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I want to use Laplace expansion to show/prove to myself formally that the rule for determinants stating that if B is a matrix obtained from A , where $A \in \mathbb M_{nxn}$, by multiplying a row or column of A by some scalar $\lambda$ then this implies $|B|=\lambda |A|$ (referring to determinant).

Il write what I have\know,

denote $A=[a_{ij}]$

from laplace expansion, we know that $|A|=\sum_{j=1}^{n}$ $a_{ij}(-1)^{i+j}|\mathbb M_{ij}|$ for any row i $\in \{1,…,n\}$ Where $\mathbb M_{ij}$ represents the minor of the element $a_{ij}$

Im thinking I could maybe do something like, suppose we are expanding on row i, and we multiply every element in i by \lambda to obtain some matrix B

then we would have $|B|= \lambda a_{ij}(-1)^{i+1}+…+\lambda a_{in}(-1)^{i+n}$

But $\lambda$ is just a scalar so we can write $|B|=\lambda( a_{ij}(-1)^{i+1}+…+ a_{in}(-1)^{i+n})$ = $\lambda |A|$

However, I am not sure if what I am thinking is valid, or makes sense. I am open to any comments/critiques/suggestions etc, thanks!

Quality
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1 Answers1

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Your way is correct. More succinctly, let $B$ be equal to $A$ with $\lambda$ multiplied to row $i$, and expand about row $i$ to get: \begin{align} \det(B) &= \sum_{j} (-1)^{i+j}b_{ij}\det(\mathbb{M}_{ij})\\ &= \sum_{j} (-1)^{i+j}\lambda a_{ij}\det(\mathbb{M}_{ij})\\ &= \lambda\sum_{j} (-1)^{i+j} a_{ij}\det(\mathbb{M}_{ij})\\ &= \lambda \det(A) \end{align} As an aside, you can also use the formula for the determinant based on the Levi-Civita symbol: \begin{align} \det(B) &= \sum_{j_1}\cdots\sum_{j_n}\varepsilon_{j_1,\ldots,j_n}\prod_k b_{kj_k}\\ &= \sum_{j_1}\cdots\sum_{j_n}\varepsilon_{j_1,\ldots,j_n} b_{ij_i}\prod_{k\ne i} b_{kj_k}\\ &= \sum_{j_1}\cdots\sum_{j_n}\varepsilon_{j_1,\ldots,j_n} \lambda a_{ij_i}\prod_{k\ne i} a_{kj_k}\\ &= \lambda\sum_{j_1}\cdots\sum_{j_n}\varepsilon_{j_1,\ldots,j_n} a_{ij_i}\prod_{k\ne i} a_{kj_k}\\ &= \lambda\det(A) \end{align} (e.g. see here).

user3658307
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