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Let $A$ be an $n\times n$ matrix and $i,j,k$ be $1\leq i,j,k\leq n$ and $\alpha,\beta \in \mathbb{R}$.

I am supposing that $\bf{a}_k$(the $k$-th row) is equal to $\alpha \bf{a}_i+\beta \bf{a}_j$. ($\bf{a}_i, \bf{a}_j \in \mathbb{R^n}$ mean the $i$-th row and $j$-th row respectively).

So I need to prove that $\det(A)=0$

I know that $\det(A)=0$, when two rows are equal, and that if we add a row of $A$ multiplied by a scalar to another row of $A$, then the determinant will not change. I'm just having trouble interpreting the proof.

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  • $\begingroup$ The result is not true as stated. You need the assumption that $k\notin\{i,j\}$. $\endgroup$ Nov 6, 2021 at 5:42

3 Answers 3

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Start with $A$. Let $B$ be the matrix obtained by adding $-\alpha$ times row $i$ to row $k$ and $-\beta$ times row $j$ to row $k$. Then $\det A = \det B$, but row $k$ of $B$ is all zeros. Thus $\det B = 0$.


Here is an illustration: row operations can be carried out by elementary matrices. For instance, starting with the matrix $\begin{bmatrix}1& 1& 1 \\ 1& 0 &1 \\ 0& 0 &1 \end{bmatrix}$ if you add $3$ times row 1 to row 2 you get $\begin{bmatrix}1& 1& 1 \\ 4& 3 &4 \\ 0& 0 &1 \end{bmatrix}$. This is carried out by the elementary matrix that performs the same row operation on the identity: $$ \begin{bmatrix}1& 0& 0 \\ 3& 1 &0 \\ 0& 0 &1 \end{bmatrix} \begin{bmatrix}1& 1& 1 \\ 1& 0 &1 \\ 0& 0 &1 \end{bmatrix} = \begin{bmatrix}1& 1& 1 \\ 4& 3 &4 \\ 0& 0 &1 \end{bmatrix}.$$ It is easy to check that this elementary matrix has determinant $1$ (having only one nonzero term off the main diagonal that consists of $1$'s) so $$\det \begin{bmatrix}1& 1& 1 \\ 1& 0 &1 \\ 0& 0 &1 \end{bmatrix} = \det \begin{bmatrix}1& 1& 1 \\ 4& 3 &4 \\ 0& 0 &1 \end{bmatrix}.$$

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  • $\begingroup$ Is it possible for you to show the proof visually i.e. using a matrix? I'm having a hard time visualizing it, sorry if it's trivial. $\endgroup$
    – Jhune
    Nov 4, 2014 at 13:51
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Under the premise that the determinant does not change upon adding a scalar multiple of one row to another, if row $a_k = \alpha_i a_i + \beta_j a_j$, we can perform the row operation $a_k \rightarrow a_k - \alpha_i a_i - \beta_j a_j = 0$ without altering the value of $\det(A)$. So now the matrix has a row containing zeros and developing the determinant along that row yields $\det(A) = 0$.

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We already known:

  1. Row vector $\bf{a}_{k} = \alpha \bf{a}_{i} + \beta \bf{a}_{j}$
  2. If a multiple of one row is added to another row, the determinant is not changed.
  3. If a row of $A$ is zero, then $det(A) = 0$.

First, add $- \beta \bf{a}_{j}$ to the $\bf{a}_{k}$, $\det(A)$ is not changed by $(2)$. Then, add $- \alpha a_{i}$ to the $\bf{a}_{k}$, $\det(A)$ is not changed by $(2)$. Finally, we get

$\bf{a}_{k} = \alpha \bf{a}_{i} + \beta \bf{a}_{j} - \beta \bf{a}_{j} - \alpha \bf{a}_{i} = 0$.

Apply $(3)$, $\det(A)=0$.

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