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Solution to Linear Algebra Hoffman & Kunze Chapter 1.4


Exercise 1.4.1

The coefficient matrix is
$$\left[\begin{array}{ccc}
\frac13 & 2 & -6\\
-4& 0& 5\\
-3&6&-13\\
-\frac73&2&-\frac83
\end{array}\right]
$$This reduces as follows:
$$\rightarrow\left[\begin{array}{ccc}
1 & 6 & -18\\
-4& 0& 5\\
-3&6&-13\\
-7&6&-8
\end{array}\right]
\rightarrow
\left[\begin{array}{ccc}
1 & 6 & -18\\
0&24& -67\\
0&24&-67\\
0&48&-134
\end{array}\right]
\rightarrow
\left[\begin{array}{ccc}
1 & 6 & -18\\
0&24& -67\\
0&0&0\\
0&0&0
\end{array}\right]
\rightarrow
$$$$\rightarrow
\left[\begin{array}{ccc}
1 & 6 & -18\\
0&1& -67/24\\
0&0&0\\
0&0&0
\end{array}\right]
\rightarrow
\left[\begin{array}{ccc}
1 & 0 & -5/4\\
0&1& -67/24\\
0&0&0\\
0&0&0
\end{array}\right]
$$Thus
$$x-\frac54z=0$$$$y-\frac{67}{24}z=0$$Thus the general solution is $(\frac54z, \frac{67}{24}z,z)$ for arbitrary $z\in F$.


Exercise 1.4.2

$A$ row-reduces as follows:
$$\rightarrow
\left[\begin{array}{cc}
1 & -i \\
1 & 1 \\
i & 1+i
\end{array}\right]
\rightarrow
\left[\begin{array}{cc}
1 & -i \\
0 & 1+i \\
0 & i
\end{array}\right]
\rightarrow
\left[\begin{array}{cc}
1 & -i \\
0 & 1 \\
0 & i
\end{array}\right]
\rightarrow
\left[\begin{array}{cc}
1 & 0 \\
0 & 1 \\
0 & 0
\end{array}\right]
$$Thus the only solution to $AX=0$ is $(0,0)$.


Exercise 1.4.3

$$
\left[\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right],\quad
\left[\begin{array}{cc}
1 & x\\
0 & 0
\end{array}\right],\quad
\left[\begin{array}{cc}
0 & 1\\
0 & 0
\end{array}\right],\quad
\left[\begin{array}{cc}
0 & 0\\
0 & 0
\end{array}\right]
$$


Exercise 1.4.4

The augmented coefficient matrix is
$$
\left[\begin{array}{ccc|c}
1 & -1 & 2 & 1\\
2 & 0 & 2 & 1\\
1 & -3 & 4 & 2
\end{array}\right]
$$We row reduce it as follows:
$$
\rightarrow
\left[\begin{array}{ccc|c}
1 & -1 & 2 & 1\\
0 & 2 & -2 & -1\\
0 & -2 & 2 & 1
\end{array}\right]
\rightarrow
\left[\begin{array}{ccc|c}
1 & -1 & 2 & 1\\
0 & 1 & -1 & -1/2\\
0 & 0 & 0 & 0
\end{array}\right]
\rightarrow
\left[\begin{array}{ccc|c}
1 & 0 & 1 & 1/2\\
0 & 1 & -1 & -1/2\\
0 & 0 & 0 & 0
\end{array}\right]
$$Thus the system is equivalent to\begin{alignat*}{1}
x_1+x_3 &= 1/2\\
x_2-x_3 &= -1/2
\end{alignat*}Thus the solutions are parameterized by $x_3$. Setting $x_3=c$ gives $x_1=1/2-c$, $x_2=c-1/2$. Thus the general solution is
$$\left(\textstyle\frac12-c,\ \ c-\frac12,\ \ c\right)$$for $c\in\mathbb R$.


Exercise 1.4.5

$$x+y=0$$$$x+y=1$$


Exercise 1.4.6

The augmented coefficient matrix is as follows
$$\left[\begin{array}{cccc|c}
1 & -2 & 1 & 2 & 1\\
1 & 1 & -1 & 1& 2\\
1 & 7 & -5 & -1 & 3
\end{array}\right]
$$This row reduces as follows:
$$
\rightarrow
\left[\begin{array}{cccc|c}
1 & -2 & 1 & 2 & 1\\
0 &3 & -2 & -1& 1\\
0 & 9 & -6 & -3 & 2
\end{array}\right]
\rightarrow
\left[\begin{array}{cccc|c}
1 & -2 & 1 & 2 & 1\\
0 &3 & -2 & -1& 1\\
0 & 0 & 0 & 0 & -1
\end{array}\right]
$$At this point there's no need to continue because the last row says $0x_1 + 0x_2+0x_3+0x_4=-1$. But the left hand side of this equation is zero so this is impossible.


Exercise 1.4.7

The augmented coefficient matrix is
$$
\left[\begin{array}{ccccc|c}
2 & -3 & -7 & 5 & 2 & -2\\
1 & -2 & -4 & 3 & 1 & -2\\
2 & 0 & -4 & 2 & 1 & 3\\
1 & -5 & -7 & 6 & 2 & -7
\end{array}\right]
$$We row-reduce it as follows
$$
\rightarrow\left[\begin{array}{ccccc|c}
1 & -2 & -4 & 3 & 1 & -2\\
2 & -3 & -7 & 5 & 2 & -2\\
2 & 0 & -4 & 2 & 1 & 3\\
1 & -5 & -7 & 6 & 2 & -7
\end{array}\right]
\rightarrow\left[\begin{array}{ccccc|c}
1 & -2 & -4 & 3 & 1 & -2\\
0 & 1 & 1 & -1 & 0 & 2\\
0 & 4 & 4 & -4 & -1 & 7\\
0 & -3 & -3 & 3 & 1 & -5
\end{array}\right]
$$$$
\rightarrow\left[\begin{array}{ccccc|c}
1 & 0 & -2 & 1 & 1 & 2\\
0 & 1 & 1 & -1 & 0 & 2\\
0 & 0 & 0 & 0 & -1 & -1\\
0 & 0 & 0 & 0 & 1 & 1
\end{array}\right]
\rightarrow\left[\begin{array}{ccccc|c}
1 & 0 & -2 & 1 & 0 & 1\\
0 & 1 & 1 & -1 & 0 & 2\\
0 & 0 & 0 & 0 & 1 & 1\\
0 & 0 & 0 & 0 & 0 & 0
\end{array}\right]
$$Thus
\begin{alignat*}{1}
x_1-2x_3+x_4 &= 1\\
x_2+x_3-x_4 &=2\\
x_5 &= 1
\end{alignat*}Thus the general solution is given by $(1+2x_3-x_4, 2+x_4-x_3, x_3, x_4, 1)$ for arbitrary $x_3,x_4\in F$.


Exercise 1.4.8

The matrix $A$ is row-reduced as follows:
$$
\rightarrow\left[\begin{array}{ccc}
1 & -3 & 0\\
0 & 7 & 1\\
0 & 8 & 2
\end{array}\right]
\rightarrow\left[\begin{array}{ccc}
1 & -3 & 0\\
0 & 7 & 1\\
0 & 1 & 1
\end{array}\right]
\rightarrow\left[\begin{array}{ccc}
1 & -3 & 0\\
0 & 1 & 1\\
0 & 7 & 1
\end{array}\right]
$$$$
\rightarrow\left[\begin{array}{ccc}
1 & -3 & 0\\
0 & 1 & 1\\
0 & 0 & -6
\end{array}\right]
\rightarrow\left[\begin{array}{ccc}
1 & -3 & 0\\
0 & 1 & 1\\
0 & 0 & 1
\end{array}\right]
\rightarrow\left[\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{array}\right]
$$Thus for every $(y_1,y_2,y_3)$ there is a (unique) solution.


Exercise 1.4.9

We row reduce as follows
$$
\left[\begin{array}{cccc|c}
3 & -6 & 2 & -1 & y_1\\
-2 & 4 & 1 & 3 & y_2\\
0 & 0 & 1 & 1 & y_3\\
1 & -2 & 1 & 0 & y_4
\end{array}\right]
\rightarrow
\left[\begin{array}{cccc|c}
1 & -2 & 1 & 0 & y_4\\
3 & -6 & 2 & -1 & y_1\\
-2 & 4 & 1 & 3 & y_2\\
0 & 0 & 1 & 1 & y_3
\end{array}\right]
\rightarrow
\left[\begin{array}{cccc|c}
1 & -2 & 1 & 0 & y_4\\
0 & 0 & -1 & -1 & y_1-3y_4\\
0 & 0 & 3 & 3 & y_2+2y_4\\
0 & 0 & 1 & 1 & y_3
\end{array}\right]
$$$$
\rightarrow
\left[\begin{array}{cccc|c}
1 & -2 & 1 & 0 & y_4\\
0 & 0 & 0 & 0 & y_1-3y_4+y_3\\
0 & 0 & 0 & 0 & y_2+2y_4+3y_3\\
0 & 0 & 1 & 1 & y_3
\end{array}\right]
\rightarrow
\left[\begin{array}{cccc|c}
1 & -2 & 1 & 0 & y_4\\
0 & 0 & 1 & 1 & y_3\\
0 & 0 & 0 & 0 & y_1-3y_4+y_3\\
0 & 0 & 0 & 0 & y_2+2y_4+3y_3
\end{array}\right]
$$Thus $(y_1,y_2,y_3,y_4)$ must satisfy
\begin{alignat*}{1}
y_1+y_3-3y_4 &= 0\\
y_2 + 3y_3 + 2y_4 &= 0
\end{alignat*}The matrix for this system is
$$
\left[\begin{array}{cccc}
1 & 0 & 1 & -3\\
0 & 1 & 3 & 2
\end{array}\right]
$$of which the general solution is
$(-y_3+3y_4, -3y_3-2y_4, y_3, y_4)$ for arbitrary $y_3,y_4\in F$. These are the only $(y_1,y_2,y_3,y_4)$ for which the system $AX=Y$ has a solution.


Exercise 1.4.10

There are seven possible $2\times3$ row-reduced echelon matrices:
\begin{equation}
R_1=\left[\begin{array}{ccc}
0 &0 &0 \\
0 & 0& 0
\end{array}\right]
\label{m1}
\end{equation}\begin{equation}
R_2=\left[\begin{array}{ccc}
1 &0 &a \\
0 &1 & b
\end{array}\right]
\label{m2}
\end{equation}\begin{equation}
R_3=\left[\begin{array}{ccc}
1 & a& 0\\
0 &0 &1
\end{array}\right]
\label{m3}
\end{equation}\begin{equation}
R_4=\left[\begin{array}{ccc}
1 &a & b\\
0& 0& 0
\end{array}\right]
\label{m4}
\end{equation}\begin{equation}
R_5=\left[\begin{array}{ccc}
0& 1& a\\
0&0 &0
\end{array}\right]
\label{m5}
\end{equation}\begin{equation}
R_6=\left[\begin{array}{ccc}
0&1 &0 \\
0&0 &1
\end{array}\right]
\label{m6}
\end{equation}\begin{equation}
R_7=\left[\begin{array}{ccc}
0& 0&1 \\
0& 0&0
\end{array}\right]
\label{m7}
\end{equation}We must show that no two of these have exactly the same solutions. For the first one $R_1$, any $(x,y,z)$ is a solution and that's not the case for any of the other $R_i$'s. Consider next $R_7$. In this case $z=0$ and $x$ and $y$ can be anything. We can have $z\not=0$ for $R_2$, $R_3$ and $R_5$. So the only ones $R_7$ could share solutions with are $R_3$ or $R_6$. But both of those have restrictions on $x$ and/or $y$ so the solutions cannot be the same. Also $R_3$ and $R_6$ cannot have the same solutions since $R_6$ forces $y=0$ while $R_3$ does not.

Thus we have shown that if two $R_i$'s share the same solutions then they must be among $R_2$, $R_4$, and $R_5$.

The solutions for $R_2$ are $(-az, -bz, z)$, for $z$ arbitrary. The solutions for $R_4$ are $(-a'y-b'z,y,z)$ for $y,z$ arbitrary. Thus $(-b',0,1)$ is a solution for $R_4$. Suppose this is also a solution for $R_2$. Then $z=1$ so it is of the form $(-a,-b,1)$ and it must be that $(-b',0,1)=(-a,-b,1)$. Comparing the second component implies $b=0$. But if $b=0$ then $R_2$ implies $y=0$. But $R_4$ allows for arbitrary $y$. Thus $R_2$ and $R_4$ cannot share the same solutions.

The solutions for $R_2$ are $(-az, -bz, z)$, for $z$ arbitrary. The solutions for $R_5$ are $(x,-a'z,z)$ for $x,z$ arbitrary. Thus $(0,-a',1)$ is a solution for $R_5$. As before if this is a solution of $R_2$ then $a=0$. But if $a=0$ then $R_2$ forces $x=0$ while in $R_5$ $x$ can be arbitrary. Thus $R_2$ and $R_5$ cannot share the same solutions.

The solutions for $R_4$ are $(-ay-bz,y,z)$ for $y,z$ arbitrary. The solutions for $R_5$ are $(x,-a'z,z)$ for $x,z$ arbitrary. Thus setting $x=1$, $z=0$ gives $(1,0,0)$ is a solution for $R_5$. But this cannot be a solution for $R_4$ since if $y=z=0$ then first component must also be zero.

Thus we have shown that no two $R_i$ and $R_j$ have the same solutions unless $i=j$.

From http://greggrant.org

Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

This Post Has One Comment

  1. Exercise 1.4.9 I believe that it is incorrect. y2 should = 3*y3 - 2*y4

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