1.5 Exercises

20 points · 4 extra credit

Solution Sets of Linear Systems

Worked solutions are available for every problem.
  1. 1. (2 pts) Solve the homogeneous system and write its solution set in parametric vector form:

    x1−2x2+3x3=02x1−3x2+4x3=0.\begin{aligned} x_1 - 2x_2 + 3x_3 &= 0 \\ 2x_1 - 3x_2 + 4x_3 &= 0. \end{aligned}

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  2. 2. (3 pts) Find the solution set of Ax=0A\mathbf{x} = \mathbf{0} in parametric vector form, where

    A=[12−1324−181205].A = \begin{bmatrix} 1 & 2 & -1 & 3 \\ 2 & 4 & -1 & 8 \\ 1 & 2 & 0 & 5 \end{bmatrix}.

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  3. 3. (3 pts) Solve the system below and express its solution in the form x=p+x2v\mathbf{x} = \mathbf{p} + x_2\mathbf{v}:

    x1+2x2+x3=42x1+4x2+x3=5.\begin{aligned} x_1 + 2x_2 + x_3 &= 4 \\ 2x_1 + 4x_2 + x_3 &= 5. \end{aligned}

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  4. 4. (3 pts) For the system in Problem 3, write the solution set of the associated homogeneous system Ax=0A\mathbf{x} = \mathbf{0}, and describe geometrically how the two solution sets are related.

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  5. 5. (3 pts) State whether each of the following is true or false, and justify each answer.

    (a) A homogeneous system Ax=0A\mathbf{x} = \mathbf{0} is always consistent.

    (b) If Ax=0A\mathbf{x} = \mathbf{0} has only the trivial solution, then Ax=bA\mathbf{x} = \mathbf{b} has a unique solution for every b\mathbf{b}.

    (c) If Ax=bA\mathbf{x} = \mathbf{b} is consistent, then its solution set is a translate of the solution set of Ax=0A\mathbf{x} = \mathbf{0}.

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  6. 6. (3 pts) Describe geometrically, in R3\mathbb{R}^3, the two solution sets, and state how they differ:

    (a) x=t[21−1]\mathbf{x} = t\begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix}

    (b) x=[102]+t[21−1]\mathbf{x} = \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} + t\begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix}

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  7. 7. (3 pts) Prove that every homogeneous linear system with more unknowns than equations has a nontrivial solution.

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  8. 8. (4 pts ••) Suppose Ax=bA\mathbf{x} = \mathbf{b} is consistent, and let p\mathbf{p} be one solution. Prove that the solution set of Ax=bA\mathbf{x} = \mathbf{b} equals {p+vh:Avh=0}\{\mathbf{p} + \mathbf{v}_h : A\mathbf{v}_h = \mathbf{0}\}.

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