MAT223 Lecture 07

MAT223 Readings 04
MAT223 PCRA 2

Activity 4 from last week
- Find $\vec{x} = [\begin{matrix} x \\ y \end{matrix}]$ and a $2 \times 2$ matrix $A$ so that $A \vec{x} = [\begin{matrix} 2 a - 3 b \\ 2 c - 3 d \end{matrix}]$
- $A \vec{x} = [\begin{matrix} 2 a - 3 b \\ 2 c - 3 d \end{matrix}]$
- $[\begin{matrix} 2 a - 3 b \\ 2 c - 3 d \end{matrix}] = [\begin{array}{cc} a & b \\ c & d \end{array}] [\begin{matrix} x \\ y \end{matrix}]$
- $[\begin{matrix} 2 a - 3 b \\ 2 c - 3 d \end{matrix}] = [\begin{array}{cc} a x + b y \\ c x + d y \end{array}]$
- $x = 2$
- $y = - 3$
2
- Let $M = [\begin{array}{cc} m_{11} & m_{12} \\ m_{21} & m_{22} \end{array}]$
- $\vec{x} = [\begin{matrix} x_{1} \\ x_{2} \end{matrix}]$
- $\vec{y} = [\begin{matrix} y_{1} \\ y_{2} \end{matrix}]$
- Compute $(M \vec{x}) \cdot \vec{y}$
  - $[\begin{array}{cc} m_{11} x_{1} + m_{12} x_{2} \\ m_{21} x_{1} + m_{22} x_{2} \end{array}] \cdot [\begin{matrix} y_{1} \\ y_{2} \end{matrix}]$
  - $= (m_{11} x_{1} + m_{12} x_{2}) y_{1} + (m_{21} x_{1} + m_{22} x_{2}) y_{2} = m_{11} x_{1} y_{1} + m_{12} x_{2} y_{1} + m_{21} x_{1} y_{2} + m_{22} x_{2} y_{2}$
- The thing is that $(M \vec{x}) \vec{y}$ is not defined. We need $a \times b, c \times d$ $b = c$ for this to work
  - With $(M \vec{x}) \vec{y}$ , we have $(M \vec{x}) = 2 \times 1$ and $\vec{y} = 2 \times 1$
  - See $1 \neq 2$
- If we have two columns, the dot product is simple:
  - $\vec{x} \cdot \vec{y}$
  - $[\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}] \cdot [\begin{matrix} 4 \\ 5 \\ 6 \end{matrix}] = (1) (4) + (2) (5) + (3) (6)$
  - It's only for rows and columns. Not $2 \times 2$ and above, or such.
  - We can also have $[\begin{array}{ccc} 1 & 2 & 3 \end{array}] \cdot [\begin{matrix} 4 \\ 5 \\ 6 \end{matrix}]$ . This is just Matrix Multiplication.
3
- $3 x_{1} y_{1} + 2 x_{2} y_{1} + 4 x_{1} y_{2} - 7 x_{2} y_{2}$
- Find a matrix $M = [\begin{array}{cc} m_{11} & m_{12} \\ m_{21} & m_{22} \end{array}]$ and fill in the missing blanks to express our sum in $Σ$ notation.
- $[\begin{array}{cc} m_{11} & m_{12} \\ m_{21} & m_{22} \end{array}] [\begin{matrix} x \\ y \end{matrix}] \cdot [\begin{matrix} a \\ b \end{matrix}] = 3 x_{1} y_{1} + 2 x_{2} y_{1} + 4 x_{1} y_{2} - 7 x_{2} y_{2}$
- $\sum_{i = ?}^{?} \sum_{j = ?}^{?} m_{?} x_{?} y_{?}$
- $\sum_{i = 1}^{2} \sum_{j = 1}^{2} m_{i j} x_{i} y_{j}$
- $M = [\begin{array}{cc} 3 & 4 \\ 2 & - 7 \end{array}]$
4
- Express the sum $\sum_{i = 1}^{n} \sum_{j = 1}^{n} a_{i j} x_{j} y_{i}$ in terms of matrix multiplication
- $[\begin{array}{cccc} a_{11} & a_{12} & \dots & a_{1 n} \\ a_{21} & a_{22} & \dots & a_{2 n} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{n 1} & a_{n 2} & \dots & a_{n n} \end{array}] [\begin{matrix} x_{1} \\ x_{2} \\ ⋮ \\ x_{n} \end{matrix}] \cdot [\begin{matrix} y_{1} \\ y_{2} \\ ⋮ \\ y_{n} \end{matrix}]$
1
- Compute $det [\begin{array}{cc} 2 & 5 \\ 4 & 8 \end{array}] = (2) (8) - (5) (4) = - 4$
- See that $det A \neq 0$ so it's invertible
2
- If $A = [\begin{array}{cccc} 1 & 0 & 1 & 2 \\ 0 & 1 & 0 & 0 \\ 2 & 1 & 3 & 4 \\ 0 & 1 & 0 & 2 \end{array}]$
- $A_{42} = [\begin{array}{ccc} 1 & 1 & 2 \\ 0 & 0 & 0 \\ 2 & 3 & 4 \end{array}]$
- #tk
  - Try creating a question where we have some information about a matrix $A$ and a cofactor $A_{i j}$ , and we have to re-derive $A$ .
- Deleting Rows and Columns in a Matrix
3
- $A = [\begin{array}{cccc} 1 & 0 & - 1 & 2 \\ 0 & 1 & 0 & 0 \\ 1 & 2 & 3 & 4 \\ 0 & 1 & 0 & 2 \end{array}]$
- Find $det A$
- $det A = (- 1)^{2 + 2} (1) det [\begin{array}{ccc} 1 & - 1 & 2 \\ 1 & 3 & 4 \\ 0 & 0 & 2 \end{array}]$
- $det A = (- 1)^{2 + 2} (1) [(- 1)^{3 + 3} (2) det [\begin{array}{cc} 1 & - 1 \\ 1 & 3 \end{array}]]$
- $det A = (- 1)^{2 + 2} (1) [(- 1)^{3 + 3} (2) [(1) (- 1) - (1) (3)]]$
  - $(- 1)^{2 + 2} (1) [(- 1)^{3 + 3} (2) [(1) (- 1) - (1) (3)]] = - 8.0$