Harrison Chapman

Linear Algebra I: Homework 1

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 $3 \ne 2$.

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   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.

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

   $$\begin{aligned} 5x - y + z + 6w &= 7 \\ 3x + 3y - 2z &= 1 \\ y - z &= 0 \\ x - y - z &= 0 \\ y &= 3 \end{aligned}$$

   $$\begin{pmatrix} 5 & -1 & 1 & 6 & 7 \\ 3 & 3 & -2 & 0 & 1 \\ 0 & 1 & -1 & 0 & 0 \\ 1 & -1 & -1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 3 \end{pmatrix}$$

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3. Draw a picture that describes a system of three linear equations that has no solution.

   There are many solutions. Here is one.

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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 $y = -5/4$.

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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 $f(2,2) = 16 \ne 2f(1,1) = 2(5) = 10$.

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