《线性代数习题集》 Chapter 1_Determinants_Sec.1.Second-and Third-Order Determinants
这是我去年做的某本苏联线性代数习题集上的几道题,没什么价值,但是毕竟算做过，丢了可惜，因此贴在这里.
1. $\begin{vmatrix}
5&2\\
7&3
\end{vmatrix}=5\times 3-7\times 2=1$
2.$\begin{vmatrix}
1&2\\
3&4
\end{vmatrix}=1\times 4-2\times 3=-2$
3.$\begin{vmatrix}
3&2\\
8&5
\end{vmatrix}=3\times 5-2\times 8=-1$
4.$\begin{vmatrix}
6&9\\
8&12
\end{vmatrix}=6\times 12-8\times 9=0$
5.$\begin{vmatrix}
a^2&ab\\
ab&b^2
\end{vmatrix}=a^2b^2-a^2b^2=0$
6.$\begin{vmatrix}
n+1&n\\
n&n-1
\end{vmatrix}=-1$
7.$\begin{vmatrix}
a+b&a-b\\
a-b&a+b
\end{vmatrix}=(a+b)^2-(a-b)^2=4ab$
8.$\begin{vmatrix}
a^2+ab+b^2&a^2-ab+b^2\\
a+b&a-b
\end{vmatrix}=(a^3-b^3)-(a^3+b^3)=-2b^3$
9.$\begin{vmatrix}
\cos\alpha&-\sin\alpha\\
\sin\alpha&\cos\alpha
\end{vmatrix}=\cos^2\alpha+\sin^2\alpha=1$
10.$\begin{vmatrix}
\sin\alpha&\cos\alpha\\
\sin\beta&\cos\beta
\end{vmatrix}=\sin\alpha\cos\beta-\sin\beta\cos\alpha=\sin(\alpha-\beta)$
11.$\begin{vmatrix}
\cos\alpha&\sin\alpha\\
\sin\beta&\cos\beta
\end{vmatrix}=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\cos(\alpha+\beta)$
12.$\begin{vmatrix}
\sin\alpha+\sin\beta&\cos\beta+\cos\alpha\\
\cos\beta-\cos\alpha&\sin\alpha-\sin\beta
\end{vmatrix}=0$
13.$\begin{vmatrix}
2\sin\phi\cos\phi&2\sin^2\phi-1\\
2\cos^2\phi-1&2\sin\phi\cos\phi
\end{vmatrix}=\sin^22\phi+\cos^22\phi=1$
14.$\begin{vmatrix}
\frac{1-t^2}{1+t^2}&\frac{2t}{1+t^2}\\
\frac{-2t}{1+t^2}&\frac{1-t^2}{1+t^2}
\end{vmatrix}$令$t=\tan\frac{\phi}{2}$.则$\begin{vmatrix}
\frac{1-t^2}{1+t^2}&\frac{2t}{1+t^2}\\
\frac{-2t}{1+t^2}&\frac{1-t^2}{1+t^2}
\end{vmatrix}=\begin{vmatrix}
\cos\phi&\sin\phi\\
-\sin\phi&\cos\phi\end{vmatrix}=1$
15.$\begin{vmatrix}
\frac{1-t^2}{1+t^2}&\frac{2t}{1+t^2}\\
\frac{2t}{1+t^2}&-\frac{(1+t)^2}{1+t^2}
\end{vmatrix}=\begin{vmatrix}
1-\frac{2t}{1+t^2}&\frac{2t}{1+t^2}\\
\frac{2t}{1+t^2}&-1-\frac{2t}{1+t^2}
\end{vmatrix}$令$t=\tan\frac{\phi}{2}$,则$\begin{vmatrix}
1-\frac{2t}{1+t^2}&\frac{2t}{1+t^2}\\
\frac{2t}{1+t^2}&-1-\frac{2t}{1+t^2}
\end{vmatrix}=\begin{vmatrix}1-\sin\phi&\sin\phi\\
\sin\phi&-1-\sin\phi\end{vmatrix}=-1$
16.$\begin{vmatrix}
\frac{1+t^2}{1-t^2}&\frac{2t}{1-t^2}\\
\frac{2t}{1-t^2}&\frac{1+t^2}{1-t^2}
\end{vmatrix}$.令$t=\tan\frac{\phi}{2}$,则$\begin{vmatrix}
\frac{1+t^2}{1-t^2}&\frac{2t}{1-t^2}\\
\frac{2t}{1-t^2}&\frac{1+t^2}{1-t^2}
\end{vmatrix}=\begin{vmatrix}
\frac{1}{\cos\phi}&\tan\phi\\
\tan\phi&\frac{1}{\cos\phi}
\end{vmatrix}=1$
17.$\begin{vmatrix}
1&\log_ba\\
\log_ab&1
\end{vmatrix}=1-\log_ab\log_ba=1-1=0$
18.$\begin{vmatrix}
a&c+di\\
c-di&b
\end{vmatrix}=ab-(c^2+d^2)$
19.$\begin{vmatrix}
a+bi&b\\
2a&a-bi
\end{vmatrix}=a^2+b^2-2ab=(a-b)^2$
20.$\begin{vmatrix}
\cos\alpha+i\sin\alpha&1\\
1&\cos\alpha-i\sin\alpha
\end{vmatrix}=1-1=0$
21.$\begin{vmatrix}
a+bi&c+di\\
-c+di&a-bi
\end{vmatrix}=a^2+b^2+c^2+d^2$
22-27
Use determinants to solve the following systems of equation:
22.
2x+5y=1
3x+7y=2
解:$$x=\frac{\begin{vmatrix}1&5\\2&7\end{vmatrix}}{\begin{vmatrix}2&5\\3&7\end{vmatrix}}=3$$
$$y=\frac{\begin{vmatrix}2&1\\3&7\end{vmatrix}}{\begin{vmatrix}2&5\\3&7 \end{vmatrix}}=-1$$
23,24,25略.
26.
$x\cos\alpha-y\sin\alpha=\cos\beta$
$x\sin\alpha+y\cos\alpha=\sin\beta$
解：$$x=\frac{\begin{vmatrix}\cos\beta&-\sin\alpha\\\sin\beta&\cos\alpha\end{vmatrix}}{\begin{vmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{vmatrix}}=\cos(\alpha-\beta)$$
同样的方法，易得$y=\sin(\beta-\alpha)$.
27.
$x\tan\alpha+y=\sin(\alpha+\beta)$
$x-y\tan\alpha=\cos(\alpha+\beta)$ where $\alpha\neq \frac{\pi}{2}+k\pi,k\in\mathbb{Z}$.
解：易得$x=\cos\alpha\cos\beta,y=\cos\alpha\sin\beta$
28-38略.
39.Prove that when $a,b,c$ are real,then the roots of the equation $$\begin{vmatrix}a-x&b\\b&c-x\end{vmatrix}=0$$are real.
Proof:$\begin{vmatrix}a-x&b\\b&c-x\end{vmatrix}=x^2-(a+c)x+ac-b^2=0$.Because $\Delta=(a-c)^2+4b^2\geq 0$,so the roots are real.
40.Prove that the quadratic trinomial $ax^2+2bx+c$ with complex coefficients is a perfect square if and only if
$$\begin{vmatrix}
a&b\\b&c
\end{vmatrix}=0$$
Proof:$\Rightarrow:$$ax^2+2bx+c=(mx+n)^2$,so $ax^2+2bx+c=m^2x^2+2mnx+n^2$.So $a=m^2,b=mn,c=n^2$.(根据的是多项式相等的定义)So $ac=b^2$.
$\Leftarrow:$When $a=0$,then $b=0$,then $c=(0x+\sqrt{c})^2$,so in this case $ax^2+2bx+c$ is a perfect square.When $a\neq 0$,then $ax^2+2bx+c=a[(x+\frac{b}{a})^2+\frac{ac-b^2}{a^2}]=a[(x+\frac{b}{a})^2]$.So $ax^2+2bx+c$ is a perfect square.
41略.
42.证明：分式$\frac{ax+b}{cx+d}$的值，此处$c$和$d$至少有一个不为零，当且仅当$\begin{vmatrix}a&b\\c&d\end{vmatrix}=0$时与$x$的值无关.
证明：$\Leftarrow$:令$\frac{ax+b}{cx+d}=t$.则当$cx+d\neq 0$时，可得$x(a-tc)=td-b$.要使$t$不随$x$的变化而变化，只能让$a=tc$,此时$b=td$.可见$ad=bc$.
$\Rightarrow$:当$ad=bc$时，若$d=0$,则$c\neq 0$,$b=0$,此时$\frac{ax+b}{cx+d}$是不变的.当$c=0$,则$d\neq 0$,则$a=0$,此时$\frac{ax+b}{cx+d}$也是不变的.当$c,d\neq 0$,则$\frac{a}{c}=\frac{b}{d}$,此时显然$\frac{ax+b}{cx+d}$也是不变的.
43-55略.
56.$\begin{vmatrix}
a&b&c\\
b&c&a\\
c&a&b
\end{vmatrix}=3abc-(a^3+b^3+c^3)$
57.$\begin{vmatrix}
a&b&c\\
c&a&b\\
b&c&a
\end{vmatrix}=a^3+b^3+c^3-3abc$
58.略.
59.$\begin{vmatrix}
a&x&x\\
x&b&x\\
x&x&c
\end{vmatrix}=2x^3-x^2(a+b+c)+abc$
