Linear-2 1

1a

\begin{matrix} Calculate determinant using permutations \\ A = (\begin{matrix} 3 & 1 & 5 \\ 4 & 0 & 6 \\ - 1 & 3 & 5 \end{matrix}) \in R^{3 \times 3} \\ S_{3} = \begin{array}{cc} 123 & + \\ 132 & - \\ 213 & - \\ 231 & + \\ 312 & + \\ 321 & - \end{array} \\ | A | = \sum_{σ \in S_{3}} s g n (σ) \prod_{i = 1}^{3} a_{i σ (i)} = \\ = a_{11} a_{22} a_{33} - a_{11} a_{23} a_{32} - a_{12} a_{21} a_{33} + a_{12} a_{23} a_{31} + a_{13} a_{21} a_{32} - a_{13} a_{22} a_{31} = \\ = 0 - 54 - 20 + (- 6) + 60 - 0 = - 20 \\ ⟹ | A | = - 20 \end{matrix}

1b

\begin{matrix} Calculate determinant using permutations \\ A = (\begin{matrix} 2 & 6 & 3 \\ 5 & 1 & 0 \\ 3 & 6 & 4 \end{matrix}) \in Z_{7}^{3 \times 3} \\ S_{3} = \begin{array}{cc} 123 & + \\ 132 & - \\ 213 & - \\ 231 & + \\ 312 & + \\ 321 & - \end{array} \\ | A | = \sum_{σ \in S_{3}} s g n (σ) \prod_{i = 1}^{3} a_{i σ (i)} = \\ = a_{11} a_{22} a_{33} - a_{11} a_{23} a_{32} - a_{12} a_{21} a_{33} + a_{12} a_{23} a_{31} + a_{13} a_{21} a_{32} - a_{13} a_{22} a_{31} = \\ = 2 \cdot 1 \cdot 4 - 2 \cdot 0 \cdot 6 - 6 \cdot 5 \cdot 4 + 6 \cdot 0 \cdot 3 + 3 \cdot 5 \cdot 6 - 3 \cdot 1 \cdot 3 = \\ = 1 - 0 - 1 + 0 + 6 - 2 = 4 \\ ⟹ | A | = 4 \end{matrix}

2a

\begin{matrix} Calculate determinant using minors \\ A = (\begin{matrix} 1 & 2 & 0 & 5 \\ 0 & 3 & 1 & 8 \\ 4 & - 1 & 6 & 9 \\ 0 & 7 & 2 & 3 \end{matrix}) \in R^{4 \times 4} \\ Let us decompose the determinant into minors by first column \\ | A | = \sum_{i = 1}^{4} (- 1)^{i + 1} a_{i 1} | M_{i 1} (A) | = \\ = 1 | M_{11} (A) | - 0 | M_{12} (A) | + 4 | M_{13} (A) | - 0 | M_{14} (A) | \\ | M_{11} (A) | = | \begin{matrix} 3 & 1 & 8 \\ - 1 & 6 & 9 \\ 7 & 2 & 3 \end{matrix} | = 54 + 63 - 16 + 3 - 54 - 336 = - 286 \\ | M_{13} (A) | = | \begin{matrix} 2 & 0 & 5 \\ 3 & 1 & 8 \\ 7 & 2 & 3 \end{matrix} | = 6 + 0 + 30 - 0 - 32 - 35 = - 31 \\ | A | = 1 (- 286) + 4 (- 31) = - 410 \end{matrix}

2b

\begin{matrix} Calculate determinant using minors \\ A = (\begin{matrix} 1 & 3 & 0 & 2 \\ 1 & 2 & i & 8 \\ 2 i & 0 & 0 & - 2 \\ 6 & 0 & 4 & 3 \end{matrix}) \in C^{4 \times 4} \\ Let us decompose the determinant into minors by third row \\ | A | = \sum_{j = 1}^{n} (- 1)^{3 + j} a_{3 j} | M_{3 j} (A) | = \\ = 2 i | M_{31} (A) | - 0 | M_{32} (A) | + 0 | M_{33} (A) | - (- 2) | M_{34} (A) | = \\ | M_{31} (A) | = | \begin{matrix} 3 & 0 & 2 \\ 2 & i & 8 \\ 0 & 4 & 3 \end{matrix} | = 9 i + 16 - 96 = 9 i - 80 \\ | M_{34} (A) | = | \begin{matrix} 1 & 3 & 0 \\ 1 & 2 & i \\ 6 & 0 & 4 \end{matrix} | = 8 + 18 i - 12 = 18 i - 4 \\ | A | = 2 i (9 i - 80) + 2 (18 i - 4) = - 18 - 160 i + 36 i - 8 = - 26 - 124 i \\ | A | = - 26 - 124 i \end{matrix}

3a

\begin{matrix} Calculate determinant using Gaussian elimination \\ A = (\begin{matrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 1 \\ 1 & 3 & 5 & 0 \\ 2 & 3 & 0 & 33 \end{matrix}) \in R^{4 \times 4} \\ (\begin{matrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 1 \\ 1 & 3 & 5 & 0 \\ 2 & 3 & 0 & 33 \end{matrix}) \overset{R_{3} - R_{1}}{\underset{R_{4} - 2 R_{1}}{\to}} (\begin{matrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 1 \\ 0 & 1 & 2 & - 4 \\ 0 & - 1 & - 6 & 25 \end{matrix}) \overset{R_{3} - R_{2}}{\underset{R_{4} + R_{2}}{\to}} (\begin{matrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & - 5 \\ 0 & 0 & - 4 & 26 \end{matrix}) \\ \overset{R_{3} \leftrightarrow R_{4}}{\to} (\begin{matrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & - 4 & 26 \\ 0 & 0 & 0 & - 5 \end{matrix}) = A_{U} ⟹ | A_{U} | = 20 \\ | A | = - | A_{U} | = - 20 \end{matrix}

3b

\begin{matrix} Calculate determinant using Gaussian elimination \\ A = (\begin{matrix} 1 & 3 & 5 \\ 7 & 9 & 11 \\ 8 & 12 & 3 \end{matrix}) \in Z_{13}^{3 \times 3} \\ (\begin{matrix} 1 & 3 & 5 \\ 7 & 9 & 11 \\ 8 & 12 & 3 \end{matrix}) \overset{R_{2} + 6 R_{1}}{\underset{R_{3} + 5 R_{1}}{\to}} (\begin{matrix} 1 & 3 & 5 \\ 0 & 1 & 2 \\ 0 & 1 & 2 \end{matrix}) \overset{R_{3} + (- 1) R_{2}}{\to} (\begin{matrix} 1 & 3 & 5 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{matrix}) = A_{U} \\ ⟹ | A_{U} | = 0 \\ ⟹ | A | = | A_{U} | = 0 \end{matrix}

4a

\begin{matrix} Prove or disprove: det (A + B) = det (A) + det (B) \\ Disproof: \\ A = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}), B = (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}) \\ det (A + B) = 1 \\ det (A) = det (B) = 0 \\ det (A + B) \neq det (A) + det (B) \end{matrix}

4b

\begin{matrix} Let A \in C^{n \times n} \\ Prove or disprove: \overset{―}{det (A)} = det (\overset{―}{A^{T}}) \\ Proof: \\ \overset{―}{(a + b i) + (c + d i)} = \overset{―}{a + c + (b + d) i} = \\ = a + c - (b + d) i = a - b i + c - d i = \overset{―}{a + b i} + \overset{―}{c + d i} \\ \overset{―}{(a + b i) (c + d i)} = \overset{―}{a c - b d + (a d + b c) i} = \\ = a c - b d - (a d + b c) i = a (c - d i) - b (d + c i) = a (c - d i) + b i (d i - c) = \\ = (a - b i) (c - d i) = \overset{―}{a + b i} \cdot \overset{―}{c + d i} \\ \overset{―}{det (A)} = \overset{―}{\sum_{σ \in S_{n}} s g n (σ) \prod_{i = 1}^{n} a_{i σ (i)}} = \sum_{σ \in S_{n}} \overset{―}{s g n (σ) \prod_{i = 1}^{n} a_{i σ (i)}} = \\ = \sum_{σ \in S_{n}} s g n (σ) \prod_{i = 1}^{n} \overset{―}{a_{i σ (i)}} = \sum_{σ \in S_{n}} s g n (σ) \prod_{i = 1}^{n} (\overset{―}{A^{T}})_{σ (i) i} \\ Let i < j : \\ σ (i) > σ (j) ⟺ σ^{- 1} (σ (i)) = i < j = σ^{- 1} (σ (j)) \\ ⟹ s g n (σ) = s g n (σ^{- 1}) \\ ⟹ \sum_{σ \in S_{n}} s g n (σ) \prod_{i = 1}^{n} (\overset{―}{A^{T}})_{σ (i) i} = \sum_{σ^{- 1} \in S_{n}} s g n (σ^{- 1}) \prod_{j = 1}^{n} (\overset{―}{A^{T}})_{j σ^{- 1} (j)} = det (\overset{―}{A^{T}}) \\ ⟹ \overset{―}{det (A)} = det (\overset{―}{A^{T}}) \end{matrix}

4c

\begin{matrix} Let A \in F^{m \times n} \\ Prove or disprove: det (A A^{T}) = det (A^{T} A) \\ Disproof: \\ A = (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix}) \\ A^{T} = (\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{matrix}) \\ A A^{T} = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) \\ A^{T} A = (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix}) \\ det (A A^{T}) = 1 \neq 0 = det (A^{T} A) \end{matrix}

4d

\begin{matrix} Let A \in F^{n \times n} \\ Prove or disprove: \forall i, j \in [1, n] : | M_{i j} (A) | = 0 ⟹ r a n k (A) \leq n - 2 \\ Disproof for n = 1 : \\ n = 1 ⟹ r a n k (A) \geq 0 = n - 1 \\ Proof for n \geq 2 : \\ Let r a n k (A) = n \\ ⟹ | A | \neq 0 ⟹ \exists i, j \in [1, n] : | M_{i j} (A) | \neq 0 - Contradiction! \\ Let r a n k (A) = n - 1 \\ ⟹ There exists exactly one column that is linearly dependant on the others \\ Let C_{k} (A) be linearly dependant on the others \\ Let A_{k} = (\begin{matrix} \overset{|}{\underset{|}{C_{1} (A)}} & \dots & \overset{|}{\underset{|}{C_{k - 1} (A)}} & \overset{|}{\underset{|}{C_{k + 1} (A)}} & \dots & \overset{|}{\underset{|}{C_{n} (A)}} \end{matrix}) \\ C (A) = s p ({C_{i} (A)}_{i \in [1, n]}) = s p ({C_{i} (A)}_{i \in [1, n]} ∖ {C_{k} (A)}) = C (A_{k}) \\ ⟹ r a n k (A_{k}) = n - 1 \\ A \in F^{n \times n - 1} ⟹ There exists exactly one row that is linearly dependant on others \\ Let R_{j} (A_{k}) be linearly dependant on the others \\ Let A_{j k} = (\begin{matrix} - R_{1} (A_{k}) - \\ ⋮ \\ - R_{j - 1} (A_{k}) - \\ - R_{j + 1} (A_{k}) - \\ ⋮ \\ - R_{n} (A_{k}) - \end{matrix}) \\ R (A_{k}) = s p ({R_{i} (A_{k})}_{i \in [1, n]} ∖ {R_{j} (A_{k})}) = R (A_{j k}) \\ ⟹ r a n k (A_{j k}) = n - 1 \\ A_{j k} \in F^{n - 1 \times n - 1} ⟹ | A_{j k} | \neq 0 \\ A_{j k} = M_{j k} (A) ⟹ | M_{j k} (A) | \neq 0 - Contradiction! \\ ⟹ r a n k (A) \leq n - 2 \end{matrix}

5a

\begin{matrix} Prove: ∄ A \in R^{3 \times 3} : A^{2} = (\begin{matrix} 1 & 0 & 1 \\ 0 & - 1 & 0 \\ 0 & 0 & 1 \end{matrix}) \\ Proof: \\ Let \exists A \in R^{3 \times 3} : A^{2} = (\begin{matrix} 1 & 0 & 1 \\ 0 & - 1 & 0 \\ 0 & 0 & 1 \end{matrix}) \\ | A^{2} | = | \begin{matrix} 1 & 0 & 1 \\ 0 & - 1 & 0 \\ 0 & 0 & 1 \end{matrix} | = 1 \cdot | \begin{matrix} - 1 & 0 \\ 0 & 1 \end{matrix} | = - 1 \\ | A^{2} | = {| A |}^{2} \geq 0 \in R \\ {| A |}^{2} = - 1 - Contradiction! \\ ⟹ ∄ A \in R^{3 \times 3} : A^{2} = (\begin{matrix} 1 & 0 & 1 \\ 0 & - 1 & 0 \\ 0 & 0 & 1 \end{matrix}) \end{matrix}

5b

\begin{matrix} Is there such matrix in C^{3 \times 3} ? \\ Solution: \\ A = (\begin{matrix} 1 & 0 & \frac{1}{2} \\ 0 & - i & 0 \\ 0 & 0 & 1 \end{matrix}) ⟹ A^{2} = (\begin{matrix} 1 & 0 & 1 \\ 0 & - 1 & 0 \\ 0 & 0 & 1 \end{matrix}) \end{matrix}

6

\begin{matrix} Let A \in R^{n \times n}, a_{i j} = {\begin{matrix} 1 & j \equiv 2 i - 1 (\mod n) \\ 0 & otherwise \end{matrix} \end{matrix}

6a

\begin{matrix} Find all n for which det (A) = 0 \\ Solution: \\ 1 \leq i \leq n ⟹ 2 \leq 2 i \leq 2 n < 2 n + 1 \\ j = n ⟺ j \equiv 0 \mod n \\ (2 i - 1) \equiv 0 \mod n ⟺ 2 i = n + 1 ⟺ {\begin{matrix} n is odd \\ i = \frac{n + 1}{2} \end{matrix} \\ ⟹ [n is even ⟹ C_{n} (A) = 0 ⟹ det (A) = 0] \\ Let n be odd \\ Let j \in [1, n] \\ ⟹ j (\mod n) \in [0, n - 1] \\ 2 i - 1 \in {1, 3, 5, \dots, n, n + 2, n + 4, n + (n - 1)} \\ ⟹ 2 i - 1 (\mod n) \in [0, n - 1] \\ ⟹ \exists i \in [1, n] : j \equiv 2 i - 1 (\mod n) \\ Let \exists i_{1} \neq i_{2} \in [1, n] : {\begin{matrix} j \equiv 2 i_{1} - 1 (\mod n) \\ j \equiv 2 i_{2} - 1 (\mod n) \end{matrix} \\ Let i_{1} < i_{2} WLOG \\ 2 i_{1} - 1 \equiv 2 i_{2} - 1 (\mod n) \\ ⟹ 2 i_{1} - 1 = 2 i_{2} - 1 + k n ⟹ 2 i_{2} = 2 i_{1} + k n, k \in N_{0} \\ i_{1} \neq i_{2} ⟹ 2 i_{1} \neq 2 i_{2} ⟹ k \neq 0 \\ 2 (i_{2} - i_{1}) > 0 ⟹ k n > 0 ⟹ k > 0 \\ 2 i_{2} = 2 i_{1} + k n ⟹ k n is even ⟹ k \geq 2 \\ Let k = 2 a, a \in N \\ i_{2} = i_{1} + a n \\ ⟹ n + 1 \leq i_{1} + a n \leq n - Contradiction! \\ ⟹ \exists! i \in [1, n] : j \equiv 2 i - 1 (\mod n) \\ ⟹ In each column there is exactly one 1 \\ Let j_{1} \equiv 2 i_{1} - 1 (\mod n) \\ Let j_{2} \equiv 2 i_{2} - 1 (\mod n) \\ j_{1} \neq j_{2} ⟹ j_{1} ≢ j_{2} (\mod n) \\ ⟹ 2 i_{1} - 1 ≢ 2 i_{2} - 1 (\mod n) ⟹ i_{1} \neq i_{2} \\ ⟹ In each row there is exacly one 1 \\ ⟹ det (A) = \pm 1 \\ ⟹ det (A) = 0 ⟺ n is even \end{matrix}

6b

\begin{matrix} Calculate det (A) for n = 3, 5, 7, 9 \\ Solution: \\ Let us find all positions where 1 ’s are \\ And calculate the sign of permutation for these positions \\ According to 6a, determinant is going to be (- 1) or 1 based on this sign \\ n = 3 ⟹ {\begin{matrix} i = 1 ⟹ j = 1 \\ i = 2 ⟹ j = 3 \\ i = 3 ⟹ j = 2 \end{matrix} ⟹ 1, 3, 2 ⟹ det (A) = - 1 \\ n = 5 ⟹ {\begin{matrix} i = 1 ⟹ j = 1 \\ i = 2 ⟹ j = 3 \\ i = 3 ⟹ j = 5 \\ i = 4 ⟹ j = 2 \\ i = 5 ⟹ j = 4 \end{matrix} ⟹ 1, 3, 5, 2, 4 ⟹ det (A) = - 1 \\ n = 7 ⟹ {\begin{matrix} i = 1 ⟹ j = 1 \\ i = 2 ⟹ j = 3 \\ i = 3 ⟹ j = 5 \\ i = 4 ⟹ j = 7 \\ i = 5 ⟹ j = 2 \\ i = 6 ⟹ j = 4 \\ i = 7 ⟹ j = 6 \end{matrix} ⟹ 1, 3, 5, 7, 2, 4, 6 ⟹ det (A) = 1 \\ n = 9 ⟹ {\begin{matrix} i = 1 ⟹ j = 1 \\ i = 2 ⟹ j = 3 \\ i = 3 ⟹ j = 5 \\ i = 4 ⟹ j = 7 \\ i = 5 ⟹ j = 9 \\ i = 6 ⟹ j = 2 \\ i = 7 ⟹ j = 4 \\ i = 8 ⟹ j = 6 \\ i = 9 ⟹ j = 8 \end{matrix} ⟹ 1, 3, 5, 7, 9, 2, 4, 6, 8 ⟹ det (A) = 1 \end{matrix}

6 bonus

\begin{matrix} Prove: det (A) = 1 ⟺ [\begin{matrix} 8 ∣ (n - 1) \\ 8 ∣ (n + 1) \end{matrix} \\ Proof: \\ n is even ⟺ det (A) = 0 \\ ⟹ det (A) \neq 0 ⟺ n is odd \\ Let n be odd \\ Let f (i) = 2 i - 1 (\mod n) \\ I m (f) = [0, n - 1] ⟹ f is not a permutation of [1, n] \\ \forall i \in [1, n] : f (i) \neq n \\ Let σ_{0} (i) = {\begin{matrix} n & 2 i - 1 (\mod n) = 0 \\ 2 i - 1 (\mod n) & otherwise \end{matrix} \\ σ_{0} (i) = {\begin{matrix} 2 i - 1 & i < \frac{n + 1}{2} \\ n & i = \frac{n + 1}{2} \\ 2 i - 1 - n & i > \frac{n + 1}{2} \end{matrix} \\ 0 \equiv n \equiv 2 \frac{n + 1}{2} - 1 (\mod n) ⟹ σ_{0} is a permutation of [1, n] \\ As proved in 6a, there is exactly one 1 in each column and row of A \\ ⟹ σ_{0} is the only permutation that yields a non-zero product \\ ⟹ det (A) = \sum_{σ \in S_{n}} s g n (σ) \prod_{i = 1}^{n} a_{i σ (i)} = s g n (σ_{0}) \underset{\prod_{i = 1}^{n} 1 = 1}{\underset{⏟}{\prod_{i = 1}^{n} a_{i σ_{0} (i)}}} = s g n (σ_{0}) \\ ⟹ It is enough to prove s g n (σ_{0}) = 1 ⟺ [\begin{matrix} 8 ∣ (n - 1) \\ 8 ∣ (n + 1) \end{matrix} \end{matrix}

\begin{matrix} Let k = \frac{n + 1}{2} \\ {\begin{matrix} σ_{0} (1) = 1 \\ σ_{0} (2) = 3 \\ ⋮ \\ σ_{0} (k - 1) = 2 (k - 1) - 1 = 2 k - 3 = n - 2 \\ σ_{0} (k) = n \\ σ_{0} (k + 1) = 2 (k + 1) - 1 - n = 2 k + 1 - n = n + 2 - n = 2 \\ σ_{0} (k + 2) = 2 (k + 2) - 1 - n = 2 k + 3 - n = n + 4 - n = 4 \\ ⋮ \\ σ_{0} (n) = 2 n - 1 - n = n - 1 \end{matrix} \\ ⟹ σ_{0} = {1, 3, 5, \dots, n, 2, 4, \dots, n - 1} \\ Let us count the number of inversions in σ_{0} \\ There are k values in {1, 3, 5, \dots, n} \\ There are k - 1 values in {2, 4, \dots, n - 1} \\ \forall i \in [2, k] : k + 1 > i and σ (i) \geq 3 > 2 = σ (k + 1) ⟹ k - 1 inversions \\ \forall i \in [3, k] : k + 2 > i and σ (i) \geq 5 > 4 = σ (k + 2) ⟹ k - 2 inversions \\ \dots \\ Other than these, there are no inversions as "halves" of permutation are ordered \\ ⟹ Total number of inversions is \\ P = \sum_{i = 1}^{k - 1} (k - i) = (k - 1) + (k - 2) + \dots + (k - (k - 1)) = \frac{k (k - 1)}{2} \\ s g n (σ_{0}) = (- 1)^{P} \\ ⟹ s g n (σ_{0}) = 1 ⟺ P is even ⟺ k (k - 1) \equiv 0 \mod 4 \\ ⟺ [\begin{matrix} k \equiv 0 \mod 4 \\ k \equiv 1 \mod 4 \end{matrix} \\ Let k \equiv 0 \mod 4 ⟺ k = 4 m \\ ⟺ n = 2 k - 1 = 8 m - 1 ⟺ n + 1 = 8 m ⟺ 8 ∣ (n + 1) \\ Let k \equiv 1 \mod 4 ⟺ k = 4 m + 1 \\ ⟺ n = 8 m + 2 - 1 = 8 m + 1 ⟺ n - 1 = 8 m ⟺ 8 ∣ (n - 1) \end{matrix}

7

\begin{matrix} Let n \geq 2 \\ Let A \in R^{n \times n} \\ Let \forall i, j \in [1, n] : a_{i j} \in {1, - 1} \\ Prove: det (A) is even \\ Proof: \\ Base case. n = 2 \\ det (A) = a_{11} a_{22} - a_{12} a_{21} = (- 1)^{x} + (- 1)^{y} \in {- 2, 0, 2} ⟹ det (A) is even \\ Induction step. Let \forall B \in R^{n \times n} : det (B) is even \\ Let C \in R^{n + 1 \times n + 1} \\ det (C) = \sum_{j = 1}^{n} (- 1)^{j + 1} c_{1 j} det (M_{1 j} (C)) \\ \forall j \in [1, n] : M_{1 j} (C) \in R^{n \times n} and c_{i j} = \pm 1 \\ ⟹ (- 1)^{j + 1} c_{1 j} det (M_{1 j} (C)) = \pm det (M_{1 j} (C)) is even \\ Sum of even numbers is even ⟹ det (C) is even \\ ⟹ By induction: \forall A \in R^{n \times n} : det (A) is even \end{matrix}

8

\begin{matrix} Find the determinant of Vandermonde matrix \\ A = (\begin{matrix} 1 & α_{1} & α_{1}^{2} & \dots & α_{1}^{n - 1} \\ 1 & α_{2} & α_{2}^{2} & \dots & α_{2}^{n - 1} \\ ⋮ & ⋮ & ⋮ & ⋱ & ⋮ \\ ⋮ & ⋮ & ⋮ & ⋱ & ⋮ \\ 1 & α_{n} & α_{n}^{2} & \dots & α_{n}^{n - 1} \end{matrix}) \\ Solution: \\ Let A_{n} = A of size n \\ (\begin{matrix} 1 & α_{1} & α_{1}^{2} & \dots & α_{1}^{n - 1} \\ 1 & α_{2} & α_{2}^{2} & \dots & α_{2}^{n - 1} \\ ⋮ & ⋮ & ⋮ & ⋱ & ⋮ \\ ⋮ & ⋮ & ⋮ & ⋱ & ⋮ \\ 1 & α_{n} & α_{n}^{2} & \dots & α_{n}^{n - 1} \end{matrix}) \overset{\forall i \in [2, n] : R_{i} - R_{1}}{\to} (\begin{matrix} 1 & α_{1} & α_{1}^{2} & \dots & α_{1}^{n - 1} \\ 0 & α_{2} - α_{1} & α_{2}^{2} - α_{1}^{2} & \dots & α_{2}^{n - 1} - α_{1}^{n - 1} \\ ⋮ & ⋮ & ⋮ & ⋱ & ⋮ \\ ⋮ & ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & α_{n} - α_{1} & α_{n}^{2} - α_{1}^{2} & \dots & α_{n}^{n - 1} - α_{1}^{n - 1} \end{matrix}) \\ \overset{\forall i \in {n, \dots, 2} : C_{i} - α_{1} C_{i - 1}}{\to} (\begin{matrix} 1 & 0 & 0 & \dots & 0 \\ 0 & α_{2} - α_{1} & α_{2} (α_{2} - α_{1}) & \dots & α_{2}^{n - 2} (α_{2} - α_{1}) \\ ⋮ & ⋮ & ⋮ & ⋱ & ⋮ \\ ⋮ & ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & α_{n} - α_{1} & α_{n} (α_{n} - α_{1}) & \dots & α_{n}^{n - 2} (α_{n} - α_{1}) \end{matrix}) \\ \exists i \in [2, n] : α_{i} = α_{1} ⟹ R_{i} (A_{n}) = 0 ⟹ det (A_{n}) = 0 \\ Let \forall i \in [2, n] : α_{i} \neq α_{1} \\ \overset{\forall i \in [2, n] : \frac{1}{α_{i} - α_{1}} R_{i}}{\to} (\begin{matrix} 1 & 0 & 0 & \dots & 0 \\ 0 & 1 & α_{2} & \dots & α_{2}^{n - 2} \\ ⋮ & ⋮ & ⋮ & ⋱ & ⋮ \\ ⋮ & ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 1 & α_{n} & \dots & α_{n}^{n - 2} \end{matrix}) \end{matrix}

\begin{matrix} ⟹ det (A_{n}) = (\prod_{i = 2}^{n} (α_{i} - α_{1})) det (M_{11} (A_{n})) = (\prod_{i = 2}^{n} (α_{i} - α_{1})) det (A_{n - 1}) = \\ = (\prod_{i_{1} = 2}^{n} (α_{i_{1}} - α_{1})) \cdot (\prod_{i_{2} = 3}^{n} (α_{i_{2}} - α_{2})) \cdot \dots \cdot (\prod_{i_{n - 1} = n}^{n} (α_{i_{n - 1}} - α_{n - 1})) \cdot \underset{1}{\underset{⏟}{det (A_{1})}} = \\ = \prod_{1 \leq i < j \leq n} (α_{j} - α_{i}) this formula also includes all cases when det (A_{n}) = 0 \end{matrix}