MA 351: Introduction to Linear Algebra and Its Applications Fall 2024, Midterm One

Instructor: Yip

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Name: Answer Key (Major: )

Question and Score
Question Score
1.(20 pts)
2.(20 pts)
3.(20 pts)
4.(20 pts)
5.(20 pts)
Total (100 pts)
  1. Consider the equation \( 2x - 3y = 5 \).

    1. Write down the solution of the above equation in parametric form as \( X = p + \mathrm{Null}(A) \) using some vector \( p \) and null space of some matrix \( A \). You should write \( \mathrm{Null}(A) \) as a span of vector(s). Give two different examples of \( p \).
    2. For each of the two forms of solution you have found, plot your solution in the following \( xy \)-plane. Clearly indicate the vector \( p \) and \( \mathrm{Null}(A) \).
    Graph of line 2x minus 3y equals 5 with vectors p1 and p2, dashed Null(A) line, and shifted Null(A).
    Figure 1
    A hand-annotated xy-coordinate grid with axes labeled x (horizontal) and y (vertical), each marked from about -3 to 3. A red dashed line through the origin with positive slope is labeled Null(A) in blue. A solid red line parallel to it, shifted down and to the right, passes through the points \( (\frac{5}{2}, 0) \) and \( (0, -\frac{5}{3}) \); it is labeled in red as \( 2x - 3y = 5 \) and, below the graph, as shifted Null(A). A red arrow from the origin along the positive x-axis to \( (\frac{5}{2}, 0) \) is labeled \( p_1 \) with the column vector \( \begin{pmatrix} \frac{5}{2} \\ 0 \end{pmatrix} \). A second red arrow from the origin straight down the negative y-axis to \( (0, -\frac{5}{3}) \) is labeled \( p_2 \) with the column vector \( \begin{pmatrix} 0 \\ -\frac{5}{3} \end{pmatrix} \). Red arrowheads along the solid line indicate the direction of the solution line through both endpoints.
    \[ 2x - 3y = 5 \iff \left[\begin{array}{cc|c} 2 & -3 & 5 \end{array}\right] \]\[ \begin{gathered} y = \alpha \ \text{free} \\[0.5em] x = \frac{5 + 3\alpha}{2} \end{gathered} \]

    (1)

    \[ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \frac{5}{2} + \frac{3\alpha}{2} \\ \alpha \end{pmatrix} = \underbrace{\begin{pmatrix} \frac{5}{2} \\ 0 \end{pmatrix}}_{p_1} + \underbrace{\alpha \begin{pmatrix} \frac{3}{2} \\ 1 \end{pmatrix}}_{\mathrm{Null}\left(\left[\begin{array}{cc} 2 & -3 \end{array}\right]\right)} \]\[ A = \left[\begin{array}{cc} 2 & -3 \end{array}\right] \]

You can use this blank page.

(2) Alternatively, \( x = \alpha \ \text{(free)} \), \( y = \frac{2\alpha - 5}{3} \)

\[ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \alpha \\ -\frac{5}{3} + \frac{2\alpha}{3} \end{pmatrix} = \underbrace{\begin{pmatrix} 0 \\ -\frac{5}{3} \end{pmatrix}}_{p_2} + \underbrace{\alpha \begin{pmatrix} 1 \\ \frac{2}{3} \end{pmatrix}}_{\mathrm{Null}\left(\left[\begin{array}{cc} 2 & -3 \end{array}\right]\right)} \]

Note:

(a) \( p \) can be any pt. on the line \( 2x - 3y = 5 \)

eg

\[ p = \begin{pmatrix} 1 \\ -1 \end{pmatrix}, \begin{pmatrix} 4 \\ 1 \end{pmatrix}, \cdots \]

(b)

\[ \mathrm{Null}\left(\left[\begin{array}{cc} 2 & -3 \end{array}\right]\right) \]\[ = \mathrm{Span}\left\{ \begin{pmatrix} \frac{3}{2} \\ 1 \end{pmatrix} \right\} = \mathrm{Span}\left\{ \begin{pmatrix} 1 \\ \frac{2}{3} \end{pmatrix} \right\} = \mathrm{Span}\left\{ \begin{pmatrix} 3 \\ 2 \end{pmatrix} \right\} \]
  1. You are given the following list of polynomials from \( \mathbb{P}_2 \), the space of polynomials of degree at most 2, \( \mathbb{P}_2 = \{p(x) = a + bx + cx^2 : a, b, c \in \mathbb{R}\} \):

    \[ \begin{array}{rcl} p_1(x) &=& 2 - x + 2x^2, \\ p_2(x) &=& 3 \quad\ - x^2, \\ p_3(x) &=& -1 + 2x - 5x^2, \\ p_4(x) &=& 19 - 8x + 15x^2, \\ p_5(x) &=& 2 + 5x - 14x^2. \end{array} \]

    Determine if the above list is linearly independent. If not, throw away all the redundant vector(s) until you get a linearly independent list. For each vector you throw away, write it as a linear combination of the vectors from the linearly independent list you have found.

    \[ \left. \begin{array}{l} c_1\left(2 - x + 2x^2\right) \\ + c_2\left(3 \quad\ - x^2\right) \\ + c_3\left(-1 + 2x - 5x^2\right) \\ + c_4\left(19 - 8x + 15x^2\right) \\ + c_5\left(2 + 5x - 14x^2\right) \\ = 0 \end{array} \right\} \Longrightarrow \begin{array}{l} 2c_1 + 3c_2 - c_3 + 19c_4 + 2c_5 = 0 \\ -c_1 \qquad\quad + 2c_3 - 8c_4 + 5c_5 = 0 \\ 2c_1 - c_2 - 5c_3 + 15c_4 - 14c_5 = 0 \end{array} \]\[ \Longrightarrow \left(\begin{array}{ccccc|c} 2 & 3 & -1 & 19 & 2 & 0 \\ -1 & 0 & 2 & -8 & 5 & 0 \\ 2 & -1 & -5 & 15 & -14 & 0 \end{array}\right) \]\[ \longrightarrow \left(\begin{array}{ccccc|c} 1 & 0 & -2 & 8 & -5 & 0 \\ 2 & 3 & -1 & 19 & 2 & 0 \\ 2 & -1 & -5 & 15 & -14 & 0 \end{array}\right) \]\[ \longrightarrow \left(\begin{array}{ccccc|c} 1 & 0 & -2 & 8 & -5 & 0 \\ 0 & 3 & 3 & 3 & 12 & 0 \\ 0 & -1 & -1 & -1 & -4 & 0 \end{array}\right) \]

You can use this blank page.

\[ \longrightarrow \left(\begin{array}{ccccc|c} 1 & 0 & -2 & 8 & -5 & 0 \\ 0 & 1 & 1 & 1 & 4 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right) \] \[ c_3 = \alpha \quad c_4 = \beta \quad c_5 = \gamma \]

free variables \( \Longrightarrow (p_1, p_2, p_3, p_4, p_5) \) lin dep.

\[ \begin{gathered} c_1 = 2\alpha - 8\beta + 5\gamma, \qquad c_2 = -\alpha - \beta - 4\gamma \\[0.5em] \left(2\alpha - 8\beta + 5\gamma\right) p_1 + \left(-\alpha - \beta - 4\gamma\right) p_2 + \alpha p_3 + \beta p_4 + \gamma p_5 = 0 \\[0.5em] \alpha p_3 + \beta p_4 + \gamma p_5 = \left(-2\alpha + 8\beta - 5\gamma\right) p_1 + \left(\alpha + \beta + 4\gamma\right) p_2 \\[0.5em] \alpha = 1, \ \beta = 0, \ \gamma = 0 \Longrightarrow p_3 = -2p_1 + p_2 \\[0.5em] \alpha = 0, \ \beta = 1, \ \gamma = 0 \Longrightarrow p_4 = 8p_1 + p_2 \\[0.5em] \alpha = 0, \ \beta = 0, \ \gamma = 1 \Longrightarrow p_5 = -5p_1 + 4p_2 \end{gathered} \]

\( p_3, p_4, p_5 \) — redundant, can be written as lin. comb. of \( p_1 \) and \( p_2 \).

  1. You are given a linearly independent list \( \mathcal{A} = \{X_1, X_2, X_3\} \) from some vector space. Consider the new list \( \mathcal{B} = \{Y_1, Y_2, Y_3\} \) where

    \[ \begin{array}{rcl} Y_1 &=& X_1 - X_2 + X_3, \\ Y_2 &=& 2X_1 + X_2, \\ Y_3 &=& X_1 + X_2 - X_3. \end{array} \]

    Is \( \mathcal{B} \) linearly independent? If so, write any linear combination of the \( X_i \)'s as a linear combination of the \( Y_i \)'s. If not, determine which of the \( Y_i \)'s are redundant.

    \( c_1 Y_1 + c_2 Y_2 + c_3 Y_3 = 0 \)

    \[ \begin{gathered} c_1\left(X_1 - X_2 + X_3\right) + c_2\left(2X_1 + X_2\right) + c_3\left(X_1 + X_2 - X_3\right) = \vec{0} \\[0.5em] \left(c_1 + 2c_2 + c_3\right) X_1 + \left(-c_1 + c_2 + c_3\right) X_2 + \left(c_1 - c_3\right) X_3 = 0 \end{gathered} \]

    lin ind, given

    \[ \Longrightarrow \left\{ \begin{array}{l} c_1 + 2c_2 + c_3 = 0 \\ -c_1 + c_2 + c_3 = 0 \\ c_1 \qquad - c_3 = 0 \end{array} \right. \Longrightarrow \left(\begin{array}{ccc|c} 1 & 2 & 1 & 0 \\ -1 & 1 & 1 & 0 \\ 1 & 0 & -1 & 0 \end{array}\right) \] \[ \longrightarrow \left(\begin{array}{ccc|c} 1 & 2 & 1 & 0 \\ 0 & 3 & 2 & 0 \\ 0 & -2 & -2 & 0 \end{array}\right) \longrightarrow \left(\begin{array}{ccc|c} 1 & 2 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 3 & 2 & 0 \end{array}\right) \] \[ \longrightarrow \left(\begin{array}{ccc|c} 1 & 2 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & -1 & 0 \end{array}\right) \]

    No free var.

    \[ c_1 = 0, \ c_2 = 0, \ c_3 = 0 \]

    \( Y_1, Y_2, Y_3 \) lin ind.

Given \( \alpha X_1 + \beta X_2 + \gamma X_3 \), an arbitrary lin comb. of \( X_1, X_2, X_3 \).

find \( c_1, c_2, c_3 \) s.t.

\[ \begin{gathered} c_1 Y_1 + c_2 Y_2 + c_3 Y_3 = \alpha X_1 + \beta X_2 + \gamma X_3 \\[0.5em] c_1\left(X_1 - X_2 + X_3\right) + c_2\left(2X_1 + X_2\right) + c_3\left(X_1 + X_2 - X_3\right) = \alpha X_1 + \beta X_2 + \gamma X_3 \end{gathered} \] \[ \left\{ \begin{array}{l} c_1 + 2c_2 + c_3 = \alpha \\ -c_1 + c_2 + c_3 = \beta \\ c_1 \qquad - c_3 = \gamma \end{array} \right. \Longrightarrow \left(\begin{array}{ccc|c} 1 & 2 & 1 & \alpha \\ -1 & 1 & 1 & \beta \\ 1 & 0 & -1 & \gamma \end{array}\right) \] \[ \longrightarrow \left(\begin{array}{ccc|c} 1 & 2 & 1 & \alpha \\ 0 & 3 & 2 & \alpha + \beta \\ 0 & -2 & -2 & \gamma - \alpha \end{array}\right) \longrightarrow \left(\begin{array}{ccc|c} 1 & 2 & 1 & \alpha \\ 0 & 1 & 1 & \frac{\alpha}{2} - \frac{\gamma}{2} \\ 0 & 3 & 2 & \alpha + \beta \end{array}\right) \] \[ \longrightarrow \left(\begin{array}{ccc|c} 1 & 2 & 1 & \alpha \\ 0 & 1 & 1 & \frac{\alpha}{2} - \frac{\gamma}{2} \\ 0 & 0 & -1 & -\frac{\alpha}{2} + \beta + \frac{3\gamma}{2} \end{array}\right) \longrightarrow \left(\begin{array}{ccc|c} 1 & 2 & 0 & \frac{\alpha}{2} + \beta + \frac{3\gamma}{2} \\ 0 & 1 & 0 & \beta + \gamma \\ 0 & 0 & -1 & -\frac{\alpha}{2} + \beta + \frac{3\gamma}{2} \end{array}\right) \] \[ \longrightarrow \left(\begin{array}{ccc|c} 1 & 0 & 0 & \frac{\alpha}{2} - \beta - \frac{\gamma}{2} \\ 0 & 1 & 0 & \beta + \gamma \\ 0 & 0 & 1 & \frac{\alpha}{2} - \beta - \frac{3\gamma}{2} \end{array}\right) \begin{array}{l} \leftarrow c_1 \\ \leftarrow c_2 \\ \leftarrow c_3 \end{array} \] \[ \alpha X_1 + \beta X_2 + \gamma X_3 = \left(\frac{\alpha}{2} - \beta - \frac{\gamma}{2}\right) Y_1 + \left(\beta + \gamma\right) Y_2 + \left(\frac{\alpha}{2} - \beta - \frac{3\gamma}{2}\right) Y_3 \]
  1. Consider a region with two main economies: M (machinery) and S (service). From historical data, it is found out that:

    1. producing 1 unit of M requires 0.1 unit of M and, 0.2 unit of S;
    2. producing 1 unit of S requires 0.1 unit of M and, 0.3 unit of S.

    Suppose the total outside demands of M and S are 40 and 50 units, respectively. Determine the actual units of M and S produced in order to satisfy the above demands.

    Flow diagram with nodes M and S, self loops 0.1 and 0.3, arrows 0.2 and 0.1, outputs 40 and 50.
    Figure 2
    A hand-drawn input-output flow diagram. Two circled nodes are drawn side by side: node M on the left, labeled in red with the variable \( x \) beneath the letter M, and node S on the right, labeled in red with the variable \( y \) beneath the letter S. A curved self-loop arrow above node M is labeled 0.1, and a curved self-loop arrow above node S is labeled 0.3. A long curved arrow runs from S leftward into M, labeled 0.2 above it. Another long curved arrow runs from M rightward into S, labeled 0.1 below it. An arrow points left out of node M toward the number 40 written at the far left, representing the outside demand for machinery. An arrow points right out of node S toward the number 50 at the far right, representing the outside demand for service.
    \[ \begin{gathered} x = 40 + 0.1x + 0.1y \\[0.5em] y = 50 + 0.2x + 0.3y \end{gathered} \] \[ \Longleftrightarrow \left\{ \begin{array}{l} 0.9x - 0.1y = 40 \\ -0.2x + 0.7y = 50 \end{array} \right. \Longleftrightarrow \begin{array}{l} 9x - y = 400 \\ -2x + 7y = 500 \end{array} \] \[ \begin{gathered} \left(7 \times 9 - 2\right) x = 7 \times 400 + 500 \implies x = \frac{3300}{61} \approx 54.10 \quad \text{M} \\[0.5em] y = 9x - 400 = \frac{9(3300)}{61} - 400 = \frac{29700 - 24400}{61} = \frac{5300}{61} \approx 86.89 \quad \text{S} \end{gathered} \]
  1. (In class we talked about the property of linear dependence as the presence of some redundant vectors. The following is another equivalent definition in terms of uniqueness of representations.) Given a list \( \mathcal{L} = \{u_1, u_2, \ldots u_n\} \) of vectors from some vector space. Prove the following two statements about \( \mathcal{L} \).

    1. Suppose you have two different, unequal sets of numbers \( \{\alpha_1, \alpha_2, \ldots \alpha_n\} \) and \( \{\beta_1, \beta_2, \ldots \beta_n\} \) such that

      \[ \alpha_1 u_1 + \alpha_2 u_2 + \cdots \alpha_n u_n = \beta_1 u_1 + \beta_2 u_2 + \cdots \beta_n u_n. \]

      Then \( \mathcal{L} \) is linearly dependent.

    2. Suppose \( \mathcal{L} \) is linearly dependent. Let \( v \) be a vector from \( \operatorname{Span}\{\mathcal{L}\} \) written as \( v = \alpha_1 u_1 + \alpha_2 u_2 + \cdots \alpha_n u_n \). Show that there is another, different, representation of \( v \) as a linear combination of the \( u_i \)'s.

      How many different representations of \( v \) can you find? How do they all look like?

    Note: parts (a) and (b) above are two separate statements.

    (a)

    \[ \begin{gathered} \alpha_1 u_1 + \alpha_2 u_2 + \cdots + \alpha_n u_n = \beta_1 u_1 + \beta_2 u_2 + \cdots + \beta_n u_n \\[0.5em] (\alpha_1 - \beta_1) u_1 + (\alpha_2 - \beta_2) u_2 + \cdots + (\alpha_n - \beta_n) u_n = 0 \\[0.5em] \underbrace{(\alpha_1 - \beta_1)}_{C_1} \quad \underbrace{(\alpha_2 - \beta_2)}_{C_2} \quad \underbrace{(\alpha_n - \beta_n)}_{C_n} \end{gathered} \]

    \( C_i \)'s not all zero as \( \{\alpha_i\} \ne \{\beta_i\} \)

    Hence \( \{u_i\} \) — lin dep.

    (b) Since \( \{u_i\} \) is lin dep, there are \( \{C_i\} \) not all zeros s.t.

    \[ C_1 u_1 + C_2 u_2 + \cdots + C_n u_n = 0 \]

You can use this blank page.

Let (1)

\[ \begin{gathered} v = \alpha_1 u_1 + \alpha_2 u_2 + \cdots + \alpha_n u_n \\[0.5em] = \alpha_1 u_1 + \alpha_2 u_2 + \cdots + \alpha_n u_n + \underbrace{\left( C_1 u_1 + C_2 u_2 + \cdots + C_n u_n \right)}_{= 0} \\[0.5em] = (\alpha_1 + C_1) u_1 + (\alpha_2 + C_2) u_2 + \cdots + (\alpha_n + C_n) u_n \end{gathered} \]

a new representation of \( v \)

( \( \{\alpha_i\} \ne \{\alpha_i + C_i\} \) )

Let \( v = \beta_1 u_1 + \beta_2 u_2 + \cdots + \beta_n u_n \) (2) be any other representation of \( V \).

(2) − (1) \( \Rightarrow \)

\[ \begin{gathered} 0 = \underbrace{(\beta_1 - a_1)}_{\sim C_1} v_1 + \cdots + \underbrace{(\beta_n - a_n)}_{\sim C_n} v_n \\[0.5em] \beta_1 - a_1 = t C_1, \ldots, \beta_n - a_n = t C_n \end{gathered} \]

\( \Rightarrow \)

\[ \beta_1 = a_1 + t C_1, \quad \cdots \quad , \beta_n = a_n + t C_n, \quad t = \text{any number} \]

infinitely many such representations.