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Exercise 11

Avg difficulty 2.5

Define the matrix A=[124−21−3317]A = \left[\begin{array}{rrr} 1 & 2 & 4 \\ -2 & 1 & -3 \\ 3 & 1 & 7 \\ \end{array}\right].

  1. Describe the solution space to the homogeneous equation Ax=0A\xvec = \zerovec using a parametric description, if appropriate. What does this solution space represent geometrically?

  2. Describe the solution space to the equation Ax=bA\xvec=\bvec where b=[−3−41]\bvec = \threevec{-3}{-4}{1}. 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 xh\xvec_h is a solution to the homogeneous equation; that is Axh=0A\xvec_h=\zerovec. Suppose also that xp\xvec_p is a solution to the equation Ax=bA\xvec = \bvec; that is, Axp=bA\xvec_p=\bvec.

    Use the Linearity Principle expressed in Linearity of matrix multiplication to explain why xh+xp\xvec_h+\xvec_p is a solution to the equation Ax=bA\xvec = \bvec. You may do this by evaluating A(xh+xp)A(\xvec_h+\xvec_p).

    That is, if we find one solution xp\xvec_p to an equation Ax=bA\xvec = \bvec, we may add any solution to the homogeneous equation to xp\xvec_p and still have a solution to the equation Ax=bA\xvec = \bvec. In other words, the solution space to the equation Ax=bA\xvec = \bvec is given by translating the solution space to the homogeneous equation by the vector xp\xvec_p.

Austin, Understanding Linear Algebra, §2.2 Matrix multiplication and linear combinations, Exercise 11

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Adapted from Understanding Linear Algebra © 2023 David Austin. Licensed under CC BY 4.0. Source: Understanding Linear Algebra. Imported at revision 3e6dced.