MAT223 Lecture 06

1
- $A = [\begin{array}{cc} 2 & 1 \\ - 1 & 2 \end{array}]$
- $b = [\begin{array}{cc} 1 & 2 \\ 1 & - 3 \end{array}]$
- $A - 2 B = [\begin{array}{cc} 0 & - 3 \\ - 3 & 8 \end{array}]$
2
- $A = [\begin{array}{ccc} 1 & - 2 & 0 \\ 1 & - 1 & 1 \end{array}]$
- $B = [\begin{array}{cc} 1 & 0 \\ - 1 & 1 \\ 1 & - 2 \end{array}]$
- $A B = [\begin{array}{cc} 3 & - 2 \\ 3 & - 3 \end{array}]$
Activity 2
- $A = [\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}]$
- $B = [\begin{array}{cc} x & y \\ z & w \end{array}]$
- Find $A B$ and $B A$
- $A B = [\begin{array}{cc} z & w \\ 0 & 0 \end{array}]$
- $B A = [\begin{array}{cc} 0 & x \\ 0 & z \end{array}]$
- Multiplying Matrices
- Find a matrix $B$ such that $A B \neq B A$
  - $B = [\begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array}]$
  - Anyways it's easy to have them not be equal.
- Setup a System of Linear Equations that must hold in order that $A B = B A$
  - $x - w = 0$
  - $z = 0$
  - $[\begin{array}{cc} z & w \\ 0 & 0 \end{array}] = [\begin{array}{cc} 0 & x \\ 0 & z \end{array}]$
  - $z = 0$
  - $w = x ⟹ w - x = 0$
  - $0 = 0$
  - $0 = z$
  - $[\begin{array}{ccccc} - 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \end{array}]$
  - $R_{1} = \frac{R_{1}}{- 1}$
  - $[\begin{array}{ccccc} 1 & 0 & 0 & - 1 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}]$
  - $R_{3} = R_{3} - R_{2}$
  - $[\begin{array}{ccccc} 1 & 0 & 0 & - 1 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}]$
  - $x = w$
  - $z = 0$
  - $y = s$
  - $w = t$
  - $[\begin{matrix} x \\ y \\ z \\ w \end{matrix}] = s [\begin{matrix} 0 \\ 0 \\ 1 \\ 0 \end{matrix}] + t [\begin{matrix} 1 \\ 0 \\ 0 \\ 1 \end{matrix}]$
- Find all matrices $B$ so that $A B = B A$
  - $x, w, z$ must be $0$
  - $B = [\begin{array}{cc} 0 & y \\ 0 & 0 \end{array}]$
- Find a non-zero matrix $C$ so that $A C = O_{2}$ . Where $O_{2}$ is the 0 matrix.
  - $A = [\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}]$
  - $C = ?$
  - $A C = [\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}]$
  - $[\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}] = [\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}] [\begin{array}{cc} x & y \\ z & w \end{array}]$
  - $[\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}] = [\begin{array}{cc} z & w \\ 0 & 0 \end{array}]$
  - $z = 0$
  - $w = 0$
  - $x = s$
  - $y = t$
8
- $A$ is an $m \times n$ matrix
- $B$ is a $k \times l$ matrix
- If $A B$ and $B A$ are both defined
- Which is true?
  - $m = k$
  - $n = l$
  - $\dots$
- $n = k$
- $m = l$
Activity 3
- Symmetric and skew symmetric
- Create two $3 \times 3$ symmetric and two skew-symmetric
- Symmetric
  - $[\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{array}]$
  - $[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$
  - $O_{3} = [\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}]$
- Skew-symmetric
  - $[\begin{array}{ccc} 0 & 0 & - 1 \\ 0 & 0 & 0 \\ - 1 & 0 & 0 \end{array}]$
  - $[\begin{array}{ccc} 0 & 1 & 1 \\ - 1 & 0 & - 1 \\ - 1 & 1 & 0 \end{array}]$
  - $O_{3} = [\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}]$
- For any $n \times n$ $A$ the matrix $A - A^{t}$ is skew-symmetric
  - Let $B = A - A^{t}$
  - $A = [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}]$
  - $A^{t} = [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}]$
  - $A - A^{t} = O_{2}$
  - It is
  - $A = [\begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array}]$
  - $A^{t} = [\begin{array}{cc} 1 & 3 \\ 2 & 4 \end{array}]$
  - $A - A^{t} = [\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array}]$
  - $2 \times 2$ case:
    - $- {[\begin{array}{cc} a & b \\ c & d \end{array}]}^{t} = [\begin{array}{cc} - a & - c \\ - b & - d \end{array}] = [\begin{array}{cc} a & b \\ c & d \end{array}] = \begin{matrix} - a = a ⟹ a = 0 \\ - d = d ⟹ d = 0 \end{matrix}$
    - $A - A^{t} = [\begin{array}{cc} a & b \\ c & d \end{array}] - {[\begin{array}{cc} a & b \\ c & d \end{array}]}^{t}$
    - $= [\begin{array}{cc} 0 & b - c \\ c - b & 0 \end{array}]$
    - $b - c = - (a - b)$ this is skew.
  - Looks like it is true
    - #tk
  - General Argument
    - Check if $B = A - A^{t}$ is skew-symmetric.
    - See if $- B^{t} = B$
    - $- (A - A^{t}) = - (A^{t} - A^{t^{t}})$
      - Based on property of transpose.
      - $A^{t^{t}} = A$
    - $= - (A^{t} - A)$
    - $= A - A^{t}$
    - So it is true.
- If $A$ is symmetric and $B$ is skew-symmetric, then $C = B - A^{t}$ is skew-symmetric
  - $A = [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}]$
  - $B = [\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array}]$
  - $C = [\begin{array}{cc} - 1 & - 1 \\ 1 & - 1 \end{array}]$
  - Not skew-symmetric.
  - Another:
    - $A = [\begin{array}{cc} 1 & 3 \\ 3 & 2 \end{array}]$
    - $B = [\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array}]$
    - $B - A^{t} = [\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array}] - [\begin{array}{cc} 1 & 3 \\ 3 & 2 \end{array}]$
    - $= [\begin{array}{cc} - 1 & - 2 \\ - 4 & - 2 \end{array}]$
9
- If $A = [\begin{array}{cc} a & b \\ c & d \end{array}]$
- Then $A + A^{t}$ is?
- $A^{t} = [\begin{array}{cc} a & c \\ b & d \end{array}]$
- $C = A + A^{t} = [\begin{array}{cc} 2 a & b + c \\ b + c & 2 d \end{array}]$
- $C^{t} = - C$
- $C^{t} = [\begin{array}{cc} 2 a & b + c \\ b + c & 2 d \end{array}]$
- $[\begin{array}{cc} 2 a & b + c \\ b + c & 2 d \end{array}] \neq [\begin{array}{cc} - 2 a & - b - c \\ - b - c & - 2 d \end{array}]$
Activity 4
- Vectorization
- Find $x = [\begin{matrix} x \\ y \end{matrix}]$ and a $2 \times 2$ $A$
- So that $A x = [\begin{matrix} 2 a - 3 b \\ 2 c - 3 d \end{matrix}]$
- $[\begin{matrix} 2 a - 3 b \\ 2 c - 3 d \end{matrix}] = [\begin{matrix} x \\ y \end{matrix}] [\begin{array}{cc} a & b \\ c & d \end{array}]$
- $[\begin{matrix} 2 a - 3 b \\ 2 c - 3 d \end{matrix}] = [\begin{array}{cc} \end{array}]$