Linear systems

**Showcase_22** · November 12th 2008, 03:27 AM

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:

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

Hence the answer is d).
I have a nagging feling about that $\delta$ though....

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).

Can someone check these answers?

**HallsofIvy** · November 12th 2008, 07:29 AM

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.

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