Quiz 1 Solution

Let \( P \) and \( Q \) be two points in space and \( PQ \) be the line segment connecting \( P \) and \( Q \). Find the point \( R \) on \( PQ \) such that the ratio between the length \( PR \) and \( RQ \) is \( 2:1 \).

Line segment from P to Q with point R between them, braces showing ratio 2 to 1
Figure 1
A hand-drawn diagram of a straight line segment rising from lower left to upper right. The left endpoint is marked with a dot and labeled \( P \); the right endpoint is marked with a dot and labeled \( Q \). A dot partway along the segment, about two-thirds of the way from \( P \) toward \( Q \), is labeled \( R \). A curly brace beneath the portion from \( P \) to \( R \) and a smaller curly brace beneath the portion from \( R \) to \( Q \) are jointly labeled \( 2 : 1 \), indicating that the length \( PR \) is twice the length \( RQ \).
\[ \begin{gathered} \vec{PR} = 2\vec{RQ} \\[0.5em] R - P = 2(Q - R) \\[0.5em] R - P = 2Q - 2R \\[0.5em] 3R = P + 2Q \\[0.5em] R = \frac{P + 2Q}{3} \end{gathered} \]

Quiz 2

Consider the following system of linear eq:

\[ \begin{array}{l} x + y + z = 2 \\ x + 2y + z = 3 \\ x + y + (c^2 - 3)z = c \end{array} \qquad c - \text{a parameter} \]

Discuss the solvability in terms of \( c \). When does the system have (i) unique solution, (ii) infinitely many solution; (iii) no solution. Solve the system when solution exists

\[ \left[\begin{array}{ccc|c} 1 & 1 & 1 & 2 \\ 1 & 2 & 1 & 3 \\ 1 & 1 & c^2-3 & c \end{array}\right] \rightarrow \left[\begin{array}{ccc|c} 1 & 1 & 1 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & c^2-4 & c-2 \end{array}\right] \]

When \( c^2 - 4 \ne 0 \) ie. \( c \ne 2, -2 \), there is unique solution.

\[ \begin{gathered} z = \frac{c-2}{c^2-4} = \frac{1}{c+2}, \qquad y = 1 \\[0.5em] x = 2 - \frac{1}{c+2} - 1 = \frac{c+1}{c+2}. \\[0.5em] \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} \frac{c+1}{c+2} \\ 1 \\ \frac{1}{c+2} \end{pmatrix} \end{gathered} \]

When \( c^2 - 4 = 0 \) and \( c - 2 = 0 \), ie. \( c = 2 \), then infinitely many solution.

\[ \left(\begin{array}{ccc|c} 1 & 1 & 1 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right) \] \[ \begin{gathered} z = \alpha \ (\text{free}), \qquad y = 1, \qquad x = 2 - \alpha - 1 = 1 - \alpha \\[0.5em] \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 - \alpha \\ 1 \\ \alpha \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + \alpha \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} \end{gathered} \]

When \( c^2 - 4 = 0 \), \( c - 2 \ne 0 \), ie. \( c = -2 \) no solution

\[ \left(\begin{array}{ccc|c} 1 & 1 & 1 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & -4 \end{array}\right) \]

Quiz 3

\[ \begin{array}{l} p_1(x) = 1 + 2x + 3x^2 \\ p_2(x) = \phantom{1 + 2x} x - x^2 \\ p_3(x) = 2 + 3x + 7x^2 \end{array} \]

Determine Spand \( \{p_1, p_2, p_3\} \), i.e. find ALL \( p(x) = a + bx + cx^2 \) such that \( p(x) \) is from Spand \( \{p_1, p_2, p_3\} \)?

\[ a + bx + cx^2 = c_1(1 + 2x + 3x^2) + c_2(x - x^2) + c_3(2 + 3x + 7x^2) \] \[ \begin{array}{rcl} c_1 \phantom{+ c_2} + 2c_3 &=& a \\ 2c_1 + c_2 + 3c_3 &=& b \\ 3c_1 - c_2 + 7c_3 &=& c \end{array} \rightarrow \left(\begin{array}{ccc|c} 1 & 0 & 2 & a \\ 2 & 1 & 3 & b \\ 3 & -1 & 7 & c \end{array}\right) \] \[ \rightarrow \left(\begin{array}{ccc|c} 1 & 0 & 2 & a \\ 0 & 1 & -1 & b - 2a \\ 0 & -1 & 1 & c - 3a \end{array}\right) \] \[ \rightarrow \left(\begin{array}{ccc|c} 1 & 0 & 2 & a \\ 0 & 1 & -1 & b - 2a \\ 0 & 0 & 0 & b + c - 5a \end{array}\right) \]

Need \( b + c - 5a = 0 \)

Quiz 4

\[ \text{Let } A = \begin{pmatrix} 1 & 2 & -1 & 3 \\ 2 & 2 & -4 & 4 \\ 1 & 3 & 0 & 4 \end{pmatrix} \]

Write Col(A) and Null(A) as spans of lin.ind vectors. Make sure you show that your lists are lin.ind.

\[ \left(\begin{array}{cccc|c} 1 & 2 & -1 & 3 & 0 \\ 2 & 2 & -4 & 4 & 0 \\ 1 & 3 & 0 & 4 & 0 \end{array}\right) \] \[ \rightarrow \left(\begin{array}{cccc|c} 1 & 2 & -1 & 3 & 0 \\ 0 & -2 & -2 & -2 & 0 \\ 0 & 1 & 1 & 1 & 0 \end{array}\right) \] \[ \rightarrow \left(\begin{array}{cccc|c} 1 & 2 & -1 & 3 & 0 \\ 0 & 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right) \] \[ \text{Col}(A) = \text{Span}\left\{ \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}, \begin{pmatrix} 2 \\ 2 \\ 3 \end{pmatrix} \right\} \]

free variables

lin ind. because there are no free var.

the same!

\[ \left(\begin{array}{cc|c} 1 & 2 & 0 \\ 2 & 2 & 0 \\ 1 & 3 & 0 \end{array}\right) \rightarrow \left(\begin{array}{cc|c} 1 & 2 & 0 \\ 0 & -2 & 0 \\ 0 & 1 & 0 \end{array}\right) \rightarrow \left(\begin{array}{cc|c} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}\right) \]\[ \text{Null}(A) = \{ X : AX = 0 \} \] \[ \left(\begin{array}{cccc|c} 1 & 2 & -1 & 3 & 0 \\ 2 & 2 & -4 & 4 & 0 \\ 1 & 3 & 0 & 4 & 0 \end{array}\right) \rightarrow \begin{array}{cccc} x_1 & x_2 & x_3 & x_4 \end{array} \left(\begin{array}{cccc|c} 1 & 2 & -1 & 3 & 0 \\ 0 & 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right) \] \[ \begin{gathered} x_3 = \alpha, \quad x_4 = \beta \ (\text{free}) \\[0.5em] x_2 = -\alpha - \beta \\[0.5em] x_1 = -2x_2 + x_3 - 3x_4 \\[0.5em] = 2\alpha + 2\beta + \alpha - 3\beta \\[0.5em] = 3\alpha - \beta \\[0.5em] \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = \begin{pmatrix} 3\alpha - \beta \\ -\alpha - \beta \\ \alpha \\ \beta \end{pmatrix} = \alpha \begin{pmatrix} 3 \\ -1 \\ 1 \\ 0 \end{pmatrix} + \beta \begin{pmatrix} -1 \\ -1 \\ 0 \\ 1 \end{pmatrix} \\[0.5em] \text{Null}(A) = \text{Span}\left\{ \begin{pmatrix} 3 \\ -1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} -1 \\ -1 \\ 0 \\ 1 \end{pmatrix} \right\} \end{gathered} \]

lin ind.

\[ c_1 \begin{pmatrix} 3 \\ -1 \\ 1 \\ 0 \end{pmatrix} + c_2 \begin{pmatrix} -1 \\ -1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \]

\( c_1 = 0 \)

\( c_2 = 0 \)