If A= [ ccc -1 & 2 & 1 -3 & 2 & -3 ] and B= [ cc 2 & 1 -3 & 2 -1 & 3 ], prove that (A+B^ )^ =A^ +B. prove that (A+B^ )^ =A^ +B

A= [ ccc -1 & 2 & 1 -3 & 2 & -3 ] and B= [ cc 2 & 1 -3 & 2 -1 & 3 ] A^T= [ cc -1 & -3 2 & 2 1 & -3 ] and B^T= [ ccc 2 & -3 & -1 1 & 2 & 3 ] A+B^T= [ ccc -1 & 2 & 1 -3 & 2 & -3 ]+ [ ccc 2 & -3 & -1 1 & 2 & 3 ] = [ ccc -1+2 & 2-3 & 1-1 -3+1 & 2+2 & -3+3 ]= [ ccc 1 & -1 & 0 -2 & 4 & 0 ] (A+B^T )^T= [ cc 1 & -2 -1 & 4 0 & 0 ] ..(i) Now, A^ T +B= [ cc -1 & -3 2 & 2 1 & -3 ]+ [ cc 2 & 1 -3 & 2 -1 & 3 ] = [ cc -1+2 & -3+1 2-3 & 2+2 1-1 & -3+3 ] = [ cc 1 & -2 -1 & 4 0 & 0 ] ...(ii)

If A= [ ccc -1 & 2 & 1 -3 & 2 & -3 ] and B= [ cc 2 & 1 -3 & 2 -1 …

Download App

Question

If $A=\left[\begin{array}{ccc}-1 & 2 & 1 \\ -3 & 2 & -3\end{array}\right]$ and $B=\left[\begin{array}{cc}2 & 1 \\ -3 & 2 \\ -1 & 3\end{array}\right]$, prove that $\left(A+B^{\top}\right)^{\top}=A^{\top}+B$.

prove that $\left(A+B^{\top}\right)^{\top}=A^{\top}+B$

✓

Answer

$A=\left[\begin{array}{ccc}-1 & 2 & 1 \\ -3 & 2 & -3\end{array}\right]$ and $B=\left[\begin{array}{cc}2 & 1 \\ -3 & 2 \\ -1 & 3\end{array}\right]$

$\therefore \quad A^T=\left[\begin{array}{cc}-1 & -3 \\ 2 & 2 \\ 1 & -3\end{array}\right]$ and $B^T=\left[\begin{array}{ccc}2 & -3 & -1 \\ 1 & 2 & 3\end{array}\right]$

$\therefore \quad A+B^T=\left[\begin{array}{ccc}-1 & 2 & 1 \\ -3 & 2 & -3\end{array}\right]+\left[\begin{array}{ccc}2 & -3 & -1 \\ 1 & 2 & 3\end{array}\right]$

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

$\therefore \quad\left(A+B^T\right)^T=\left[\begin{array}{cc}1 & -2 \\ -1 & 4 \\ 0 & 0\end{array}\right]$

$\ldots$..(i)

Now, $A^{\mathrm{T}}+\mathrm{B}=\left[\begin{array}{cc}-1 & -3 \\ 2 & 2 \\ 1 & -3\end{array}\right]+\left[\begin{array}{cc}2 & 1 \\ -3 & 2 \\ -1 & 3\end{array}\right]$

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

$=\left[\begin{array}{cc}1 & -2 \\ -1 & 4 \\ 0 & 0\end{array}\right]$

...(ii)