqrt{2} i}{1 - 2 \sqrt{i}}

Find the general value of ${l o g}_{3}$ (3i). Where i= $\sqrt{- 1}$

z = \frac{(1 + i) (1 + 2 i) (1 + 3 i) \dots \dots \dots (1 + n i)}{(1 - i) (2 - i) (3 - i) \dots \dots \dots (n - i)}

i = \sqrt{- 1}, n \in N

, then principal argument of

z

Q. Find the value of the following determinants, where

i = \sqrt{- 1} .

∣ ∣ ∣ \begin{matrix} 2 i & - 3 i i^{3} & {- 2 i}^{5} \end{matrix} ∣ ∣ ∣

∣ ∣ ∣ \begin{matrix} 1 + 3 i & i - 2 - i - 2 & 1 - 3 i \end{matrix} ∣ ∣ ∣

Q. If for any square matrix

A^{2}

(I + A)^{n}

is equal to (where I is identity matrix)

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Logarithmic Differentiation

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Logarithmic Differentiation

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon