Question

( 2 A-B=left[begin{array}{rrr}2 & -1 & 5 2 & -1 & 6 0 & 1 & 2end{array}right] . . ) is
in ( A+2 B=left[begin{array}{ccc}1 & 2 & 0 6 & -3 & 3 -5 & 3 & 1end{array}right] )
( therefore frac{2(i)+(11)}{5}=A=left[begin{array}{ccc}1 & 0 & 1 2 & -1 & 3 -1 & 1 & 1end{array}right] )
( frac{2|i|-(i)}{5}=B=left[begin{array}{ccc}0 & 61 & -1 2 & -1 & 0 -2 & 1 & 0end{array}right] )
( T_{r}(A)+operatorname{Tr}(B)=mid-1+1-1 )
( =0 )

# T12 07 12 -1 57 Let A +2B= 6 -3 3 and 24-R-2 -1 6 2-5 3 1 0 1 2 then Tr (A) - Tr (B) has the value is equal to (a) 0 (b) 1 (c) 2 (d) None of these

Solution