Answer to Question #150754 in Linear Algebra for Ashweta Padhan

Question

Let T: R^2 to R^3 be defined by (a, b)=(a+b, a-2b, 3a+b). Show that T is nonsingular. Hence find T inverse.

Answer 1

$T(a,b)=(a+b,a-2b,3a+b)$

Let $T(a,b)=0\\$ then (a+b,a-2b,3a+b)=0

$\Rightarrow a=0, b=0$

Therefore $(a,b)=(0,0).$

Hence T is nonsingular.

Therefore T$^{-1}$ exists from $R^{3}$ to $R^{2}$.

Now let, $(a+b,a-2b,3a+b)=(x,y,z)$

$\Rightarrow$ a+b=x

a-2b=y

3a+b=z

Now equating these equations one can get $a=\dfrac{2z+y}{7}, b=\dfrac{z-3y}{7}$

Therefore the inverse of T is $T^{-1}: R^{3}\rightarrow R^{2}\\ T^{-1}(x,y,z)=(\dfrac{2z+y}{7}, \dfrac{z-3y}{7})$