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

Instructor: Yip

• This test booklet has five questions, totaling 100 points for the whole test. You have 75 minutes to do this test. Plan your time well. Read the questions carefully.
• This test is closed book, closed note, with no electronic devices.
• In order to get full credits, you need to give correct, simplified, and complete answers and explain in a comprehensible way how you arrive at them.
• As a rule of thumb, you should give explicit and useful answers. No points will be given for just writing down some generically true statements. In other words, your solutions and answers should be relevant to the information given in the question.
• As a rule of thumb, you should only use those methods that have been covered in class. If you use some other methods “for the sake of convenience”, at our discretion, we might not give you any credit. You have the right to contest. In that event, you will be asked to explain your answer using only what has been covered in class up to the point of time of this exam.

Read the above instructions!

Name: Answer Key (Major: )

1. Consider the equation \( 2x + 3y = 5 \).

   1. Write down the solution of the above equation in parametric form as \( X = p + \text{Null}(A) \) in terms of a vector \( p \) and the null space of a matrix \( A \). You should write \( \text{Null}(A) \) as a span of vector(s). Give three different examples of \( p \).

   2. In the following \( xy \)-plane, clearly plot \( \text{Null}(A) \) and also each of the three forms of the solution you have found.

   

You can use this blank page.

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

1. \[ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \frac{5}{2} - \frac{3}{2}\alpha \\ \alpha \end{pmatrix} = \begin{pmatrix} \frac{5}{2} \\ 0 \end{pmatrix} + \begin{pmatrix} -\frac{3}{2} \\ 1 \end{pmatrix} \alpha \]

   \( p_1 \) labels

   \[ \begin{pmatrix} \frac{5}{2} \\ 0 \end{pmatrix} \]

   .

2. \( x = \alpha, \qquad y = \frac{5 - 2\alpha}{3} = \frac{5}{3} - \frac{2}{3}\alpha \)

   \[ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \alpha \\ \frac{5}{3} - \frac{2}{3}\alpha \end{pmatrix} = \begin{pmatrix} 0 \\ \frac{5}{3} \end{pmatrix} + \alpha \begin{pmatrix} 1 \\ -\frac{2}{3} \end{pmatrix} \]

   \( p_2 \) labels

   \[ \begin{pmatrix} 0 \\ \frac{5}{3} \end{pmatrix} \]

   .

3. Or choose any \( p \) on the line, e.g. \( p_3 = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \)

   \[ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} + \begin{pmatrix} -\frac{3}{2} \\ 1 \end{pmatrix} \alpha \]

   \( p_3 \) labels

   \[ \begin{pmatrix} 1 \\ 1 \end{pmatrix} \]

   .

2. You are given the following list of \( 2 \times 2 \) matrices:

   \[ \left\{ A = \begin{pmatrix} 1 & 2 \\ 1 & 3 \end{pmatrix},\ B = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix},\ C = \begin{pmatrix} 1 & 2 \\ 3 & 5 \end{pmatrix},\ D = \begin{pmatrix} 1 & 2 \\ 0 & 2 \end{pmatrix} \right\} \]

   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.

   \[ c_1 A + c_2 B + c_3 C + c_4 D = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \] \[ \left(\begin{array}{cccc|c} 1 & 1 & 1 & 1 & 0 \\ 2 & 2 & 2 & 2 & 0 \\ 1 & 2 & 3 & 0 & 0 \\ 3 & 4 & 5 & 2 & 0 \end{array}\right) \longrightarrow \left(\begin{array}{cccc|c} 1 & 1 & 1 & 1 & 0 \\ 0 & 1 & 2 & -1 & 0 \\ 0 & 1 & 2 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right) \longrightarrow \left(\begin{array}{cccc|c} 1 & 0 & -1 & 2 & 0 \\ 0 & 1 & 2 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right) \] \[ \begin{gathered} c_3 = \alpha, \quad c_4 = \beta \\[0.5em] c_1 = \alpha - 2\beta, \quad c_2 = -2\alpha + \beta \end{gathered} \]

You can use this blank page.

\( C \), \( D \) are redundant.

\( A \), \( B \) are lin. ind.

3. Given that the following lists of functions are all linearly dependent, for each list, express one of the vectors as a linear combination of the rest.

   \[ \begin{array}{rcl} \mathcal{A} &=& \{\sinh(x),\ \cosh(x),\ 3e^{x} - 5e^{-x}\} \\ \mathcal{B} &=& \{1,\ x-1,\ (x-1)^{2},\ 2x^{2} + 5x + 3\} \\ \mathcal{C} &=& \left\{\cos x,\ \sin\left(x + \frac{\pi}{4}\right),\ \sin\left(x - \frac{\pi}{4}\right)\right\} \end{array} \]

   (Note: \( \sinh(x) = \frac{e^{x} - e^{-x}}{2} \), \( \cosh(x) = \frac{e^{x} + e^{-x}}{2} \), \( \sin(x+y) = \sin x \cos y + \sin y \cos x \).)

   \[ \begin{gathered} \mathcal{A}: \quad \sinh(x) = \frac{e^{x} - e^{-x}}{2}, \quad \cosh(x) = \frac{e^{x} + e^{-x}}{2} \\[0.5em] \implies \cosh(x) + \sinh(x) = e^{x} \\[0.5em] \cosh(x) - \sinh(x) = e^{-x} \\[0.5em] \implies \left(3e^{x} - 5e^{-x}\right) = 3\left(\cosh(x) + \sinh(x)\right) - 5\left(\cosh(x) - \sinh(x)\right) = -2\cosh(x) + 8\sinh(x) \\[0.5em] \mathcal{B}: \quad 2x^{2} + 5x + 3 = c_1 \cdot 1 + c_2 (x-1) + c_3 (x-1)^{2} \\[0.5em] = c_1 + c_2 x - c_2 + c_3 x^{2} - 2c_3 x + c_3 \\[0.5em] = \left(c_1 - c_2 + c_3\right) + \left(c_2 - 2c_3\right)x + c_3 x^{2} \\[0.5em] c_3 = 2, \quad c_2 - 2c_3 = 5 \implies c_2 = 9 \\[0.5em] c_1 - c_2 + c_3 = 3 \implies c_1 = 3 - 2 + 9 = 10 \end{gathered} \]

You can use this blank page.

\( y = \frac{\pi}{4} \)

\( y = -\frac{\pi}{4} \)

4. You are given the following two lists of vectors:

   \[ \mathcal{A} = \left\{ X_1 = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix},\ X_2 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \right\} \quad \text{and} \quad \mathcal{B} = \left\{ Y_1 = \begin{pmatrix} 4 \\ -2 \\ 4 \end{pmatrix},\ Y_2 = \begin{pmatrix} 0 \\ -1 \\ 0 \end{pmatrix} \right\}. \]

   Is \( \operatorname{Span}\{\mathcal{A}\} = \operatorname{Span}\{\mathcal{B}\} \)?

   If so, write every element from \( \operatorname{Span}\{\mathcal{A}\} \) as a linear combination of \( Y_1 \) and \( Y_2 \) and write every element from \( \operatorname{Span}\{\mathcal{B}\} \) as a linear combination of \( X_1 \) and \( X_2 \).

   If not, determine which vectors from \( \operatorname{Span}\{\mathcal{A}\} \) that cannot be written as a linear combination of \( Y_1 \) and \( Y_2 \) and determine which vectors from \( \operatorname{Span}\{\mathcal{B}\} \) that cannot be written as a linear combination of \( X_1 \) and \( X_2 \).

   Consider: \( s_1 X_1 + s_2 X_2 = t_1 Y_1 + t_2 Y_2 \)

   given \( t_1, t_2 \), find \( s_1, s_2 \):

   \[ \left[\begin{array}{cc|c} 1 & 1 & 4t_1 \\ -1 & 0 & -2t_1 - t_2 \\ 1 & 1 & 4t_1 \end{array}\right] \rightarrow \left[\begin{array}{cc|c} 1 & 1 & 4t_1 \\ 0 & 1 & 2t_1 - t_2 \\ 0 & 0 & 0 \end{array}\right] \rightarrow \left[\begin{array}{cc|c} 1 & 0 & 2t_1 + t_2 \\ 0 & 1 & 2t_1 - t_2 \\ 0 & 0 & 0 \end{array}\right] \]

   Solvable for any \( t_1, t_2 \)

   \( s_1 = 2t_1 + t_2 \)

   \( s_2 = 2t_1 - t_2 \)

   \[ \underbrace{\left(2t_1 + t_2\right)}_{s_1} X_1 + \underbrace{\left(2t_1 - t_2\right)}_{s_2} X_2 = t_1 Y_1 + t_2 Y_2 \]

Check:

Given \( s_1, s_2 \), find \( t_1, t_2 \)

Solvable for any \( s_1, s_2 \)

\( t_1 = \frac{s_1 + s_2}{4} \)

\( t_2 = +\frac{s_1 - s_2}{2} \)

Check:

5. Consider the following configuration of a network of heat conducting wires,

   

   At thermal equilibrium, the temperature at each vertex – intersection between two or more wires – is the average of the temperatures at the adjacent vertices. For example,

   \[ T_2 = \frac{T_3 + T_1 + 200 + 0}{4}. \]

   Find the values of \( T_1, T_2 \) and \( T_3 \).

   \[ \begin{gathered} T_1 = \frac{200 + T_2 + 0 + 0}{4} \\[0.5em] T_2 = \frac{T_1 + 200 + 0 + T_3}{4} \\[0.5em] T_3 = \frac{T_2 + 400 + 0 + 0}{4} \\[0.5em] \Rightarrow \quad 4T_1 - T_2 = 200 \\[0.5em] T_1 - 4T_2 + T_3 = -200 \\[0.5em] T_2 - 4T_3 = -400 \end{gathered} \] \[ \left(\begin{array}{ccc|c} 4 & -1 & 0 & 200 \\ 1 & -4 & 1 & -200 \\ 0 & 1 & -4 & -400 \end{array}\right) \rightarrow \left(\begin{array}{ccc|c} 1 & -4 & 1 & -200 \\ 4 & -1 & 0 & 200 \\ 0 & 1 & -4 & -400 \end{array}\right) \]

You can use this blank page.