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Thread: Linear systems

  1. #1
    Super Member Showcase_22's Avatar
    Sep 2006
    The raggedy edge.

    Linear systems

    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}

    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....

    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).

    Can someone check these answers?
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  2. #2
    MHF Contributor

    Apr 2005
    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.
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