## 考題
[NCKU_CS_Math112](https://drive.google.com/file/d/1VzhM8lTBSerMHNJZkjUlIt6wGTSEWINI/view?usp=drive_link)

## 1.
Fix the second digital : $1X1\text{~}~9X9$
Sum : $\frac{9(101+909)}{2}=4545$
Since $X$ from 0 to 9, the sum of this part is 45450
Fix the first and third digital : $X0X\text{~}X9X$
Sum : $\frac{9(10+90)}{2}=450$
Since $X$ from 1 to 9, this part is 4050
Thus, the total sum is 49500.

## 2. Generating function
$f(x)=1+x+x^2+\cdots+x^n$
## 3. The number of bijection function
Answer : $n!$
## 4. Recursive Relation
$x^2-5x+6=0\implies (x-2)(x-3)=0$
Thus, $a_n^{(h)}=c_02^n+c_13^n$ and $a^{(p)}_n=d_0$
Put the particular solution into original equation : $d_0-5d_0+6d_0=2\implies d_0=1$
Thus, $a_n=a_n^{(h)}+a^{(p)}_n=c_02^n+c_13^n+1$
Put $a_0$ and $a_1$ into equation : $c_0=1, c_1=2$
Thus , $a_n=2\cdot3^n+1$

## 5. Number Theory
Assume $2a+3b=17k$, for some $k\in \mathbb{Z}$
Assume the statement is False.
$\implies \exists s\in \mathbb{N}$ with $s<17$ such that $9a+5b=17q+s$
Thus, $(9a+5b)-(2a+3b)=17(q-s)+s=7a+2b$
$\implies 21a+6b=17(3(q-k))+3s$
Since $17$ is prime and $~3\not\mid17~\land s\not\mid17$, $3s\not\mid17$.
$\implies 17\not\mid 21a+6b$
On the other hands, $4a+6b=17(2k)$
$\implies 21a+6b-(4a+6b)=17a=17(3(q-k)-2k)+3s$
It's a contradiction. Thus, the statement is true.
## 6. Coordinate Vector
Slove $w=(8,-4,-12)=a(1,1,1)+b(1,5,-3)+c(2,2,1)$
with $B=(1,1,1),(1,5,-3),(2,2,1)$
$\implies[w]_B=[-53,-3,32]$
## 7. Orthonormal Basis
$x=\frac{<x,u_1>}{\lVert u_1 \rVert^2}u_1+\frac{<x,u_1>}{\lVert u_2 \rVert^2}u_2+\frac{<x,u_3>}{\lVert u_3 \rVert^2}u_1=4u_1+0+<x,u_3>u_3$
$\implies \lVert x \rVert=5=\sqrt{16+(<x,u_3>)^2}$
Thus, $<x,u_3>=\pm 3$
Finally, we have $c_1=4,c_2=0,c_3=\pm3$
## 8. Dimension Theory
### a.
By Dimension Theory, $Rank(A)+Nullity(A)=7$.
Thus, $Nullity(A)=3$.
### b.
No.
Since $b$ is combination of Columns of $A$ and $dim(Col(A))=4<5$, $\exists b\not\in Col(A)$ such that $Ax=b$ has no solution.
## 9. Eigenvalue
$det(A)=p(0)=1(-2)^2(3)^2=36$
Thus, $det(A^{-1})=\frac{1}{36}$
## 10. True or False
### a. Singular and Eigenvalue
True.
$det(A)=\displaystyle\prod^n_{i=1}\lambda_i\not=0$
Thus, $A$ is nonsingular.
### b. Diagonalizable
True.
$Rank(A)=1\implies Nullity(A)=4=dim(E(0))$
$\implies am(0)=gm(0)$

### c. Determinant
False.
Row exchange is forbidden.
### d. Invertible
False.
$A$ is invertible $\implies Rank(A)=n=Rank(A^T)$