Question

Given $2A^{\ast }+B^{\ast }=\left[\begin{array}{cc}-2i& -4i\\ -6i& -8i\end{array}\right]$ and all the elements of B are 2. Then matrix A is given by. ($\ast$ represents transpose conjugate operator)

• A. $\left[\begin{array}{cc}i& 2i\\ 3i& 4i\end{array}\right]$
• B. $\left[\begin{array}{cc}-1+i& -1+2i\\ -1+3i& -1+4i\end{array}\right]$
• C. $\left[\begin{array}{cc}2+2i& 2+4i\\ 2+6i& 2+8i\end{array}\right]$
• D. $\left[\begin{array}{cc}i-1& 3i-1\\ 2i-1& 4i-1\end{array}\right]$

Answer 1

The correct option is B

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

Given that $2A^{\ast }+B^{\ast }=\left[\begin{array}{cc}-2i& -4i\\ -6i& -8i\end{array}\right]andB=\left[\begin{array}{cc}2& 2\\ 2& 2\end{array}\right]$

Then $2A^{\ast }+2\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}-2i& -4i\\ -6i& -8i\end{array}\right]$

( SInce elements of B are all same and is equal to a real number $\to B^{\ast }=B$ )

$A^{\ast }=\left[\begin{array}{cc}-2i-1& -2i-1\\ -3i-1& -4i-1\end{array}\right]A=\therefore (A^{\ast }{)}^{\ast }=\left[\begin{array}{cc}i-1& 3i-1\\ 2i-1& 4i-1\end{array}\right]$

(This is a property. If conjugate transpose is done twice on a matrix you will get the original matrix)

Hence the option (b)