Linear Algebra I: Homework 1

Due: Friday, January 26, 2018
  1. For the following parts, answer whether the given tuple describes a solution to the system of linear equations. If it is not a solution, explain why!

    1. For the system,

      \[\begin{aligned} 2x + y + 2z &= 3 \\ 3x + y - z &= 2 \end{aligned}\]

      is \((1, -1, 1)\) a solution?

      No. Plugging the point into the second equation yields \(3+1-1=2\) which is false, since \(3 \ne 2\).

    2. For the system,

      \[\begin{aligned} 5x - y + z &= 7 \\ 3x + y - z &= 1 \end{aligned}\]

      is \((1, s-2, s)\) a solution (for every real number \(s\))?

      Yes. Plugging the point into the first equation yields \(7=7\) and the second equation yields \(1=1\), both of which are true.

  2. Rewrite the following system of equations as an augmented matrix. You do not actually have to solve the system.

    \[\begin{aligned} 5x - y + z + 6w &= 7 \\ 3x + 3y - 2z &= 1 \\ y - z &= 0 \\ x - y - z &= 0 \\ y &= 3 \end{aligned}\]
    \[\begin{pmatrix} 5 & -1 & 1 & 6 & 7 \\ 3 & 3 & -2 & 0 & 1 \\ 0 & 1 & -1 & 0 & 0 \\ 1 & -1 & -1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 3 \end{pmatrix}\]
  3. Draw a picture that describes a system of three linear equations that has no solution.

    There are many solutions. Here is one.

  4. Find all solutions to the following system of linear equations:

    \[\begin{aligned} 3x + y &= 1 \\ x - y &= 2 \end{aligned}\]

    As an augmented matrix, the system reduces to:

    \[\begin{pmatrix} 1 & 0 & 3/4 \\ 0 & 1 & -5/4 \\ \end{pmatrix}\]

    So, \(x = 3/4\) and \(y = -5/4\).

  5. For the following parts, answer whether the function described is linear or not.

    1.  \(f(x) = 3x + 1\)

    2.  \(f(x_1, x_2) = 3x_1 - 6x_2\)

    3.  \(f(x_1, x_2, x_3) = 3x_1 + x_3\)

    4.  \(f(x, y) = 3xy + x + y\)

    Originally, we had to solve this using the form of a linear function. Now, we can use the definition.

    1. No, as \(f(2) = 7 \ne 2f(1) = 2(4)=8\).

    2. Yes, as \(f(a+b,c+d) = 3(a+b)-6(c+d) = (3a-6c)+(3b-6d) = f(a,c)+f(b,d)\).

    3. Yes, as \(f(a+b,c+d,x+y) = 3(a+b)+(x+y) = (3a+x) + (3b+y) = f(a,c,x) + f(b,d,y)\).

    4. No, as \(f(2,2) = 16 \ne 2f(1,1) = 2(5) = 10\).