Let T be a linear transformation from R2 into R2 such that T(1,0)=(1,1) and T(0,1)=(−1,1).

Concept: property of linear transformation from ${\displaystyle R^n\text{to}{R}^m}$ states that ${\displaystyle T(v1+v2)=T\left(v1\right)+T\left(v2\right)}$ where T is linear transformation and v1, v2 are vectors from ${\displaystyle R^n}$
${\displaystyle T(a\cdot V1)=aT\left(V1\right)}$ where a is scalar, V1 is vector
given ${\displaystyle T(1,0)=(1,1)}$
${\displaystyle T(0,1)=(-1,1)}$
${\displaystyle T(6,1)=?\text{and}T(-5,1)=?}$
${\displaystyle T(6,1)=T(6+0,0+1)=T((6,0)+(0,1))}$
using properties of linear transformation we get
${\displaystyle =T(6,0)+T(0,1)}$
${\displaystyle =T\left(6(1,0)\right)+T(0,1)}$
${\displaystyle =6T(1,0)+T(0,1)}$
${\displaystyle =6(1,1)+(-1,1)}$
${\displaystyle =(6,6)+(-1,1)}$
${\displaystyle =(6-1,6+1)=(5,7)}$
linear transformation of (6,1) is (5,7)
${\displaystyle T(-5,1)=T(-5+0,0+1)=T((-5,0)+(0,1))}$
using properties of linear transformation we get
${\displaystyle =T(-5,0)+T(0,1)}$
${\displaystyle =T(-5(1,0))+T(0,1)}$
${\displaystyle =-5T(1,0)+T(0,1)}$
${\displaystyle =-5(1,1)+(-1,1)}$
${\displaystyle =(-5,-5)+(-1,1)}$
${\displaystyle =(-5-1,-5+1)=(-6,-4)}$
linear transformation of (-5,1) is (-6,-4)