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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.

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