Linear systems

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

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). $\displaystyle \mu \neq \delta$
What I did:

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

The determinant of the matrix cannot be zero so:

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

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

Quote:

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
I know that for a transformation to be linear:

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

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

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 $\displaystyle \delta$. The right hand side, and $\displaystyle \delta$, would be important if they asked whether there were no solutions or an infinite number of solutions.