MA 351 Fall 2026 (Aaron N. K. Yip)
Homework 4 Due: Thursday, Sept. 24, in class

Penney, Linear Algebra: Ideas and Applications (4th edition)

p.17 EXERCISES: 1.3, 1.5, 1.9, 1.10, 1.24(a,b,e,f,g), 1.25, 1.26, 1.27, 1.33

Additional Problem.

Find the "simplest" linear system such that its solution is given by the following:

\[ \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix} + \alpha \begin{pmatrix} -2 \\ 1 \\ 0 \\ 0 \end{pmatrix} + \beta \begin{pmatrix} -3 \\ 0 \\ 2 \\ 1 \end{pmatrix}, \quad \alpha, \beta \text{ are free variables.} \]

(Hint: try a \( 2 \times 4 \) matrix that is already in Reduced Row Echelon Form. Is the system homogeneous or inhomogeneous?)

Remarks/Comments/Hints for #1.3, #1.5, #1.26, #1.27, #1.33.

You can first try to find constants \( c_1, c_2, c_3, \ldots \) such that \( c_1 u_1 + c_2 u_2 + c_3 u_3 + \cdots = \mathbf{0} \)! Sometimes you can find them by guessing. Otherwise, set up some system of linear equations. Then you can express one of the vectors as a linear combination of the rest.

  1. #1.3(a): \( -A - B + C = 0 \). Hence,

    \[ A = -B + C, \text{ or } B = -A + C, \text{ or } C = A + B. \]

  2. #1.5(a): \( -\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} - 2 \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} + \begin{pmatrix} 1 \\ 1 \\ 4 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \). Hence

    \[ \begin{pmatrix} 1 \\ 1 \\ 4 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} + 2 \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}, \text{ or } \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} = -2 \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} + \begin{pmatrix} 1 \\ 1 \\ 4 \end{pmatrix}, \text{ and so forth.} \]

  3. #1.33(a): \( \boxed{\tfrac{1}{3}}(3\sin^2 x) + \boxed{-\tfrac{1}{5}}(-5\cos^2 x) + \boxed{-\tfrac{1}{119}}(119) \equiv 0 \). Hence

    \[ \boxed{1}(3\sin^2 x) \equiv \boxed{\tfrac{3}{5}}(-5\cos^2 x) + \boxed{\tfrac{3}{119}}(119) \text{ and so forth.} \]

    (The \( \square \) is just for clarity. It is not necessary that you use \( \square \) in your homework.)

  4. #1.33(g): \( \boxed{1}(x^2 + 3x + 3) + \boxed{-3}(x + 1) + \boxed{-\tfrac{1}{2}}(2x^2) \equiv 0 \). Hence

    \[ (x^2 + 3x + 3 \equiv 3(x + 1) + \frac{1}{2}(2x^2), \text{ and so forth.} \]