(1) The value of the determinant remains unchanged if its rows and columns are interchanged.
\[
\begin{vmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{vmatrix}
=
\begin{vmatrix}
a & d & g \\
b & e & h \\
c & f & i
\end{vmatrix}
\] Verification
LHS=a(ei - fh) -b(di - fg) + c(dh - eg)
RHS (expanding along column C1)=a(ei - fh) -b(di - fg) + c(dh - eg)
(2)If we interchange any two rows (or columns), then sign of determinant changes.
\[
\begin{vmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{vmatrix}
=
-\begin{vmatrix}
d & e & f \\
a & b & c \\
g & h & i
\end{vmatrix}
\] Verification
We can easily proof this by expanding
LHS=a(ei - fh) -b(di - fg) + c(dh - eg)=aei-afh=bdi+bfg+cdh-ceg
RHS= d(bi - ch) -e(ai - cg) + f(ah - bg)=dbi-dch-aei+ecg+fah-fbg=-(aei-afh=bdi+bfg+cdh-ceg)
(2)If any two rows (or columns) of a determinant are identical (all corresponding elements are same), then value of determinant is zero. Proof If we interchange the identical rows (or columns) of the determinant $\Delta$, then $\Delta$ does not change. However, by above property, it follows that $\Delta$ has changed its sign
$\Delta= - \Delta$
hence $\Delta=0$
(3)If we multiply each element of a row or a column of a determinant by constant k, then value of determinant is multiplied by k.
\[
A= \begin{vmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{vmatrix}
\]
\[
B= \begin{vmatrix}
ka & kb & kc \\
d & e & f \\
g & h & i
\end{vmatrix}
= k \begin{vmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{vmatrix}
=kA
\] Verification
A=a(ei - fh) -b(di - fg) + c(dh - eg)=aei-afh=bdi+bfg+cdh-ceg
B=ka(ei - fh) -kb(di - fg) + kc(dh - eg)=k(aei-afh=bdi+bfg+cdh-ceg)
Similarly
\[
C=|kA|= \begin{vmatrix}
ka & kb & kc \\
kd & ke & kf \\
kg & kh & ki
\end{vmatrix}
= k^3 \begin{vmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{vmatrix}
=k^3A
\]
(4) If some or all elements of a row or column of a determinant are expressed as sum of two (or more) terms, then the determinant can be expressed as sum of two (or more) determinants
\[
\begin{vmatrix}
a+k & b+k & c+k \\
d & e & f \\
g & h & i
\end{vmatrix}
= \begin{vmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{vmatrix}
+
\begin{vmatrix}
k & k & k \\
d & e & f \\
g & h & i
\end{vmatrix}
\] Verification
We can easily proof this by expanding
LHS=(a+k)(ei - fh) -(b+k)(di - fg) + (c+h)(dh - eg)=a(ei - fh) -b(di - fg) + c(dh - eg) + k(ei - fh) -k(di - fg) + k(dh - eg)
RHS=a(ei - fh) -b(di - fg) + c(dh - eg) + k(ei - fh) -k(di - fg) + k(dh - eg)
(5)If to each element of a row or a column of a determinant the equimultiples of corresponding elements of other rows or columns are added, then value of determinant remains same
\[
A= \begin{vmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{vmatrix}
=
\begin{vmatrix}
a+kd & b+ke & c+kf \\
d & e & f \\
g & h & i
\end{vmatrix}
\] Verification
RHS= \[
\begin{vmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{vmatrix}
+
\begin{vmatrix}
kd & ke & kf \\
d & e & f \\
g & h & i
\end{vmatrix}
\]
$=\begin{vmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{vmatrix}
+
k\begin{vmatrix}
d & e & f \\
d & e & f \\
g & h & i
\end{vmatrix}
$
As second determinant is having same values, so it is zero
Hence
$=\begin{vmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{vmatrix}
$
