# Linear systems

• Nov 12th 2008, 03:27 AM
Showcase_22
Linear systems
Quote:

The linear system in x,y:

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

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

has a unique solution precisely when:

a). $\lambda=\pm \mu \delta$
b). $\mu \neq \delta$
c). $\lambda \mu =2\delta$
d). $\mu \neq \delta$
What I did:

$\begin{pmatrix}
{1+\lambda}&{-\mu}\\
{1-\lambda}&{\mu}
\end{pmatrix}\begin{pmatrix}
{x}\\
{y}
\end{pmatrix}=\begin{pmatrix}
{\delta}\\
{2}
\end{pmatrix}$

The determinant of the matrix cannot be zero so:

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

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

Quote:

Let $T_1, T_2, T _3 : \Re^3 \rightarrow \Re ^2$ be given by:

$T_1(x,y,z)=(x+1,y+z),$
$T_2(x,y,z)=(2x,y)$
$T_3(x,y,z)=(x^2,y+z)$

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

a). $T_1$
b). $T_2$
c). $T_3$
d). none of them
I know that for a transformation to be linear:

$T(x+y)=T(x)+T(y)$
$T(kx)=kT(x)$

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

A matrix equation Ax= b has a unique solution if and only if det(A) is not 0. That has nothing to do with the right hand side and so nothing to do with $\delta$. The right hand side, and $\delta$, would be important if they asked whether there were no solutions or an infinite number of solutions.