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# System prepended metadata

title: Linear Algebra hw3

---

# Linear Algebra hw3

## 11
![圖片](https://hackmd.io/_uploads/SJUXreszA.png)

### a

consider two 2d matrix $A=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$　and $B=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$, $det(A+B)=4, det(A)=det(B)=1$, 

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

### b

consider A is nonsingular, A can be perform as a series of elementary matrix be multiplied, $\therefore A=E, E = \{E_nE_{n-1}\cdots E_1\}, det(A)=det(E)$, 

since $det(EB)=det(E)det(B)$, we have $det(AB)=det(EB)=det(E)det(B)=det(A)det(B)$

A is singular, AB is also singular, therefore we have $det(AB)=0=det(A)det(B)$

### c
by b. equation above, $det(AB)=det(A)det(B)$, since det(A), det(B)$\in R$, $det(A)det(B)=det(B)det(A)$, therefore, we have $det(AB)=det(A)det(B)=det(B)det(A)=det(BA)$


## 2.
![圖片](https://hackmd.io/_uploads/BJeXJ-jzR.png)

### a.

$$A=\begin{bmatrix}
     2 &  1 &  2 &  2 \\
    -3 &  3 &  1 &  3 \\
     2 &  1 & -1 & -4 \\
     1 & -3 &  2 &  0 \\ 
\end{bmatrix}$$

$a_2=a_2+\frac{3}{2}a_1$
$a_3=a_3-a_1$
$a_4=a_4-\frac{1}{2}a_1$

$$A=\begin{bmatrix}
     2 &  1 &  2 &  2 \\
     0 & \frac{9}{2} &  4 &  6 \\
     0 &  0 & -3 & -6 \\
     0 & -\frac{7}{2} &  1 &  -1 \\ 
\end{bmatrix}$$

$a_4=a_4+\frac{7}{9}a_2+\frac{1}{27}a_3$

$$A=\begin{bmatrix}
     2 &  1 &  2 &  2 \\
     0 & \frac{9}{2} &  4 &  6 \\
     0 &  0 & -3 & -6 \\
     0 & 0 &  0 &  \frac{31}{9} \\ 
\end{bmatrix}$$

$det(A)=2\cdot \frac{9}{2}\cdot-3\cdot-\frac{31}{9}=-93$

### b
the first det is exchange twice so det is $det(A)$
second one is adding a row to another one so is also $det(A)$
so ans is 2 * det(A) = -186

## 18

![圖片](https://hackmd.io/_uploads/Hke43bjM0.png)

### a
$det(E)=det(I_k)det(B)=det(B)$

### b
$det(F)=det(A)det(I_{n-k})=det(A)$

### c
$det(C)=det(A)det(B)$ 

![圖片](https://hackmd.io/_uploads/HkenUaWiMC.png)


### a
$$
A=\begin{bmatrix}
    2 & 3 \\
    8 & -6
\end{bmatrix}
$$
$x_1=\frac{det(A_1)}{det(A)}=\frac{-18}{-36}=2$

$x_2=\frac{det(A_2)}{det(A)}=\frac{-12}{-36}=-3$

### b
$$
A=\begin{bmatrix}
    6 & 3 \\
    5 & 7
\end{bmatrix}
$$
$x_1=\frac{det(A_1)}{det(A)}=\frac{-1}{27}$

$x_2=\frac{det(A_2)}{det(A)}=\frac{47}{27}$


### c
$$
A=\begin{bmatrix}
    2 & -2 & 3 \\
    1 & 2 & 3 \\
    -2 & 4 &１
\end{bmatrix}
$$
$x_1=\frac{det(A_1)}{det(A)}=\frac{228}{114}=2$

$x_2=\frac{det(A_2)}{det(A)}=\frac{0}{114}=0$

$x_３=\frac{det(A_３)}{det(A)}=\frac{-1026}{114}=9$



### d
$$
A=\begin{bmatrix}
    4 & 0 & -2 \\
    7 & 7 & 2 \\
    3 & -2 &１
\end{bmatrix}
$$
$x_1=\frac{det(A_1)}{det(A)}=\frac{0}{18}=0$

$x_2=\frac{det(A_2)}{det(A)}=\frac{-16}{18}=-\frac{8}{9}$

$x_3=\frac{det(A_3)}{det(A)}=\frac{60}{18}=\frac{10}{3}$

### e
![圖片](https://hackmd.io/_uploads/BJSaQMiz0.png)


$$
A=\begin{bmatrix}
    -1 & 2 & 1  \\
    2 & 1 & -2 & 1 \\
    3 & 3 & 1 & 0 \\
    0 & 2 & 1 & -1
\end{bmatrix}
$$
$x_1=\frac{det(A_1)}{det(A)}=\frac{-16}{32}=-2$

$x_2=\frac{det(A_2)}{det(A)}=\frac{72}{32}=\frac{9}{4}$

$x_3=\frac{det(A_3)}{det(A)}=\frac{24}{32}=\frac{3}{4}$

$x_4 =\frac{det(A_4)}{det(A)}=\frac{104}{32}=\frac{13}{4}$



### 4 
![圖片](https://hackmd.io/_uploads/rkDABGiz0.png)

$$
A=\begin{bmatrix}
    2 & 1 & 2 \\
    0 & 3 & 4 \\
    1 & 1 & 2 
\end{bmatrix}
$$
$x_1=\frac{det(A_1)}{det(A)}=\frac{0}{2}=0$

$x_2=\frac{det(A_2)}{det(A)}=\frac{2}{2}=1$

$x_3=\frac{det(A_3)}{det(A)}=\frac{-1}{2}$

Ans: $\begin{bmatrix}0 \\ 1 \\ -\frac{1}{2}\end{bmatrix}$