Linear-2 1

\begin{matrix} det : F^{n \times n} \to F \\ det (A) = | A | = \sum_{σ \in S_{n}} s g n (σ) \prod_{i = 1}^{n} a_{i σ (i)} \\ For example: | \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix} | = a_{11} a_{22} - a_{12} a_{21} \\ Let i \in [1, n] \\ det (A) = | A | = \sum_{j = 1}^{n} (- 1)^{i + j} a_{i j} | M_{i j} (A) | \\ Let j \in [1, n] \\ det (A) = | A | = \sum_{i = 1}^{n} (- 1)^{i + j} a_{i j} | M_{i j} (A) | \\ For example: \\ | \begin{matrix} 1 & 0 & 2 \\ 3 & 1 & 0 \\ 5 & 7 & 6 \end{matrix} | = 1 | \begin{matrix} 1 & 0 \\ 7 & 6 \end{matrix} | - 0 | \begin{matrix} 3 & 0 \\ 5 & 6 \end{matrix} | + 2 | \begin{matrix} 3 & 1 \\ 5 & 7 \end{matrix} | = 6 + 32 = 38 \end{matrix}

Exercises

\begin{matrix} Let A, B \in R^{n \times n} \\ | A | = 7, | B | = 4 \\ C = A^{T} (B^{- 1})^{5} \\ Find | C | \\ Solution: \\ | C | = | A^{T} (B^{- 1})^{5} | = | A^{T} | \cdot | (B^{- 1})^{5} | = | A | \cdot {| B^{- 1} |}^{5} = \\ | A | \cdot {| B |}^{- 5} = 7 \cdot \frac{1}{4^{5}} = \frac{7}{4^{5}} \end{matrix}

\begin{matrix} Let n \in N be odd \\ Let A \in R^{n \times n} be anti-symmetric \\ Prove: A is non-invertible \\ Proof: \\ A = - A^{T} \\ | A | = | - A^{T} | = (- 1)^{n} | A^{T} | = - | A | \\ ⟹ | A | = 0 ⟹ A is non-invertible \end{matrix}

\begin{matrix} | \begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix} | = 2 \\ Find | \begin{matrix} i - 4 c & f & 2 i + f \\ g - 4 a & d & 2 g + d \\ h - 4 b & e & 2 h + e \end{matrix} | = | \begin{matrix} i - 4 c & f & 2 i \\ g - 4 a & d & 2 g \\ h - 4 b & e & 2 h \end{matrix} | = 2 | \begin{matrix} i - 4 c & f & i \\ g - 4 a & d & g \\ h - 4 b & e & h \end{matrix} | = \\ = 2 | \begin{matrix} - 4 c & f & i \\ - 4 a & d & g \\ - 4 b & e & h \end{matrix} | = - 8 | \begin{matrix} c & f & i \\ a & d & g \\ b & e & h \end{matrix} | = - 8 | \begin{matrix} c & a & b \\ f & d & e \\ i & g & h \end{matrix} | = \\ 8 | \begin{matrix} a & c & b \\ d & f & e \\ g & i & h \end{matrix} | = - 8 | \begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix} | = - 16 \end{matrix}

\begin{matrix} Let A = {\begin{matrix} 2 & i = j \\ - 1 & i = j + 1 \\ - 1 & i = j - 1 \\ 0 & otherwise \end{matrix} \\ Find | A | \\ Let A_{n} = A \in R^{n \times n} \\ | A_{n} | = 2 \cdot | A_{n - 1} | - (- 1) | M_{12} (A_{n}) | = \\ = 2 | A_{n - 1} | + (- 1) | A_{n - 2} | = 2 | A_{n - 1} | - | A_{n - 2} | \\ | A_{1} | = 2 \\ | A_{2} | = 3 \\ | A_{3} | = 4 \\ 2 | A_{n - 1} | - | A_{n - 2} | = 2 (2 A_{n - 2} - A_{n - 3}) - A_{n - 2} = 3 A_{n - 2} - 2 A_{n - 3} = 4 A_{n - 3} - 3 A_{n - 4} = \dots = \\ = (n - 1) A_{2} - (n - 2) A_{1} = 3 n - 3 - 2 n + 4 = n + 1 \end{matrix}

\begin{matrix} Let A = {\begin{matrix} 2 i & i \neq j \\ 0 & i = j \end{matrix} \\ Let A^{'} = {\begin{matrix} 1 & i \neq j \\ 0 & i = j \end{matrix} \\ | A | = (2 n)!! | A^{'} | = 2^{n} n! | A^{'} | \\ | A^{'} | = | \begin{matrix} 0 & 1 & \dots & \dots & 1 \\ 1 & 0 & 1 & \dots & 1 \\ 1 & 1 & ⋱ & ⋱ & ⋮ \\ ⋮ & ⋮ & ⋱ & ⋱ & 1 \\ 1 & 1 & \dots & 1 & 0 \end{matrix} | \overset{\forall i \in [2, n] : R_{i} - R_{1}}{=} | \begin{matrix} 0 & 1 & \dots & \dots & 1 \\ 1 & - 1 & 0 & \dots & 0 \\ 1 & 0 & ⋱ & ⋱ & ⋮ \\ ⋮ & ⋮ & ⋱ & ⋱ & 0 \\ 1 & 0 & \dots & 1 & - 1 \end{matrix} | = \\ \overset{\forall i \in [2, n] : C_{1} + C_{i}}{=} | \begin{matrix} n - 1 & 1 & \dots & \dots & 1 \\ 0 & - 1 & 0 & \dots & 0 \\ 0 & 0 & ⋱ & ⋱ & ⋮ \\ ⋮ & ⋮ & ⋱ & ⋱ & 0 \\ 0 & 0 & \dots & 1 & - 1 \end{matrix} | = (n - 1) \cdot (- 1)^{n - 1} \\ ⟹ | A | = (- 1)^{n} (n - 1) 2^{n} n! \end{matrix}