Exercise11

Define the matrix

\begin{equation*} A = \left[\begin{array}{rrr} 1 \amp 2 \amp 4 \\ -2 \amp 1 \amp -3 \\ 3 \amp 1 \amp 7 \\ \end{array}\right] \text{.} \end{equation*}
  1. Describe the solution space to the homogeneous equation \(A\xvec = \zerovec\text{.}\) What does this solution space represent geometrically?

  2. Describe the solution space to the equation \(\bvec = \threevec{-3}{-4}{1}\text{.}\) What does this solution space represent geometrically and how does it compare to the previous solution space?

  3. We will now explain the relationship between the previous two solution spaces. Suppose that \(\xvec_h\) is a solution to the homogeneous equation; that is \(A\xvec_h=\zerovec\text{.}\) We will also suppose that \(\xvec_p\) is a solution to the equation \(A\xvec = \bvec\text{;}\) that is, \(A\xvec_p=\bvec\text{.}\)

    Use the Linearity Principle expressed in Proposition 3 to explain why \(\xvec_h+\xvec_p\) is a solution to the equation \(A\xvec = \bvec\text{.}\) You may do this by evaluating \(A(\xvec_h+\xvec_p)\text{.}\)

    That is, if we find one solution \(\xvec_p\) to an equation \(A\xvec = \bvec\text{,}\) we may add any solution to the homogeneous equation to \(\xvec_p\) and still have a solution to the equation \(A\xvec = \bvec\text{.}\) In other words, the solution space to the equation \(A\xvec = \bvec\) is given by translating the solution space to the homogeneous equation by the vector \(\xvec_p\text{.}\)

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