hcju/proofwikiqa huggingface dataset:
Let $\map e {\mathbf A}$ be the elementary row operation: :$e := r_k \to \lambda r_k$ where $\lambda \ne 0$. Then $r'_k$ is such that: :$\forall a'_{k i} \in r'_k: a'_{k i} = \lambda a_{k i}$ Now let $\map {e'} {\mathbf A'}$ be the elementary row operation which transforms $\mathbf A'$ to $\mathbf A''$: :$e' := r_k \to \dfrac 1 \lambda r_k$ Because it is stipulated in the definition of an elementary row operation that $\lambda \ne 0$, it follows by definition of a field that $\dfrac 1 \lambda$ exists. Hence $e'$ is defined. So applying $e'$ to $\mathbf A'$ we get: {{begin-eqn}} {{eqn | lo= \forall a''_{k i} \in r''_k:       | l = a''_{k i}       | r = \dfrac 1 \lambda a'_{k i}       | c =  }} {{eqn | r = \dfrac 1 \lambda \paren {\lambda a_{k i} }       | c =  }} {{eqn | r = a_{k i}       | c =  }} {{eqn | ll= \leadsto       | lo= \forall a''_{k i} \in r''_k:       | l = a''_{k i}       | r = a_{k i}       | c =  }} {{eqn | ll= \leadsto       | l = r''_k       | r = r_k       | c =  }} {{eqn | ll= \leadsto       | l = \mathbf A''       | r = \mathbf A       | c =  }} {{end-eqn}} It is noted that for $e'$ to be an elementary row operation, the only possibility is for it to be as defined.
Let $\map e {\mathbf A}$ be the elementary row operation: :$e := r_k \to r_k + \lambda r_l$ Then $r'_k$ is such that: :$\forall a'_{k i} \in r'_k: a'_{k i} = a_{k i} + \lambda a_{l i}$ Now let $\map {e'} {\mathbf A'}$ be the elementary row operation which transforms $\mathbf A'$ to $\mathbf A''$: :$e' := r'_k \to r'_k - \lambda r'_l$ Applying $e'$ to $\mathbf A'$ we get: {{begin-eqn}} {{eqn | lo= \forall a''_{k i} \in r''_k:       | l = a''_{k i}       | r = a'_{k i} - \lambda a'_{l i}       | c =  }} {{eqn | r = \paren {a_{k i} + \lambda a_{l i} } - \lambda a'_{l i}       | c =  }} {{eqn | r = \paren {a_{k i} + \lambda a_{l i} } - \lambda a_{l i}       | c = as $\lambda a'_{l i} = \lambda a_{l i}$: row $l$ was not changed by $e$ }} {{eqn | r = a_{k i}       | c =  }} {{eqn | ll= \leadsto       | lo= \forall a''_{k i} \in r''_k:       | l = a''_{k i}       | r = a_{k i}       | c =  }} {{eqn | ll= \leadsto       | l = r''_{k i}       | r = r_{k i}       | c =  }} {{eqn | ll= \leadsto       | l = \mathbf A''       | r = \mathbf A       | c =  }} {{end-eqn}} It is noted that for $e'$ to be an elementary row operation, the only possibility is for it to be as defined.
Let $\map e {\mathbf A}$ be the elementary row operation: :$e := r_k \leftrightarrow r_l$ Thus we have: {{begin-eqn}} {{eqn | l = r'_k       | r = r_l       | c =  }} {{eqn | lo= \text {and}       | l = r'_l       | r = r_k       | c =  }} {{end-eqn}} Now let $\map {e'} {\mathbf A'}$ be the elementary row operation which transforms $\mathbf A'$ to $\mathbf A''$: :$e' := r'_k \leftrightarrow r'_l$ Applying $e'$ to $\mathbf A'$ we get: {{begin-eqn}} {{eqn | l = r''_k       | r = r'_l       | c =  }} {{eqn | lo= \text {and}       | l = r''_l       | r = r'_k       | c =  }} {{eqn | ll= \leadsto       | l = r''_k       | r = r_k       | c =  }} {{eqn | lo= \text {and}       | l = r''_l       | r = r_l       | c =  }} {{eqn | ll= \leadsto       | l = \mathbf A''       | r = \mathbf A       | c =  }} {{end-eqn}} It is noted that for $e'$ to be an elementary row operation, the only possibility is for it to be as defined.
We will demonstrate this for each of the $3$ types of elementary row operation. Let $\sim$ denote row equivalence. Let $\mathbf I$ denote the unit matrix. === $\text {ERO} 1$: Scalar Product of Row === Let $e_1$ be the elementary row operation: :$e_1 := r_i \to \lambda r_i$ for an arbitrary row $r_i$ of $\mathbf I$. Let $\mathbf E_1$ be the elementary row matrix created by applying $e_1$ to $\mathbf I$. From Existence of Inverse Elementary Row Operation: Scalar Product of Row, the inverse $e_1'$ of $e_1$ is given by: :$e_1' := r_i \to \dfrac 1 \lambda r_i$ Thus applying $e_1'$ to $\mathbf E_1$ transforms $\mathbf E_1$ back to $\mathbf I$. From Elementary Row Operations as Matrix Multiplications, for every elementary row operation there exists a corresponding elementary row matrix.  Let $\mathbf E_1'$ be the elementary row matrix created by applying $e_1'$ to $\mathbf I$. Then: :$\mathbf E_1 \mathbf E_1' = \mathbf E_1' \mathbf E_1 = \mathbf I$ === $\text {ERO} 2$: Add Scalar Product of Row to Another === Let $e_2$ be the elementary row operation: :$e_2 := r_i \to r_i + \lambda r_j$ for arbitrary row $r_i$ and $r_j$ of $\mathbf I$ such that $i \ne j$. Let $\mathbf E_2$ be the elementary row matrix created by applying $e_2$ to $\mathbf I$. From Existence of Inverse Elementary Row Operation: Add Scalar Product of Row to Another, the inverse $e_2'$ of $e_2$ is given by: :$e_2' := r_i \to r_i - \lambda r_j$ Thus applying $e_2'$ to $\mathbf E_2$ transforms $\mathbf E_2$ back to $\mathbf I$. From Elementary Row Operations as Matrix Multiplications, for every elementary row operation there exists a corresponding elementary row matrix.  Let $\mathbf E_2'$ be the elementary row matrix created by applying $e_2'$ to $\mathbf I$. Then: :$\mathbf E_2 \mathbf E_2' = \mathbf E_2' \mathbf E_2 = \mathbf I$ === $\text {ERO} 3$: Exchange Rows === Let $e_3$ be the elementary row operation: :$e_3 := r_i \leftrightarrow r_j$ for arbitrary row $r_i$ and $r_j$ of $\mathbf I$ such that $i \ne j$. Let $\mathbf E_3$ be the elementary row matrix created by applying $e_3$ to $\mathbf I$. From Existence of Inverse Elementary Row Operation: Exchange Rows, the inverse $e_3'$ of $e_3$ is $e_3$ itself: :$e_3' := r_i \leftrightarrow r_j = e_3$ Thus applying $e_3$ to $\mathbf E_3$ transforms $\mathbf E_3$ back to $\mathbf I$. :$\mathbf E_3 \mathbf E_3 = \mathbf I$ Hence the result, from Proof by Cases. {{qed}}
Let $\map e {\mathbf A}$ be the elementary column operation: :$e := \kappa_k \to \lambda \kappa_k$ where $\lambda \ne 0$. Then $\kappa'_k$ is such that: :$\forall a'_{k i} \in \kappa'_k: a'_{k i} = \lambda a_{k i}$ Now let $\map {e'} {\mathbf A'}$ be the elementary column operation which transforms $\mathbf A'$ to $\mathbf A''$: :$e' := \kappa_k \to \dfrac 1 \lambda \kappa_k$ Because it is stipulated in the definition of an elementary column operation that $\lambda \ne 0$, it follows by definition of a field that $\dfrac 1 \lambda$ exists. Hence $e'$ is defined. So applying $e'$ to $\mathbf A'$ we get: {{begin-eqn}} {{eqn | lo= \forall a''_{i k} \in \kappa''_k:       | l = a''_{i k}       | r = \dfrac 1 \lambda a'_{i k}       | c =  }} {{eqn | r = \dfrac 1 \lambda \paren {\lambda a_{i k} }       | c =  }} {{eqn | r = a_{i k}       | c =  }} {{eqn | ll= \leadsto       | lo= \forall a''_{i k} \in \kappa''_k:       | l = a''_{i k}       | r = a_{i k}       | c =  }} {{eqn | ll= \leadsto       | l = \kappa''_k       | r = \kappa_k       | c =  }} {{eqn | ll= \leadsto       | l = \mathbf A''       | r = \mathbf A       | c =  }} {{end-eqn}} It is noted that for $e'$ to be an elementary column operation, the only possibility is for it to be as defined.
Let $\map e {\mathbf A}$ be the elementary column operation: :$e := \kappa_k \to \kappa_k + \lambda r_l$ Then $\kappa'_k$ is such that: :$\forall a'_{i k} \in \kappa'_k: a'_{i k} = a_{i k} + \lambda a_{i l}$ Now let $\map {e'} {\mathbf A'}$ be the elementary column operation which transforms $\mathbf A'$ to $\mathbf A''$: :$e' := \kappa'_k \to \kappa'_k - \lambda \kappa'_l$ Applying $e'$ to $\mathbf A'$ we get: {{begin-eqn}} {{eqn | lo= \forall a''_{i k} \in \kappa''_k:       | l = a''_{i k}       | r = a'_{i k} - \lambda a'_{i l}       | c =  }} {{eqn | r = \paren {a_{i k} + \lambda a_{i l} } - \lambda a'_{i l}       | c =  }} {{eqn | r = \paren {a_{i k} + \lambda a_{i l} } - \lambda a_{i l}       | c = as $\lambda a'_{i l} = \lambda a_{i l}$: column $l$ was not changed by $e$ }} {{eqn | r = a_{i k}       | c =  }} {{eqn | ll= \leadsto       | lo= \forall a''_{i k} \in \kappa''_k:       | l = a''_{i k}       | r = a_{i k}       | c =  }} {{eqn | ll= \leadsto       | l = \kappa''_{i k}       | r = \kappa_{i k}       | c =  }} {{eqn | ll= \leadsto       | l = \mathbf A''       | r = \mathbf A       | c =  }} {{end-eqn}} It is noted that for $e'$ to be an elementary column operation, the only possibility is for it to be as defined.
Let $\map e {\mathbf A}$ be the elementary column operation: :$e := \kappa_k \leftrightarrow \kappa_l$ Thus we have: {{begin-eqn}} {{eqn | l = \kappa'_k       | r = \kappa_l       | c =  }} {{eqn | lo= \text {and}       | l = \kappa'_l       | r = \kappa_k       | c =  }} {{end-eqn}} Now let $\map {e'} {\mathbf A'}$ be the elementary column operation which transforms $\mathbf A'$ to $\mathbf A''$: :$e' := \kappa'_k \leftrightarrow \kappa'_l$ Applying $e'$ to $\mathbf A'$ we get: {{begin-eqn}} {{eqn | l = \kappa''_k       | r = \kappa'_l       | c =  }} {{eqn | lo= \text {and}       | l = \kappa''_l       | r = \kappa'_k       | c =  }} {{eqn | ll= \leadsto       | l = \kappa''_k       | r = \kappa_k       | c =  }} {{eqn | lo= \text {and}       | l = \kappa''_l       | r = \kappa_l       | c =  }} {{eqn | ll= \leadsto       | l = \mathbf A''       | r = \mathbf A       | c =  }} {{end-eqn}} It is noted that for $e'$ to be an elementary column operation, the only possibility is for it to be as defined.
Let $s, t \in \closedint 1 m$ such that $s \ne t$. === Case $1$ === Let $e$ be the elementary column operation $\kappa_s \to \lambda \kappa_s$: :$E_{k i} = \begin{cases} \delta_{k i} & : i \ne s \\ \lambda \delta_{k i} & : i = s \end{cases}$ where $\delta$ denotes the Kronecker delta. Then: {{begin-eqn}} {{eqn | l = \sqbrk {A E}_{i j}       | r = \sum_{k \mathop = 1}^m A_{j k} E_{k i}        | c =  }} {{eqn | r = \begin {cases} A_{j i} & : i \ne r \\ \lambda A_{j i} & : i = r \end{cases}       | c =  }} {{eqn | ll= \leadsto       | l = \mathbf {A E}       | r = e \paren {\mathbf A}       | c =  }} {{end-eqn}} {{qed|lemma}} === Case $2$ === Let $e$ be the elementary column operation $\kappa_s \to \kappa_s + \lambda \kappa_t$: :$E_{k i} = \begin {cases} \delta_{k i} & : i \ne s \\ \delta_{k s} + \lambda \delta_{k t} & : i = s \end {cases}$ where $\delta$ denotes the Kronecker delta. Then: {{begin-eqn}} {{eqn | l = \sqbrk {A E}_{j i}       | r = \sum_{k \mathop = 1}^m  A_{j k} E_{k i}       | c =  }} {{eqn | r = \begin {cases} A_{j i} & : j \ne s \\ A_{i j} + \lambda A_{j t} & : i = s \end {cases}       | c =  }} {{eqn | ll= \leadsto       | l = \mathbf {A E}       | r = e \paren {\mathbf A}       | c =  }} {{end-eqn}} {{qed|lemma}} === Case $3$ === Let $e$ be the elementary column operation $\kappa_s \leftrightarrow \kappa_t$: By Exchange of Columns as Sequence of Other Elementary Column Operations, this elementary column operation can be expressed as: :$\paren {e_1 e_2 e_3 e_4 \mathbf A} = e \paren {\mathbf A}$ where the $e_i$ are elementary column operation of the other two types. For each $e_i$, let $\mathbf E_i = e_i \paren {\mathbf I}$. Then: {{begin-eqn}} {{eqn | l = e \paren {\mathbf A}       | r = e_1 e_2 e_3 e_4 \paren {\mathbf A}       | c = Definition of $e$ }} {{eqn | r = \mathbf A \mathbf E_4 \mathbf E_3 \mathbf E_2 \mathbf E_1       | c = Cases $1$ and $2$ }} {{eqn | r = \mathbf A e_4 \paren {\mathbf I} \mathbf E_3 \mathbf E_2 \mathbf E_1 }} {{eqn | r = \mathbf A e_3 e_4 \paren {\mathbf I} \mathbf E_2 \mathbf E_1 }} {{eqn | r = \mathbf A e_2 e_3 e_4 \paren {\mathbf I} \mathbf E_1 }} {{eqn | r = \mathbf A e_1 e_2 e_3 e_4 \paren {\mathbf I} }} {{eqn | r = \mathbf A e \paren {\mathbf I} }} {{eqn | r = \mathbf A \mathbf E }} {{end-eqn}} {{qed}}
We are given that $a^2 - b > 0$. Then: :$a > \sqrt b$ and so $\displaystyle \sqrt {a + \sqrt b}$ is defined on the real numbers. Let $\displaystyle \sqrt {a + \sqrt b} = \sqrt x + \sqrt y$ where $x, y$ are (strictly) positive real numbers. Squaring both sides gives: {{begin-eqn}} {{eqn | l = a + \sqrt b       | r = \paren {\sqrt x + \sqrt y}^2       | c =  }} {{eqn | r = x + y + 2 \sqrt {x y}       | c =  }} {{end-eqn}} Set $x + y = a$ and $\sqrt b = 2 \sqrt {x y}$ {{explain|How do you know that the $a$ and $b$ which are $x + y$ and $2 \sqrt {x y}$ are the same $a$ and $b$ that you started with? $x$ and $y$ are free to choose. They were introduced by hand, and can be set to any value provided that they satisfy the constraint above. This is similar to the proof of Cardano's Formula. In that case it needs to be explained. As it stands, it looks as though $x$ and $y$ are pulled out of thin air, with no actual indication that having picked them, they bear the relations given to $a$ and $b$ as presented. It's incredibly confusing for beginners, and others whose abilities and understanding are limited, like me.}} From $\sqrt b = 2 \sqrt {x y}$ we get: {{begin-eqn}} {{eqn | l = \sqrt b       | r = 2 \sqrt {x y}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = b       | r = 4 x y       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x y       | r = \frac b 4       | c =  }} {{end-eqn}} By Viète's Formulas, $x$ and $y$ are solutions to the quadratic equation: :$z^2 - a z + \dfrac b 4 = 0$ From Solution to Quadratic Equation: :$z_{1, 2} = \dfrac {a \pm \sqrt {a^2 - b} } 2$ where $a^2 - b > 0$ (which is a given) {{explain|the notation $z_{1, 2}$}} {{WLOG}}: {{begin-eqn}} {{eqn | l = x = z_1       | r = \dfrac {a + \sqrt {a^2 - b} } 2 }} {{eqn | l = y = z_2       | r = \dfrac {a - \sqrt {a^2 - b} } 2 }} {{end-eqn}} Subsituting into $\displaystyle \sqrt {a + \sqrt b} = \sqrt x + \sqrt y$: {{begin-eqn}} {{eqn | l = \sqrt {a + \sqrt b}       | r = \sqrt x + \sqrt y       | c =  }} {{eqn | r = \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} + \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \paren {\sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} + \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2} }^2       | r = \dfrac {a + \sqrt {a^2 - b} } 2 + \dfrac {a - \sqrt {a^2 - b} } 2 + 2 \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}       | c = multiplying out }} {{eqn | r = a + \sqrt {a + \sqrt {a^2 - b} } \sqrt {a - \sqrt {a^2 - b} }       | c = simplifying }} {{eqn | r = a + \sqrt {a^2 - \paren {a^2 - b} }       | c = Difference of Two Squares }} {{eqn | r = a + \sqrt b       | c = simplifying }} {{eqn | ll= \leadsto       | l = \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} + \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}       | r = \sqrt {a + \sqrt b}       | c = taking square root of both sides }} {{end-eqn}} {{finish|Report on the matter of the signs and magnitudes of $a$ and $b$ according to the constraints given}} {{qed}}
{{begin-eqn}} {{eqn | l = \paren {\sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} - \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2} }^2       | r = \dfrac {a + \sqrt {a^2 - b} } 2 + \dfrac {a - \sqrt {a^2 - b} } 2 - 2 \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}       | c = multiplying out }} {{eqn | r = a - \sqrt {a + \sqrt {a^2 - b} } \sqrt {a - \sqrt {a^2 - b} }       | c = simplifying }} {{eqn | r = a - \sqrt {a^2 - \paren {a^2 - b} }       | c = Difference of Two Squares }} {{eqn | r = a - \sqrt b       | c = simplifying }} {{eqn | ll= \leadsto       | l = \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} - \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}       | r = \sqrt {a - \sqrt b}       | c = taking square root of both sides }} {{end-eqn}} {{finish|Report on the matter of the signs and magnitudes of $a$ and $b$ according to the constraints given}} {{qed}}
We are given that $a^2 - b > 0$. Then: :$a > \sqrt b$ and so $\displaystyle \sqrt {a - \sqrt b}$ is defined on the real numbers. Let $\displaystyle \sqrt {a - \sqrt b} = \sqrt x - \sqrt y$ where $x, y$ are (strictly) positive real numbers. Observe that: :$0 < \sqrt {a - \sqrt b} = \sqrt x - \sqrt y \implies x > y$ Squaring both sides gives: {{begin-eqn}} {{eqn | l = a - \sqrt b       | r = \paren {\sqrt x - \sqrt y}^2       | c =  }} {{eqn | r = x + y - 2 \sqrt {x y}       | c =  }} {{end-eqn}} Set $x + y = a$ and $\sqrt b = 2 \sqrt {x y}$ From $\sqrt b = 2 \sqrt {x y}$ we get: {{begin-eqn}} {{eqn | l = \sqrt b       | r = 2 \sqrt {x y}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = b       | r = 4 x y       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x y       | r = \frac b 4       | c =  }} {{end-eqn}} By Viète's Formulas, $x$ and $y$ are solutions to the quadratic equation: :$z^2 - a z + \dfrac b 4 = 0$ From Solution to Quadratic Equation: :$z_{1, 2} = \dfrac {a \pm \sqrt {a^2 - b} } 2$ where $a^2 - b > 0$ (which is a given) Because we have that $x > y$: {{begin-eqn}} {{eqn | l = x = z_1       | r = \dfrac {a + \sqrt {a^2 - b} } 2 }} {{eqn | l = y = z_2       | r = \dfrac {a - \sqrt {a^2 - b} } 2 }} {{end-eqn}} Subsituting into $\displaystyle \sqrt {a - \sqrt b} = \sqrt x - \sqrt y$: {{begin-eqn}} {{eqn | l = \sqrt {a - \sqrt b}       | r = \sqrt x - \sqrt y       | c =  }} {{eqn | r = \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} - \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}       | c =  }} {{end-eqn}} {{qed}}
In the below: :$\kappa_i$ denotes the initial state of column $i$ :$\kappa_j$ denotes the initial state of column $j$ :$\kappa_i'$ denotes the state of column $i$ after having had the latest elementary column operation applied :$\kappa_j'$ denotes the state of column $j$ after having had the latest elementary column operation applied. $(1)$: Apply $\text {ECO} 2$ to column $j$ for $\lambda = 1$: :$\kappa_j \to \kappa_j + \kappa_i$ After this operation: {{begin-eqn}} {{eqn | l = \kappa_i'       | r = \kappa_i }} {{eqn | l = \kappa_j'       | r = \kappa_i + \kappa_j }} {{end-eqn}} {{qed|lemma}} $(2)$: Apply $\text {ECO} 2$ to column $i$ for $\lambda = -1$: :$\kappa_i \to \kappa_i + \paren {-\kappa_j}$ After this operation: {{begin-eqn}} {{eqn | l = \kappa_i'       | r = \kappa_i - \paren {\kappa_i + \kappa_j} }} {{eqn | r = -\kappa_j }} {{eqn | l = \kappa_j'       | r = \kappa_i + \kappa_j }} {{end-eqn}} {{qed|lemma}} $(3)$: Apply $\text {ECO} 2$ to column $j$ for $\lambda = 1$: :$\kappa_j \to \kappa_j + \kappa_i$ After this operation: {{begin-eqn}} {{eqn | l = \kappa_i'       | r = -\kappa_j }} {{eqn | l = \kappa_j'       | r = \kappa_i + \kappa_j - \kappa_j }} {{eqn | r = \kappa_i }} {{end-eqn}} {{qed|lemma}} $(4)$: Apply $\text {ECO} 1$ to column $i$ for $\lambda = -1$: :$\kappa_i \to -\kappa_i$ After this operation: {{begin-eqn}} {{eqn | l = \kappa_i'       | r = -\paren {-\kappa_j} }} {{eqn | r = \kappa_j }} {{eqn | l = \kappa_j'       | r = \kappa_i }} {{end-eqn}} {{qed|lemma}} Thus, after all the $4$ elementary column operations have been applied, we have: {{begin-eqn}} {{eqn | l = \kappa_i'       | r = \kappa_j }} {{eqn | l = \kappa_j'       | r = \kappa_i }} {{end-eqn}} Hence the result. {{qed}}
From Elementary Column Operations as Matrix Multiplications, an elementary column operation on $\mathbf A$ is equivalent to matrix multiplication by the elementary column matrices corresponding to the elementary column operations. From Determinant of Elementary Column Matrix, the determinants of those elementary column matrices are as follows: === Scale Column === {{:Determinant of Elementary Column Matrix/Scale Column}} === Add Scalar Product of Column to Another === {{:Determinant of Elementary Column Matrix/Scale Column and Add}} === Exchange Columns === {{:Determinant of Elementary Column Matrix/Exchange Columns}} Hence the result. {{qed}}
By Elementary Matrix corresponding to Elementary Column Operation: Scale Column, the elementary column matrix corresponding to $e_1$ is of the form: :$E_{a b} = \begin {cases} \delta_{a b} & : a \ne k \\ \lambda \cdot \delta_{a b} & : a = k \end{cases}$ where: :$E_{a b}$ denotes the element of $\mathbf E_1$ whose indices are $\tuple {a, b}$ :$\delta_{a b}$ is the Kronecker delta: ::$\delta_{a b} = \begin {cases} 1 & : \text {if $a = b$} \\ 0 & : \text {if $a \ne b$} \end {cases}$ Thus when $a \ne b$, $E_{a b} = 0$. This means that $\mathbf E_1$ is a diagonal matrix. {{begin-eqn}} {{eqn | l = \displaystyle \map \det {\mathbf E_1}       | r = \prod_i E_{i i}       | c = Determinant of Diagonal Matrix       | cc= where the index variable $i$ ranges over the order of $\mathbf E_1$ }} {{eqn | r = \prod_i \paren {\begin {cases} 1 & : i \ne k \\ \lambda & : a = k \end{cases} }       | c =  }} {{eqn | r = \prod_{i \mathop \ne k} 1 \times \prod_{i \mathop = k} \lambda       | c =  }} {{eqn | r = 1 \times \lambda       | c =  }} {{eqn | r = \lambda       | c =  }} {{end-eqn}} {{qed}}
By Elementary Matrix corresponding to Elementary Column Operation: Scale Column and Add, $\mathbf E_2$ is of the form: :$E_{a b} = \delta_{a b} + \lambda \cdot \delta_{b i} \cdot \delta_{j a}$ where: :$E_{a b}$ denotes the element of $\mathbf E$ whose indices are $\tuple {a, b}$ :$\delta_{a b}$ is the Kronecker delta: ::$\delta_{a b} = \begin {cases} 1 & : \text {if $a = b$} \\ 0 & : \text {if $a \ne b$} \end {cases}$ Because $i \ne j$ it follows that: :if $a = i$ and $b = j$ then $a \ne b$ Hence when $a = b$ we have that: :$\delta_{b i} \cdot \delta_{j a} = 0$ Hence the diagonal elements of $\mathbf E_2$ are all equal to $1$. We also have that $\delta_{b i} \cdot \delta_{j a} = 1$ {{iff}} $a = i$ and $b = j$. Hence, all elements of $\mathbf E_2$ apart from the diagonal elements and $a_{i j}$ are equal to $0$. Thus $\mathbf E_2$ is a triangular matrix (either upper or lower). From Determinant of Triangular Matrix, $\map \det {\mathbf E_2}$ is equal to the product of all the diagonal elements of $\mathbf E_2$. But as we have seen, these are all equal to $1$. Hence the result. {{qed}}
We have that: :$\mathbf A^\intercal = \begin {bmatrix} a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\  \vdots &  \vdots & \ddots &  \vdots \\ a_{r 1} & a_{r 2} & \cdots & a_{r n} \\  \vdots &  \vdots & \ddots &  \vdots \\ a_{s 1} & a_{s 2} & \cdots & a_{s n} \\  \vdots &  \vdots & \ddots &  \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \\ \end {bmatrix}$ where $\mathbf A^\intercal$ denotes the transpose of $\mathbf A$. Similarly, we have that: :$\mathbf B^\intercal = \begin{bmatrix} a_{1 1} & a_{1 2} & \ldots & a_{1 n} \\  \vdots &  \vdots & \ddots &  \vdots \\ a_{r 1} + \lambda a_{s 1} & a_{r 2} + \lambda a_{s 2} & \cdots & a_{r n} + \lambda a_{s n} \\  \vdots &  \vdots & \ddots &  \vdots \\ a_{s 1} & a_{s 2} & \cdots & a_{s n} \\  \vdots &  \vdots & \ddots &  \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \\ \end {bmatrix}$ From Multiple of Row Added to Row of Determinant: :$\map \det {\mathbf B^\intercal} = \map \det {\mathbf A^\intercal}$ From from Determinant of Transpose: :$\map \det {\mathbf B^\intercal} = \map \det {\mathbf B}$ :$\map \det {\mathbf A^\intercal} = \map \det {\mathbf A}$ and the result follows. {{qed}}