# Subtract 11011001 from (00100011 + 00001101) using 8-bit signed magnitude arithmetic

Subtract using 8-bit signed magnitude arithmetic 11011001 from (00100011 + 00001101 ) The result also should be signed magnitude format. I do it like this

(+0 0001101 ) + (+0 0100011)=+0 0110000

then

(+0 0110000)- (-1 1011001)= -1 0001010

is this correct??

Doesn't the first bit itself want to be the sign? (If the leading bit is $1$, it means the number wants to be negative.)
Note that $11111111$ plays the role of $-1$, and similarly, for any byte $x$ in binary, $$-x = (x\ {\rm xor}\ 11111111)+1$$ Using this, $-1 1011001 = (00100110)+1 = 00100111$. Hope it helps.
The leading bit being one means the number is negative. Since we are to subtract it, we can change the leading bit to $0$ and add all three, getting $$\ \ \ 0001101\\ \ \ \ 0100011\\ \underline{+1011001}\\ 10001001$$ which is an overflow. In base $10$ this is $13+35+89=137$. As the largest positive number in signed 8-bit is $+127$ this is consistent.