Solution to Linear Algebra Hoffman & Kunze Chapter 1.3 Exercise 1.3.7 Solution: 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 $R_1$…
Solution to Linear Algebra Hoffman & Kunze Chapter 1.3 Exercise 1.3.6 Solution: 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 $c=0$,…
Solution to Linear Algebra Hoffman & Kunze Chapter 1.3 Exercise 1.3.5 Solution: 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…
Solution to Linear Algebra Hoffman & Kunze Chapter 1.3 Exercise 1.3.4 Solution: We have $$A\rightarrow\left[\begin{array}{ccc} 1 & -2 & 1\\ i & -(1+i) & 0\\ 1 & 2i & -1…
Solution to Linear Algebra Hoffman & Kunze Chapter 1.3 Exercise 1.3.3 Solution: The system $AX=2X$ is $$\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]$$which is the same as \begin{alignat*}{1} 6x-4y&=2x\\ 4x-2y&=2y\\ -x+3z&=2z \end{alignat*}which is equivalent to \begin{alignat*}{1}…
Solution to Linear Algebra Hoffman & Kunze Chapter 1.3 Exercise 1.3.2 Solution: We have$$\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&1/7\\0&0&1\end{array}\right].$$Thus $A$ is row-equivalent to the identity matrix. It follows that the only solution to the…
Solution to Linear Algebra Hoffman & Kunze Chapter 1.3 Exercise 1.3.1 Solution: The matrix of coefficients is $$\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] $$Thus $2x_1+(1-i)x_2=0$. Thus for any $x_2\in\mathbb C$, $(\frac12(i-1)x_2,x_2)$ is…
Solution to Linear Algebra Hoffman & Kunze Chapter 1.2 Exercise 1.2.8 Solution: Call the additive and multiplicative identities of $\mb F$ $0_{\mb F}$ and $1_{\mb F}$ respectively. Define $n_{\mb F}$…
Solution to Linear Algebra Hoffman & Kunze Chapter 1.2 Exercise 1.2.7 Solution: Every subfield of $\C$ has characterisitc zero since if $\mb F$ is a subfield then $1\in \mb F$…
Solution to Linear Algebra Hoffman & Kunze Chapter 1.2 Exercise 1.2.6 Solution: Write the two systems as follows: $$ \begin{array}{c} a_{11}x+a_{12}y=0\\ a_{21}x+a_{22}y=0\\ \vdots\\ a_{m1}x+a_{m2}y=0 \end{array} \quad\quad \begin{array}{c} b_{11}x+b_{12}y=0\\ b_{21}x+b_{22}y=0\\ \vdots\\…