PX153 - K7 - matrix inverse

some definitions

minor: $m_{i j}$ is determinant of $(n - 1) \times (n - 1)$ matrix where the $i^{t h}$ row and $j^{t h}$ column of $A$ are removed
co-factor: $(- 1)^{i + j} m_{i j} = c_{i j}$
- $det A = a_{11} c_{11} + \dots + a_{1 n} c_{1 n}$
- this formula suggests that the $1^{s t}$ row is special
- the rows and columns can be exchanged
  - $(i) : det A = \sum_{j = 1} a_{i j} c_{i j}$ (co-factor rule)
  - $(i i) : \sum_{j = 1} a_{i j} c_{k j} = 0 \forall i \neq k$ (*false co-factor rule)
co-factor rule for $3 \times 3 :$

det [\begin{matrix} e_{1} & e_{2} & e_{3} \\ f_{1} & f_{2} & f_{3} \\ g_{1} & g_{2} & g_{3} \end{matrix}] = - det [\begin{matrix} f_{1} & f_{2} & f_{3} \\ e_{1} & e_{2} & e_{3} \\ g_{1} & g_{2} & g_{3} \end{matrix}] = f_{1} c_{21} + f_{2} c_{22} + f_{3} c_{23}

- switching row 1 and row 2 is equivalent to expanding from the $2^{n d} r o w$

false co-factor rule: $A = [\begin{matrix} e_{1} & e_{2} & e_{3} \\ f_{1} & f_{2} & f_{3} \\ g_{1} & g_{2} & g_{3} \end{matrix}]$
- $det A = \vec{e} \cdot (\vec{f} \times \vec{g})$
- using false co-factor rule: $\sum_{j} a_{i j} c_{k j} \overset{?}{=} 0$
- using $i = 2, k = 1 :$

\sum_{j} a_{2 j} c_{1 j} = f_{1} c_{11} + f_{2} c_{12} + f_{3} c_{13} = \vec{f} \cdot (\vec{f} \times \vec{g}) = 0

- in general, $det A = 0$ if two rows are the same

if $A = (a_{i j})_{n \times m}$ , the adjugate (AKA adjoint) of $A$ , $A d j (A) = (c_{i j}^{T})$ , where $c_{i j}$ are co-factors of matrix $A :$

A d j (A) = [\begin{matrix} c_{11} & \dots & c_{n 1} \\ ⋮ & ⋱ & ⋮ \\ c_{1 n} & \dots & c_{n n} \end{matrix}]

A^{- 1} = \frac{1}{det A} A d j (A)

\begin{aligned} B_{i j} & = \sum_{k = 1}^{n} a_{i k} (A d j (A))_{k j} \\ = \sum_{k = 1}^{n} a_{i k} c_{j k} \\ = {\begin{cases} det A & i f i = j & c o f a c t o r r u l e \\ 0 & i f i \neq j & f a l s e c o f a c t o r r u l e \end{cases} \end{aligned}

\begin{aligned} A A d j (A) & = det A I \\ A^{- 1} & = \frac{1}{det A} A d j (A) & i f det A & \neq 0 \end{aligned}

the inverse method is not quick because the determinant and co-factors need to be calculated
$A A d j (A) = det A I$ can be used to infer $det A$
eg: find $A d j A$ , then $A^{- 1}$ , for $A = [\begin{matrix} 1 & 0 & 2 \\ - 1 & 1 & 3 \\ 0 & 2 & 1 \end{matrix}]$
$c_{11} = (- 1)^{1 + 1} | \begin{matrix} 1 & 3 \\ 2 & 1 \end{matrix} | = - 5$
$c_{12} = (- 1)^{1 + 2} | \begin{matrix} - 1 & 3 \\ 0 & 1 \end{matrix} | = 1$
$c_{13} = (- 1)^{1 + 3} | \begin{matrix} - 1 & 1 \\ 0 & 2 \end{matrix} | = - 2$
$c_{21} = (- 1)^{2 + 1} | \begin{matrix} 0 & 2 \\ 2 & 1 \end{matrix} | = 4$
$⋮$
$A d j A = [\begin{matrix} - 5 & 4 & - 2 \\ 1 & 1 & - 5 \\ - 2 & - 2 & 1 \end{matrix}]$
- using $A d j A \cdot A = | A | I :$

\begin{aligned} A d j A \cdot A & = [\begin{array}{c} - 5 & 4 & - 2 \\ 1 & 1 & - 5 \\ - 2 & - 2 & 1 \end{array}] [\begin{array}{c} 1 & 0 & 2 \\ - 1 & 1 & 3 \\ 0 & 2 & 1 \end{array}] \\ = [\begin{array}{c} - 9 & 0 & 0 \\ 0 & - 9 & 0 \\ 0 & 0 & - 9 \end{array}] \\ = - 9 I \\ o r, | A | & = - 9 \\ ∴ A^{- 1} & = \frac{1}{- 9} A \end{aligned}

eg: $[\begin{matrix} 1 & 6 & - 4 \\ 3 & - 20 & 1 \\ - 1 & 3 & 5 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} 8 \\ 12 \\ 3 \end{matrix}]$

\begin{aligned} A d j (A) & = [\begin{array}{c} (- 100 - 3) & - (30 + 12) & (6 - 8 -) \\ - (15 + 1) & (5 - 4) & - (1 + 12) \\ (9 - 20) & - (3 + 6) & (- 20 - 18) \end{array}] \\ = [\begin{array}{c} - 103 & - 42 & - 74 \\ - 16 & 1 & - 13 \\ - 11 & - 9 & - 38 \end{array}] \end{aligned}

\begin{aligned} A d j (A) \cdot A & = | A | \cdot I \\ A d j (A) \cdot A & = [\begin{array}{c} - 103 & - 42 & - 74 \\ - 16 & 1 & - 13 \\ - 11 & - 9 & - 38 \end{array}] [\begin{array}{c} 1 & 6 & - 4 \\ 3 & - 20 & 1 \\ - 1 & 3 & 5 \end{array}] \\ = [\begin{array}{c} - 155 & 0 & 0 \\ 0 & - 155 & 0 \\ 0 & 0 & - 155 \end{array}] \\ = - 155 [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \end{aligned}

\begin{aligned} \vec{x} & = A^{- 1} \vec{b} \\ \vec{x} & = \frac{1}{| A |} A d j (A) \vec{b} \\ \vec{x} & = \frac{1}{- 155} [\begin{array}{c} - 103 & - 42 & - 74 \\ - 16 & 1 & - 13 \\ - 11 & - 9 & - 38 \end{array}] [\begin{array}{c} 8 \\ 12 \\ 3 \end{array}] \\ = [\begin{array}{c} 10 \\ 1 \\ 2 \end{array}] \end{aligned}