# Thread: Proof that det(A) = -det(B) if B is A after row interchange

1. ## Proof that det(A) = -det(B) if B is A after row interchange

Hello everyone,

I am having trouble understanding my instructor's proof due to two questions that I haven't been able to figure out.

Please note that for my elementary linear algebra course, the instructor has said that permutations (and thus any proofs involving this concept) will not be covered in the course.

I have pasted the proof and my questions below. Thank you very much!

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2. $A=\begin{bmatrix}
a_{11} & a_{12}\\
a_{21} & a_{22}
\end{bmatrix}\rightarrow det(A)=a_{11}a_{22}-a_{21}a_{12}$

$det(E_1A)=\begin{bmatrix}
a_{21} & a_{22}\\
a_{11} & a_{12}
\end{bmatrix}=a_{12}a_{12}-a_{11}a_{22}=-1(a_{11}a_{22}-a_{21}a_{12})=-det(A)$

In regards to adding determinants, it is possible to associate a real number, scalar, with each non-singular square matrix; therefore, since the determinant is just a constant, adding determinants yields: $det(A)+det(B)=k+c \ \mbox{where} \ k,c\in\mathbb{Z}$

$det(A)+det(B)\neq det(A+B)$

3. Determinants are linear in any row (column) of the matrix.

$\det \left( \begin{array}{ccc}
& \vdots & \\
\textrm{---} & r+s & \textrm{---} \\
& \vdots & \end{array} \right) = \det \left( \begin{array}{ccc}
& \vdots & \\
\textrm{---} & r & \textrm{---} \\
& \vdots & \end{array} \right) + \det \left( \begin{array}{ccc}
& \vdots & \\
\textrm{---} & s & \textrm{---} \\
& \vdots & \end{array} \right)$

The line starting 0=det(A)+det(B) follows because the matrix after the second equal sign contains linearly dependant rows. Then you just apply linearity in the rows until you get det(A)+det(B) out.