设(A = (a_{ij}))，将(A)的元素行列对换即：
[egin{vmatrix}
a_{11} & a_{21} & cdots & a_{n1} \
a_{12} & a_{22} & cdots & a_{n2} \
vdots & vdots & & vdots \
a_{1n} & a_{2n} & cdots & a_{nn}
end{vmatrix}
]
称为(A)的转置，记为(A')或(A^T)
性质 1：(|A'| = |A|)
证明：利用 2.2 节最后的推论可得：
[egin{aligned}
|A'| &= sum_{i_1i_2dots i_n}(-1)^{ au(i_1i_2dots i_n)}a_{1i_1}a_{2i_2}dots a_{ni_n} \
|A| &= sum_{i_1i_2dots i_n}(-1)^{ au(i_1i_2dots i_n)}a_{1i_1}a_{2i_2}dots a_{ni_n} \
end{aligned}
]
[ herefore |A'| = |A|
]
性质 2：
[egin{vmatrix}
a_{11} & a_{12} & cdots & a_{1n} \
vdots & vdots & & vdots \
ka_{i1}& ka_{i2} & cdots & ka_{in} \
vdots & vdots & & vdots \
a_{n1} & a_{n2} & cdots & a_{nn}
end{vmatrix}
= k
egin{vmatrix}
a_{11} & a_{12} & cdots & a_{1n} \
vdots & vdots & & vdots \
a_{i1}& a_{i2} & cdots & a_{in} \
vdots & vdots & & vdots \
a_{n1} & a_{n2} & cdots & a_{nn}
end{vmatrix}
]
证明：等号左侧的行列式记为(|B|)，右侧的行列式记为(|A|)
[egin{aligned}
|A'| &= sum_{j_1j_2dots j_n}(-1)^{ au(j_1j_2dots j_n)}a_{1j_1}dots (ka_{ij_i})dots a_{nj_n} \
&= ksum_{j_1j_2dots j_n}(-1)^{ au(j_1j_2dots j_n)}a_{1j_1}dots a_{ij_i}dots a_{nj_n} \
&= k|A|（k = 0时仍成立）
end{aligned}
]
推论：行列式出现(0)行，则行列式为(0)
性质 3：
[egin{vmatrix}
a_{11} & a_{12} & cdots & a_{1n} \
vdots & vdots & & vdots \
a_{i1}+b_1& a_{i2} + b_2 & cdots & a_{in}+b_n \
vdots & vdots & & vdots \
a_{n1} & a_{n2} & cdots & a_{nn}
end{vmatrix}
=
egin{vmatrix}
a_{11} & a_{12} & cdots & a_{1n} \
vdots & vdots & & vdots \
a_{i1}& a_{i2} & cdots & a_{in} \
vdots & vdots & & vdots \
a_{n1} & a_{n2} & cdots & a_{nn}
end{vmatrix}
+
egin{vmatrix}
a_{11} & a_{12} & cdots & a_{1n} \
vdots & vdots & & vdots \
b_1& b_2 & cdots & b_n \
vdots & vdots & & vdots \
a_{n1} & a_{n2} & cdots & a_{nn}
end{vmatrix}
]
证明：
[egin{aligned}
左式 &= sum_{j_1j_2dots j_n}(-1)^{ au(j_1j_2dots j_n)}a_{1j_1}dots (a_{ij_i} + b_{j_i})dots a_{nj_n} \
&= sum_{j_1j_2dots j_n}(-1)^{ au(j_1j_2dots j_n)}a_{1j_1}dots a_{ij_i}dots a_{nj_n} \
&+ sum_{j_1j_2dots j_n}(-1)^{ au(j_1j_2dots j_n)}a_{1j_1}dots b_{j_i}dots a_{nj_n} \
&= 右式
end{aligned}
]
性质 4：
[(-1) cdot
egin{vmatrix}
a_{11} & a_{12} & cdots & a_{1n} \
vdots & vdots & & vdots \
a_{i1}& a_{i2} & cdots & a_{in} \
vdots & vdots & & vdots \
a_{k1}& a_{k2} & cdots & a_{kn} \
vdots & vdots & & vdots \
a_{n1} & a_{n2} & cdots & a_{nn}
end{vmatrix}
=
egin{vmatrix}
a_{11} & a_{12} & cdots & a_{1n} \
vdots & vdots & & vdots \
a_{k1}& a_{k2} & cdots & a_{kn} \
vdots & vdots & & vdots \
a_{i1}& a_{i2} & cdots & a_{in} \
vdots & vdots & & vdots \
a_{n1} & a_{n2} & cdots & a_{nn}
end{vmatrix}
]
证明：
[egin{aligned}
右边 &= sum_{j_1j_2dots j_n}(-1)^{ au(j_1dots j_idots j_k dots j_n)}a_{1j_1}dots a_{kj_i}dots a_{ij_k} dots a_{nj_n} \
&= sum_{j_1j_2dots j_n}(-1)^{ au(j_1dots j_idots j_k dots j_n) + 1}a_{1j_1}dots a_{ij_i}dots a_{kj_k} dots a_{nj_n} \
&= 左边
end{aligned}
]
性质 5：若行列式两行相等，则行列式为(0)。
证明：根据性质 4，两行对换，(|A| = -|A| Rightarrow |A| = 0)
性质 6：两行成比例，行列式为(0)
性质 7：若(A stackrel{ⓚ+ⓘcdot l}{longrightarrow} D)，则(|A| = |D|)
例 1：
[egin{aligned}
egin{vmatrix}
k & lambda & cdots & lambda \
lambda & k & cdots & lambda \
vdots & vdots & ddots & vdots \
lambda & lambda & cdots & k
end{vmatrix}
&=
egin{vmatrix}
k + (n - 1)lambda & k + (n - 1)lambda & cdots & k + (n - 1)lambda \
lambda & k & cdots & lambda \
vdots & vdots & ddots & vdots \
lambda & lambda & cdots & k
end{vmatrix} \
&=
[k + (n - 1)lambda ]
egin{vmatrix}
1 & 1 & cdots & 1 \
lambda & k & cdots & lambda \
vdots & vdots & ddots & vdots \
lambda & lambda & cdots & k
end{vmatrix} \
&=
[k + (n - 1)lambda ]
egin{vmatrix}
1 & 1 & cdots & 1 \
0 & k - lambda & cdots & 0 \
vdots & vdots & ddots & vdots \
0 & 0 & cdots & k - lambda
end{vmatrix} \
&= (k - lambda)^{n - 1} [k + (n-1)lambda]
end{aligned}
]