Let $A,B \in K^{n,n}$ and suppose $B$ is obtained from $A$ by adding $\lambda$ times row $j$ to row $i$. Prove $det(A)=det(B)$.
My Attempt
I tried to use proof by induction for this .
Take the base case n=2.
$$A=\left( \begin{array}{ccc}
a_{11} & a_{12} \\
a_{21} & a_{22} \end{array} \right)$$ det(A)=$a_{11}a_{22}-a_{12}a_{21}$
$$B=\left( \begin{array}{ccc}
a_{11}+ \lambda a_{21} & a_{12}+ \lambda a_{22} \\
a_{21} & a_{22} \end{array} \right)$$ det(B)=$(a_{11}+ \lambda a_{21})a_{22}-(a_{12}+ \lambda a_{22})a_{21}=a_{11}a_{22}-a_{12}a_{21}$
So it is true for n=2, assume true for $n=k$ ,
I must now show that it holds for $n=k+1$ but this is where I get stuck , how do I show this??
Any help would be much appreciated
First notice that this is only true for $i\neq j$. Here's a counterexample to show that:
$$A = \begin{pmatrix}1&0\\0&2\end{pmatrix} \qquad B = \begin{pmatrix}3&0\\0&2\end{pmatrix} \qquad\det(A) \neq \det(B)$$ I took $i = j = 1$ and $\lambda = 2$.
Knowing that $i \neq j$ you can easily prove the theorem by using the following property of determinants: $$\det(AB) = \det(A)\det(B)$$ Because $i\neq j$ we have $B = SA$ where $S$ is identity matrix with one entry being $\lambda$ somewhere off the main diagonal. Could be something like this one:
$$ S = \begin{pmatrix} 1&0&0&0&0\\ 0&1&0&0&0\\ 0&0&1&0&\lambda\\ 0&0&0&1&0\\ 0&0&0&0&1\\ \end{pmatrix} $$
Using determinant definition you can show that $\det(S) = 1$, because all components except product of the diagonal diappear, and therefore $\det(A) = \det(B)$. Position of $\lambda$ in the matrix $S$ does not matter.