Complex Number elements MCQ Quiz in मल्याळम - Objective Question with Answer for Complex Number elements - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Calculation:

Determinant Δ = $a(ei-fh)-b(di-fg)+c(dh-eg)$

Now, For our matrix, 

$a=2,b=3+i,c=-1,d=3-i,e=0,f=i,g=-1,h=-i,i=1$

calculate the subdeterminants

⇒ $ei-fh=(0)(1)-(i)(-i)=0-(-1)=1$

⇒ $di-fg=(3-i)(1)-(i)(-1)=3-i+i=3$

⇒ $dh-eg=(3-i)(-i)-(0)(-1)=-3i+i2=-3i-1=-1-3i$

⇒ Δ = $2(1)-(3+i)(3)+(-1)(-1-3i)$

⇒ Δ = $2-9-3i+1+3i$

⇒ $\Delta =-6+0i$

Since we are given that $\Delta =A+iB$ comparing the real and imaginary parts, we find:

A  = -6 and B = 0

Thus A + B = -6 + 0 = - 6

Hence, the Correct answer is Option 2.

Solution:

For a complex number to be equal they must have equal magnitude as well as equal amplitude.

For two complex numbers to be negative to each other then the sum of their real parts must be 0. i.e.,

For z = a + ib

its negative complex number is -a + ib

For two complex number to be conjugate to each other, their magnitude should be equal and their amplitude must be of opposite polarity. i.e., amp z1 = - amp z2

⇒ amp z1 + amp z2 = 0

Hence, the given condition is of two complex number to be conjugate of each other.

Calculation:

We are given the following matrix:

$A(\theta )=\left[\begin{array}{cc}\sin \theta & i\cos \theta \\ i\cos \theta & \sin \theta \end{array}\right]$

The determinant of $A(\theta )$ is calculated as follows:

$\text{det}(A(\theta ))={\sin }^2\theta -(i\cos \theta )(i\cos \theta )={\sin }^2\theta +{\cos }^2\theta =1$

Since the determinant is 1 (non-zero), matrix Aθ  is invertible for all θ in $\mathbb{R}$

Therefore, Statement I is correct.

We first compute the inverse of A(θ)^-1 using the formula for the inverse of a 2x2 matrix:

$A(\theta {)}^{-1}=\frac{1}{\text{det}(A(\theta ))}\left[\begin{array}{cc}\sin \theta & -i\cos \theta \\ -i\cos \theta & \sin \theta \end{array}\right]$

Since the determinant is 1, we have:

$A(\theta {)}^{-1}=\left[\begin{array}{cc}\sin \theta & -i\cos \theta \\ -i\cos \theta & \sin \theta \end{array}\right]$

Now, let’s compute A(-θ) :

$A(-\theta )=\left[\begin{array}{cc}\sin (-\theta )& i\cos (-\theta )\\ i\cos (-\theta )& \sin (-\theta )\end{array}\right]$

Using the trigonometric identities$\sin (-\theta )=-\sin (\theta )$ and $\cos (-\theta )=\cos (\theta )$ we get:

$A(-\theta )=\left[\begin{array}{cc}-\sin \theta & i\cos \theta \\ i\cos \theta & -\sin \theta \end{array}\right]$

This is not equal to$A(\theta {)}^{-1}$, because the signs are different.

Therefore, Statement II is not correct.

Hence, the correct answer is: Option (1)

Calculation:

We are given two matrices:

$A(\theta )=\left[\begin{array}{cc}\sin \theta & i\cos \theta \\ i\cos \theta & \sin \theta \end{array}\right]$

$B(\theta )=A(\frac{\pi }{2}-\theta )=\left[\begin{array}{cc}\sin (\frac{\pi }{2}-\theta )& i\cos (\frac{\pi }{2}-\theta )\\ i\cos (\frac{\pi }{2}-\theta )& \sin (\frac{\pi }{2}-\theta )\end{array}\right]$

Using the trigonometric identities:

$\sin (\frac{\pi }{2}-\theta )=\cos (\theta )$

$\cos (\frac{\pi }{2}-\theta )=\sin (\theta )$

The matrix $B(\theta )$ becomes:

$B(\theta )=\left[\begin{array}{cc}\cos \theta & i\sin \theta \\ i\sin \theta & \cos \theta \end{array}\right]$

Now, compute the matrix product AB :

$AB=\left[\begin{array}{cc}\sin \theta & i\cos \theta \\ i\cos \theta & \sin \theta \end{array}\right]\left[\begin{array}{cc}\cos \theta & i\sin \theta \\ i\sin \theta & \cos \theta \end{array}\right]$

Performing the matrix multiplication:

$AB=\left[\begin{array}{cc}\sin \theta \cos \theta +i\cos \theta i\sin \theta & \sin \theta i\sin \theta +i\cos \theta \cos \theta \\ i\cos \theta \cos \theta +\sin \theta i\sin \theta & i\cos \theta i\sin \theta +\sin \theta \cos \theta \end{array}\right]$

Simplifying the terms:

$AB=\left[\begin{array}{cc}i{\cos }^2\theta +{\sin }^2\theta & i\sin \theta \cos \theta +\sin \theta \cos \theta \\ i\sin \theta \cos \theta +\cos \theta \sin \theta & i{\cos }^2\theta +{\sin }^2\theta \end{array}\right]$

Now, simplify further:

$AB=\left[\begin{array}{cc}0& i\\ i& 0\end{array}\right]$

Hence, the correct answer is Option (1).

Calculation:

Statement I

Expanding along the first row

Δ = 2[0 - (2 + i) (2 - i)] - (1 + i)[(1 - i) - 3(2 + i)] + 3 (1 - i)(2 - i) 

⇒ Δ =  - 2 (4 - i²) - (1 + i) (- 5 - 4i) + 3 (2 + i² - 3i)

⇒ Δ = - 10 + 5 + 9i - 4 + 3 - 9i = - 6 which is real.

Statement II

Expanding along the first row

Δ = 1 (- 4 - i²) - i(2i) + 2(- i²)

⇒ Δ = - 3 - 2 + 2 = - 3 which is real.

Statement III

Expanding along the first row

Δ = √6 [-(3 - i√8)(3 + i√8)] - (3 - i√2) [√11(3 + i√2) - (2 - i√6)(3 - i√8)] + (2 + i√6)(3 + i√2)(3 + i√8)

⇒ Δ = -√6 [9 - 8i²] - √11 (9 - 2i²) + (3 - i√2)(2 - i√6)(3 - i√8) + (2 + i√6)(3 + i√2)(3 + i√8)

⇒ Δ = - 17√6 - 11√11 + (2 - i√6) (9 - 4 - i6√2 - i3√2) + (2 + i√6) (9 - 4 + i6√2 + i3√2)

⇒ Δ = - 17√6 - 11√11 + (2 - i√6) (5 - 9i√2) + (2 + i√6) (5 + 9i√2) 

⇒ Δ = - 17√6 - 11√11 + 10 - 9√12 - 18i√2 - 5i√6 + 10 - 9√12 + 18i√2 + 5i√6 

⇒ Δ = - 17√6 - 11√11 + 20 - 18√12 which is real.

∴ Determinant of all the 3 statement are purely real.