Question

Let $A,B,C$ and $D$ are square matrices of order $2$, such that $A{D}^T=B{C}^T+I$ and $D{A}^T=I+C{B}^T$. If $A{B}^T$ and $C{D}^T$ are symmetric, then which of the following(s) is/are always true ?

(where $T$ denotes the transpose of matrix and $I$ be the identity matix)

• A. $D^{T}A-B^{T}C=I$
• B. $A^{T}D-C^{T}B=I$
• C. $A^{T}C$ is symmetric
• D. $B^{T}D$ is symmetric

Answer 1

Answer: (A), (B), (C), (D)

$B^{T}D$ is symmetric

Given $A{D}^T=B{C}^T+I$
$\Rightarrow A{D}^T-B{C}^T=I...\left(1\right)$
$\Rightarrow D{A}^T-C{B}^T=I...\left(2\right)$
And $A{B}^T$ and $C{D}^T$ are symmetric
$\Rightarrow A{B}^T=B{A}^T$
$\Rightarrow A{B}^T-B{A}^T=O...\left(3\right)$
Similarly,
$C{D}^T-D{C}^T=O...\left(4\right)$

Now, write these 4 equatins in matrix form
$\left[\begin{array}{cc}A{D}^T-B{C}^T& A{B}^T-B{A}^T\\ C{D}^T-D{C}^T& D{A}^T-C{B}^T\end{array}\right]=\left[\begin{array}{cc}I& O\\ O& I\end{array}\right]$

Now, split the L.H.S. of matrix as product of two matrix , we get

$\left[\begin{array}{cc}A& B\\ C& D\end{array}\right]\left[\begin{array}{cc}{D}^T& -B^T\\ -C^T& A^T\end{array}\right]=\left[\begin{array}{cc}A{D}^T-B{C}^T& -A{B}^T+B{A}^T\\ C{D}^T-D{C}^T& -C{B}^T+D{A}^T\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}A& B\\ C& D\end{array}\right]\left[\begin{array}{cc}{D}^T& -B^T\\ -C^T& A^T\end{array}\right]=\left[\begin{array}{cc}I& O\\ O& I\end{array}\right]$

Since, $\text{R.H.S}$ is the identity matrix of order $4$. Therefore, the above equation can be re-written as
$\left[\begin{array}{cc}{D}^T& -B^T\\ -C^T& A^T\end{array}\right]\left[\begin{array}{cc}A& B\\ C& D\end{array}\right]=\left[\begin{array}{cc}I& O\\ O& I\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}{D}^{T}A-B^{T}C& D^{T}B-B^{T}D\\ -C^{T}A+A^{T}C& -C^{T}B+A^{T}D\end{array}\right]=\left[\begin{array}{cc}I& O\\ O& I\end{array}\right]$

$\Rightarrow D^{T}A-B^{T}C=I$
$\Rightarrow A^{T}D-C^{T}B=I$
$\Rightarrow D^{T}B=B^{T}D$
$\Rightarrow C^{T}A=A^{T}C$