Question

A
=
⎡
⎢
⎣
1
2
−
3
i
3
+
4
i
2
+
3
i
0
4
−
5
i
3
−
4
i
4
+
5
i
2
⎤
⎥
⎦
A=[12−3i3+4i2+3i04−5i3−4i4+5i2]
then A is


      
symmetric matrix

      
skew - symmetric matrix

      
hermitian matrix

      
skew-hermitian matrix

Answer 1

Step-by-step explanation:

Given that:

$\left[\begin{array}{ccc}1&2-3i&3+4i\\2+3i&0&4-5i\\3-4i&4+5i&2\end{array}\right]$

To find:

A is

a) symmetric matrix

b) skew - symmetric matrix

c) hermitian matrix

d) skew-hermitian matrix

Solution:

A matrix is called hermitian matrix,if transpose of A and it's conjugate is equal to matrix itself.

$A=\left[\begin{array}{ccc}1&2-3i&3+4i\\2+3i&0&4-5i\\3-4i&4+5i&2\end{array}\right]...eq1$

$A^T=\left[\begin{array}{ccc}1&2+3i&3-4i\\2-3i&0&4+5i\\3+4i&4-5i&2\end{array}\right]$

Take conjugate of Complex numbers

if

$z = a + ib$

then its conjugate is

$z' = a - ib$

$A^{TC}=\left[\begin{array}{ccc}1&2-3i&3+4i\\2+3i&0&4-5i\\3-4i&4+5i&2\end{array}\right]...eq2$

From eq1 and eq2

it is clear that

$A=A^{TC}$

Thus,

Matrix A is Hermitian matrix.

Option C is correct.

Hope it helps you