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Removed error in determinants.

This commit is contained in:
Luc Bijl 2024-01-10 22:04:58 +01:00
parent ec3613bf68
commit fb5828a787

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@ -124,22 +124,22 @@ $$
??? note "*Proof*:" ??? note "*Proof*:"
If $i = j$ then we obtain the cofactor expansion of $\det(A)$ along the $i$th row of $A$. If $i = j$ then we obtain the cofactor expansion of $\det(A)$ along the $i$th row of $A$.
If $i \neq j$, let $A^*$ be the matrix obtained by replacing the $j$th row of $A$ by the $i$th row of $A$ If $i \neq j$, let $A^*$ be the matrix obtained by replacing the $j$th row of $A$ by the $i$th row of $A$
$$ $$
A^* = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ \vdots \\ a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots \\ a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{pmatrix} \begin{array}{ll} j\text{th row}\\ \\ \\ \\\end{array} A^* = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ \vdots \\ a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots \\ a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{pmatrix} \begin{array}{ll} j\text{th row}\\ \\ \\ \\\end{array}
$$ $$
since two rows of $A^*$ are the same its determinant must be zero. It follows from the cofactor expansion of $\det(A^*)$ along the $j$th row that since two rows of $A^*$ are the same its determinant must be zero. It follows from the cofactor expansion of $\det(A^*)$ along the $j$th row that
$$ $$
\begin{align*} \begin{align*}
0 &= \det(A^*) = a_{i1} A_{j1}^* + a_{i2} A_{j2}^* + \dots + a_{in} A_{jn}^*, \\ 0 &= \det(A^*) = a_{i1} A_{j1}^* + a_{i2} A_{j2}^* + \dots + a_{in} A_{jn}^*, \\
&= a_{i1} A_{j1} + a_{i2} A_{j2} + \dots + a_{in} A_{jn}. &= a_{i1} A_{j1} + a_{i2} A_{j2} + \dots + a_{in} A_{jn}.
\end{align*} \end{align*}
$$ $$
> *Theorem*: let $E$ be an $n \times n$ elementary matrix and $A$ an $n \times n$ matrix with $n \in \mathbb{N}$ then we have > *Theorem*: let $E$ be an $n \times n$ elementary matrix and $A$ an $n \times n$ matrix with $n \in \mathbb{N}$ then we have
> >