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## Solution to Linear Algebra Hoffman & Kunze Chapter 1.3

#### Exercise 1.3.1

The matrix of coefficients is
$$\left[\begin{array}{cc}1-i&-i\\2&1-i\end{array}\right].$$Row reducing
$$\rightarrow \left[\begin{array}{cc}2&1-i\\1-i&-i\end{array}\right]\rightarrow\left[\begin{array}{cc}2&1-i\\0&0\end{array}\right]$$Thus $2x_1+(1-i)x_2=0$. Thus for any $x_2\in\mathbb C$, $(\frac12(i-1)x_2,x_2)$ is a solution and these are all solutions.

#### Exercise 1.3.2

We have$$\rightarrow\left[\begin{array}{ccc}1&-3&0\\2&1&1\\3&-1&2\end{array}\right]\rightarrow\left[\begin{array}{ccc}1&-3&0\\0&7&1\\0&8&2\end{array}\right]\rightarrow\left[\begin{array}{ccc}1&-3&0\\0&1&1/7\\0&8&2\end{array}\right]$$ $$\rightarrow\left[\begin{array}{ccc}1&0&3/7\\0&1&1/7\\0&0&6/7\end{array}\right]\rightarrow\left[\begin{array}{ccc}1&0&3/7\\0&1&1/7\\0&0&1\end{array}\right]\rightarrow\left[\begin{array}{ccc}1&0&0\\0&1&10\\0&0&1\end{array}\right].$$Thus $A$ is row-equivalent to the identity matrix. It follows that the only solution to the system is $(0,0,0)$.

#### Exercise 1.3.3

The system $AX=2X$ is
$$\left[\begin{array}{ccc}6&-4&0\\4&-2&0\\-1&0&3\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]=2\left[\begin{array}{c}x\\y\\z\end{array}\right]$$which is the same as
\begin{alignat*}{1}
6x-4y&=2x\\
4x-2y&=2y\\
-x+3z&=2z
\end{alignat*}which is equivalent to
\begin{alignat*}{1}
4x-4y&=0\\
4x-4y&=0\\
-x+z&=0
\end{alignat*}The matrix of coefficients is
$$\left[\begin{array}{ccc}4&-4&0\\4&-4&0\\-1&0&1\end{array}\right]$$which row-reduces to
$$\left[\begin{array}{ccc}1&0&-1\\0&1&-1\\0&0&0\end{array}\right]$$Thus the solutions are all elements of $F^3$ of the form $(x,x,x)$ where $x\in F$.

The system $AX=3X$ is
$$\left[\begin{array}{ccc}6&-4&0\\4&-2&0\\-1&0&3\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]=3\left[\begin{array}{c}x\\y\\z\end{array}\right]$$which is the same as
\begin{alignat*}{1}
6x-4y &=3x\\
4x-2y&=3y\\
-x+3z&=3z
\end{alignat*}
which is equivalent to
\begin{alignat*}{1}
3x-4y&=0\\
x-2y&=0\\
-x&=0
\end{alignat*}The matrix of coefficients is
$$\left[\begin{array}{ccc}3&-4&0\\1&-2&0\\-1&0&0\end{array}\right]$$which row-reduces to$$\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&0\end{array}\right]$$Thus the solutions are all elements of $F^3$ of the form $(0,0,z)$ where $z\in F$.

#### Exercise 1.3.4

We have $$A\rightarrow\left[\begin{array}{ccc} 1 & -2 & 1\\ i & -(1+i) & 0\\ 1 & 2i & -1 \end{array}\right]\rightarrow \left[\begin{array}{ccc} 1 & -2 & 1\\ 0 & -1+i & -i\\ 0 & 2+2i & -2 \end{array}\right]\rightarrow \left[\begin{array}{ccc} 1 & -2 & 1\\ 0 & 1 & \frac{1-i}2\\ 0 &2+ 2i & -2 \end{array}\right]$$ $$\rightarrow \left[\begin{array}{ccc} 1 & -2 & 1\\ 0 & 1 & \frac{i-1}2\\ 0 & 0 & 0 \end{array}\right]\rightarrow \left[\begin{array}{ccc} 1 & 0 & i\\ 0 & 1 & \frac{i-1}2\\ 0 & 0 & 0 \end{array}\right].$$

#### Exercise 1.3.5

Call the first matrix $A$ and the second matrix $B$. The matrix $A$ is row-equivalent to $$A’=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$$ and the matrix $B$ is row-equivalent to $$B’=\left[\begin{array}{ccc}1&0&1/2\\0&1&3/2\\0&0&0\end{array}\right].$$

By Theorem 3 page 7 $AX=0$ and $A’X=0$ have the same solutions. Similarly $BX=0$ and $B’X=0$ have the same solutions. Now if $A$ and $B$ are row-equivalent then $A’$ and $B’$ are row equivalent. Thus if $A$ and $B$ are row equivalent then $A’X=0$ and $B’X=0$ must have the same solutions. But $B’X=0$ has infinitely many solutions and $A’X=0$ has only the trivial solution $(0,0,0)$. Thus $A$ and $B$ cannot be row-equivalent.

#### Exercise 1.3.6

Case $a\not=0$: Then to be in row-reduced form it must be that $a=1$ and $A=\left[\begin{array}{cc}1&b\\c&d\end{array}\right]$ which implies $c=0$, so $A=\left[\begin{array}{cc}1&b\\0&d\end{array}\right]$. Suppose $d\not=0$. Then to be in row-reduced form it must be that $d=1$ and $b=0$, so $A=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]$ which implies $a+b+c+d\not=0$. So it must be that $d=0$, and then it follows that $b=-1$. So $a\not=0$ $\Rightarrow$ $A=\left[\begin{array}{cc}1&-1\\0&0\end{array}\right]$.

Case $a=0$: Then $A=\left[\begin{array}{cc}0&b\\c&d\end{array}\right]$. If $b\not=0$ then $b$ must equal $1$ and $A=\left[\begin{array}{cc}0&1\\c&d\end{array}\right]$ which forces $d=0$. So $A=\left[\begin{array}{cc}0&1\\c&0\end{array}\right]$ which implies (since $a+b+c+d=0$) that $c=-1$. But $c$ cannot be $-1$ in row-reduced form. So it must be that $b=0$. So $A=\left[\begin{array}{cc}0&0\\c&d\end{array}\right]$. If $c\not=0$ then $c=1$, $d=-1$ and $A=\left[\begin{array}{cc}0&0\\1&-1\end{array}\right]$. Otherwise $c=0$ and $A=\left[\begin{array}{cc}0&0\\0&0\end{array}\right]$.

Thus the three possibilities are:
$$\left[\begin{array}{cc}0&0\\0&0\end{array}\right],\quad\left[\begin{array}{cc}1&-1\\0&0\end{array}\right],\quad\left[\begin{array}{cc}0&0\\1&-1\end{array}\right].$$

#### Exercise 1.3.7

Write the matrix as
$$\left[\begin{array}{c} R_1\\ R_2\\ R_3\\ \vdots\\ R_n \end{array}\right].$$WOLOG we’ll show how to exchange rows $R_1$ and $R_2$. First add $R_2$ to $R_1$:
$$\left[\begin{array}{c} R_1+R_2\\ R_2\\ R_3\\ \vdots\\ R_n \end{array}\right].$$Next subtract row one from row two:
$$\left[\begin{array}{c} R_1+R_2\\ -R_1\\ R_3\\ \vdots\\ R_n \end{array}\right].$$Next add row two to row one again
$$\left[\begin{array}{c} R_2\\ -R_1\\ R_3\\ \vdots\\ R_n \end{array}\right].$$Finally multiply row two by $-1$:
$$\left[\begin{array}{c} R_2\\ R_1\\ R_3\\ \vdots\\ R_n \end{array}\right].$$

#### Exercise 1.3.8

(a) In this case the system of equations is
\begin{alignat*}{1}
0\cdot x_1 + 0\cdot x_2 &= 0\\
0\cdot x_1 + 0\cdot x_2 &= 0
\end{alignat*}Clearly any $(x_1,x_2)$ satisfies this system since $0\cdot x=0$ $\forall$ $x\in F$.

(b) Let $(u,v)\in F^2$. Consider the system:
\begin{alignat*}{1}
a\cdot x_1 + b\cdot x_2 &= u\\
c\cdot x_1 + d\cdot x_2 &= v
\end{alignat*}If $ad-bc\not=0$ then we can solve for $x_1$ and $x_2$ explicitly as
$$x_1=\frac{du-bv}{ad-bc}\quad x_2=\frac{av-cu}{ad-bc}.$$Thus there’s a unique solution for all $(u,v)$ and in partucular when $(u,v)=(0,0)$.

(c) Assume WOLOG that $a\not=0$. Then $ad-bc=0$ $\Rightarrow$ $d=\frac{cb}{a}$. Thus if we multiply the first equation by $\frac ca$ we get the second equation. Thus the two equations are redundant and we can just consider the first one $a x_1+bx_2=0$. Then any solution is of the form $(-\frac bay,y)$ for arbitrary $y\in F$. Thus letting $y=1$ we get the solution $(-b/a,1)$ and the arbitrary solution is of the form $y(-b/a,1)$ as desired.