What I did:The linear system in x,y:

$\displaystyle (1+\lambda)x-\mu y=\delta$

$\displaystyle (1-\lambda)x+ \mu y=2$

has a unique solution precisely when:

a). $\displaystyle \lambda=\pm \mu \delta$

b). $\displaystyle \mu \neq \delta$

c). $\displaystyle \lambda \mu =2\delta$

d). $\displaystyle \mu \neq \delta$

$\displaystyle \begin{pmatrix}

{1+\lambda}&{-\mu}\\

{1-\lambda}&{\mu}

\end{pmatrix}\begin{pmatrix}

{x}\\

{y}

\end{pmatrix}=\begin{pmatrix}

{\delta}\\

{2}

\end{pmatrix}$

The determinant of the matrix cannot be zero so:

$\displaystyle \mu+\mu \lambda+\mu -\mu \lambda=2 \mu$

Hence the answer is d).

I have a nagging feling about that $\displaystyle \delta$ though....

I know that for a transformation to be linear:Let $\displaystyle T_1, T_2, T _3 : \Re^3 \rightarrow \Re ^2$ be given by:

$\displaystyle T_1(x,y,z)=(x+1,y+z),$

$\displaystyle T_2(x,y,z)=(2x,y)$

$\displaystyle T_3(x,y,z)=(x^2,y+z)$

which of $\displaystyle T_1,T_2$ and $\displaystyle T_3$ is a linear transformation?

a). $\displaystyle T_1$

b). $\displaystyle T_2$

c). $\displaystyle T_3$

d). none of them

$\displaystyle T(x+y)=T(x)+T(y)$

$\displaystyle T(kx)=kT(x)$

applying these, $\displaystyle T_2$ is the only linear transformation so the answer is b).

Can someone check these answers?