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\ihead{Fall 2019 Math 201-003 at Drexel: blackboard notes of 2019-10-30}
\ohead{page \thepage}
\cfoot{}
\begin{document}

\title{Fall 2019 Math 201-003 at Drexel: blackboard notes of 2019-10-30 (updated version)}
\author{Darij Grinberg}
\date{\today}
\maketitle

\section{Determinants}

\subsection{Some determinant identities (recall)}

Recall how we defined determinants:

\begin{definition}
Let $A$ be an $n\times n$-matrix. Then, the \textbf{determinant} $\det A$ of
$A$ is defined to be the sum%
\[
\sum_{\sigma\text{ is a permutation of }\left[  n\right]  }%
\operatorname*{sign}\left(  \sigma\right)  \cdot A_{1,\sigma\left(  1\right)
}A_{2,\sigma\left(  2\right)  }\cdots A_{n,\sigma\left(  n\right)  }.
\]

\end{definition}

Last time we proved the following properties of determinants:

\begin{theorem}
\label{thm.det.triang}If an $n\times n$-matrix $A$ is triangular (i.e.,
upper-triangular or lower-triangular), then its determinant is the product of
its diagonal elements:
\[
\det A=A_{1,1}A_{2,2}\cdots A_{n,n}.
\]

\end{theorem}

\begin{theorem}
\label{thm.det.transpose}If $A$ is any $n\times n$-matrix, then $\det\left(
A^{T}\right)  =\det A$.
\end{theorem}

\begin{theorem}
\label{thm.det.equal-rows}If an $n\times n$-matrix $A$ has two equal rows,
then $\det A=0$.
\end{theorem}

\begin{theorem}
\label{thm.det.scale-row}If we scale a row of an $n\times n$-matrix $A$ by a
number $\lambda$, then $\det A$ gets multiplied by $\lambda$.
\end{theorem}

\subsection{More determinant identities}

\begin{corollary}
\label{cor.det.zero-row}If an $n\times n$-matrix $A$ has a zero row (i.e., a
row full of zeroes), then $\det A=0$.
\end{corollary}

\begin{proof}
Assume that the $n\times n$-matrix $A$ has a zero row. Then, scaling this row
by $0$ does not change $A$, but (by Theorem \ref{thm.det.scale-row}) it
multiplies $\det A$ by $0$. Hence,%
\[
\det A=0\cdot\det A=0.
\]

\end{proof}

\begin{example}
Here is how this argument works if $n=3$ and if the second row of $A$ is zero:%
\[
\det\left(
\begin{array}
[c]{ccc}%
a & b & c\\
0 & 0 & 0\\
g & h & i
\end{array}
\right)  =0\cdot\det\left(
\begin{array}
[c]{ccc}%
a & b & c\\
0 & 0 & 0\\
g & h & i
\end{array}
\right)  =0.
\]

\end{example}

The next theorem goes under the name \textquotedblleft multilinearity of the
determinant with respect to the rows\textquotedblright\ or \textquotedblleft
the determinant is linear in each row\textquotedblright.

\begin{theorem}
\label{thm.det.muli-rows}Let $n$ be a nonnegative integer and $k\in\left[
n\right]  $. (Recall that $\left[  n\right]  $ stands for the set $\left\{
1,2,\ldots,n\right\}  $.)

Let $A$, $B$ and $C$ be three $n\times n$-matrices that differ from each other
only in their $k$-th rows: i.e., that satisfy%
\[
\operatorname{row}_{i}A=\operatorname{row}_{i}B=\operatorname{row}%
_{i}C\ \ \ \ \ \ \ \ \ \ \text{for all }i\neq k.
\]
Further assume that%
\[
\operatorname{row}_{k}C=\operatorname{row}_{k}A+\operatorname{row}_{k}B.
\]
Then,%
\[
\det C=\det A+\det B.
\]

\end{theorem}

\begin{example}
\textbf{(a)} If $n=3$ and $k=2$, then three $n\times n$-matrices $A$, $B$, $C$
that differ from each other only in their $k$-th rows look as follows:%
\[
A=\left(
\begin{array}
[c]{ccc}%
a & b & c\\
d & e & f\\
g & h & i
\end{array}
\right)  ,\ \ \ \ \ \ \ \ \ \ B=\left(
\begin{array}
[c]{ccc}%
a & b & c\\
d^{\prime} & e^{\prime} & f^{\prime}\\
g & h & i
\end{array}
\right)  ,\ \ \ \ \ \ \ \ \ \ C=\left(
\begin{array}
[c]{ccc}%
a & b & c\\
d^{\prime\prime} & e^{\prime\prime} & f^{\prime\prime}\\
g & h & i
\end{array}
\right)
\]
for some reals $a,b,c,d,e,f,d^{\prime},e^{\prime},f^{\prime},d^{\prime\prime
},e^{\prime\prime},f^{\prime\prime},g,h,i$. If they furthermore satisfy
$\operatorname{row}_{k}C=\operatorname{row}_{k}A+\operatorname{row}_{k}B$,
then we must have $d^{\prime\prime}=d+d^{\prime}$ and $e^{\prime\prime
}=e+e^{\prime}$ and $f^{\prime\prime}=f+f^{\prime}$.

Hence, if $n=3$ and $k=2$, then Theorem \ref{thm.det.muli-rows} is saying that%
\[
\det\underbrace{\left(
\begin{array}
[c]{ccc}%
a & b & c\\
d+d^{\prime} & e+e^{\prime} & f+f^{\prime}\\
g & h & i
\end{array}
\right)  }_{\text{this is }C}=\det\underbrace{\left(
\begin{array}
[c]{ccc}%
a & b & c\\
d & e & f\\
g & h & i
\end{array}
\right)  }_{\text{this is }A}+\det\underbrace{\left(
\begin{array}
[c]{ccc}%
a & b & c\\
d^{\prime} & e^{\prime} & f^{\prime}\\
g & h & i
\end{array}
\right)  }_{\text{this is }B}.
\]


\textbf{(b)} If $n=3$ and $k=1$, then Theorem \ref{thm.det.muli-rows} is
saying that%
\[
\det\left(
\begin{array}
[c]{ccc}%
a+a^{\prime} & b+b^{\prime} & c+c^{\prime}\\
d & e & f\\
g & h & i
\end{array}
\right)  =\det\left(
\begin{array}
[c]{ccc}%
a & b & c\\
d & e & f\\
g & h & i
\end{array}
\right)  +\det\left(
\begin{array}
[c]{ccc}%
a^{\prime} & b^{\prime} & c^{\prime}\\
d & e & f\\
g & h & i
\end{array}
\right)  .
\]

\end{example}

\textbf{Warning:} Theorem \ref{thm.det.muli-rows} does \textbf{not} say that
$\det\left(  A+B\right)  =\det A+\det B$. Indeed, if $A$ and $B$ are two
$n\times n$-matrices with $n>1$, then $\det A$ and $\det B$ do not determine
$\det\left(  A+B\right)  $. For example, for $n=2$, if $A=\left(
\begin{array}
[c]{cc}%
a & b\\
c & d
\end{array}
\right)  $ and $B=\left(
\begin{array}
[c]{cc}%
a^{\prime} & b^{\prime}\\
c^{\prime} & d^{\prime}%
\end{array}
\right)  $, then%
\begin{align*}
\det\left(  A+B\right)   &  =\det\left(
\begin{array}
[c]{cc}%
a+a^{\prime} & b+b^{\prime}\\
c+c^{\prime} & d+d^{\prime}%
\end{array}
\right) \\
&  =\det\left(
\begin{array}
[c]{cc}%
a & b\\
c+c^{\prime} & d+d^{\prime}%
\end{array}
\right)  +\det\left(
\begin{array}
[c]{cc}%
a^{\prime} & b^{\prime}\\
c+c^{\prime} & d+d^{\prime}%
\end{array}
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Theorem \ref{thm.det.muli-rows},
applied to }n=2\text{ and }k=2\right) \\
&  =\det\underbrace{\left(
\begin{array}
[c]{cc}%
a & b\\
c & d
\end{array}
\right)  }_{=A}+\det\left(
\begin{array}
[c]{cc}%
a & b\\
c^{\prime} & d^{\prime}%
\end{array}
\right)  +\det\left(
\begin{array}
[c]{cc}%
a^{\prime} & b^{\prime}\\
c & d
\end{array}
\right)  +\det\underbrace{\left(
\begin{array}
[c]{cc}%
a^{\prime} & b^{\prime}\\
c^{\prime} & d^{\prime}%
\end{array}
\right)  }_{=B}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Theorem \ref{thm.det.muli-rows},
applied to }n=2\text{ and }k=1\right) \\
&  \neq\det A+\det B\ \ \ \ \ \ \ \ \ \ \left(  \text{in general}\right)  .
\end{align*}


\begin{proof}
[Proof of Theorem \ref{thm.det.muli-rows}.]Let us first prove Theorem
\ref{thm.det.muli-rows} on an example: Set $n=3$ and $k=2$. Then, the
definition of a determinant of a $3\times3$-matrix yields%
\begin{align*}
&  \det\left(
\begin{array}
[c]{ccc}%
a & b & c\\
d+d^{\prime} & e+e^{\prime} & f+f^{\prime}\\
g & h & i
\end{array}
\right) \\
&  =a\left(  e+e^{\prime}\right)  i+b\left(  f+f^{\prime}\right)  g+c\left(
d+d^{\prime}\right)  h-a\left(  f+f^{\prime}\right)  h-b\left(  d+d^{\prime
}\right)  i-c\left(  e+e^{\prime}\right)  g\\
&  =aei+ae^{\prime}i+bfg+bf^{\prime}g+cdh+cd^{\prime}h-afh-af^{\prime
}h-bdi-bd^{\prime}i-ceg-ce^{\prime}g\\
&  =\underbrace{\left(  aei+bfg+cdh-afh-bdi-ceg\right)  }_{=\det\left(
\begin{array}
[c]{ccc}%
a & b & c\\
d & e & f\\
g & h & i
\end{array}
\right)  }+\underbrace{\left(  ae^{\prime}i+bf^{\prime}g+cd^{\prime
}h-af^{\prime}h-bd^{\prime}i-ce^{\prime}g\right)  }_{=\det\left(
\begin{array}
[c]{ccc}%
a & b & c\\
d^{\prime} & e^{\prime} & f^{\prime}\\
g & h & i
\end{array}
\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have sorted all products that involve one of }d,e,f\\
\text{into one pair of parentheses, and all products that involve}\\
\text{one of }d^{\prime},e^{\prime},f^{\prime}\text{ into another}%
\end{array}
\right) \\
&  =\det\left(
\begin{array}
[c]{ccc}%
a & b & c\\
d & e & f\\
g & h & i
\end{array}
\right)  +\det\left(
\begin{array}
[c]{ccc}%
a & b & c\\
d^{\prime} & e^{\prime} & f^{\prime}\\
g & h & i
\end{array}
\right)  .
\end{align*}
This proves Theorem \ref{thm.det.muli-rows} when $n=3$ and $k=2$.

A similar argument applies in general (but of course, much more bookkeeping is
required in the general case): The definition of a determinant yields%
\begin{align}
\det A  &  =\sum_{\sigma\text{ is a permutation of }\left[  n\right]
}\operatorname*{sign}\left(  \sigma\right)  \cdot\underbrace{A_{1,\sigma
\left(  1\right)  }A_{2,\sigma\left(  2\right)  }\cdots A_{n,\sigma\left(
n\right)  }}_{\text{let me call this product }A_{\sigma}}\nonumber\\
&  =\sum_{\sigma\text{ is a permutation of }\left[  n\right]  }%
\operatorname*{sign}\left(  \sigma\right)  \cdot A_{\sigma},
\label{pf.thm.det.muli-rows.detA=}%
\end{align}
where we set $A_{\sigma}:=A_{1,\sigma\left(  1\right)  }A_{2,\sigma\left(
2\right)  }\cdots A_{n,\sigma\left(  n\right)  }$ for any permutation $\sigma
$. Similarly,%
\begin{equation}
\det B=\sum_{\sigma\text{ is a permutation of }\left[  n\right]
}\operatorname*{sign}\left(  \sigma\right)  \cdot B_{\sigma},
\label{pf.thm.det.muli-rows.detB=}%
\end{equation}
where we set $B_{\sigma}=B_{1,\sigma\left(  1\right)  }B_{2,\sigma\left(
2\right)  }\cdots B_{n,\sigma\left(  n\right)  }$. Similarly,%
\begin{equation}
\det C=\sum_{\sigma\text{ is a permutation of }\left[  n\right]
}\operatorname*{sign}\left(  \sigma\right)  \cdot C_{\sigma},
\label{pf.thm.det.muli-rows.detC=}%
\end{equation}
where we set $C_{\sigma}=C_{1,\sigma\left(  1\right)  }C_{2,\sigma\left(
2\right)  }\cdots C_{n,\sigma\left(  n\right)  }$.

Now, fix a permutation $\sigma$ of $\left[  n\right]  $. Consider the product
$A_{\sigma}=A_{1,\sigma\left(  1\right)  }A_{2,\sigma\left(  2\right)  }\cdots
A_{n,\sigma\left(  n\right)  }$. One of its factors is $A_{k,\sigma\left(
k\right)  }$, which is an entry of the $k$-th row of $A$. All its other
factors come from the other $n-1$ rows of $A$. Thus, we can write it as
$A_{\sigma}=A_{k,\sigma\left(  k\right)  }\widetilde{A}_{\sigma}$, where
\begin{equation}
\widetilde{A}_{\sigma}=A_{1,\sigma\left(  1\right)  }A_{2,\sigma\left(
2\right)  }\cdots A_{k-1,\sigma\left(  k-1\right)  }A_{k+1,\sigma\left(
k+1\right)  }\cdots A_{n,\sigma\left(  n\right)  }
\label{pf.thm.det.muli-rows.Atil}%
\end{equation}
is the product of all factors of $A_{1,\sigma\left(  1\right)  }%
A_{2,\sigma\left(  2\right)  }\cdots A_{n,\sigma\left(  n\right)  }$ other
than $A_{k,\sigma\left(  k\right)  }$. Similarly, we can write $B_{\sigma}$ as
$B_{\sigma}=B_{k,\sigma\left(  k\right)  }\widetilde{B}_{\sigma}$, where
\begin{equation}
\widetilde{B}_{\sigma}=B_{1,\sigma\left(  1\right)  }B_{2,\sigma\left(
2\right)  }\cdots B_{k-1,\sigma\left(  k-1\right)  }B_{k+1,\sigma\left(
k+1\right)  }\cdots B_{n,\sigma\left(  n\right)  }
\label{pf.thm.det.muli-rows.Btil}%
\end{equation}
is a product similar to $\widetilde{A}_{\sigma}$ but using the entries of $B$
instead of $A$. Similarly, we can write $C_{\sigma}$ as $C_{\sigma
}=C_{k,\sigma\left(  k\right)  }\widetilde{C}_{\sigma}$, where
\begin{equation}
\widetilde{C}_{\sigma}=C_{1,\sigma\left(  1\right)  }C_{2,\sigma\left(
2\right)  }\cdots C_{k-1,\sigma\left(  k-1\right)  }C_{k+1,\sigma\left(
k+1\right)  }\cdots C_{n,\sigma\left(  n\right)  }
\label{pf.thm.det.muli-rows.Ctil}%
\end{equation}
is a product similar to $\widetilde{A}_{\sigma}$ but using the entries of $C$
instead of $A$.

But we assumed that the matrices $A$, $B$ and $C$ differ from each other only
in their $k$-th rows. Thus, the entries $A_{i,\sigma\left(  i\right)  }$ that
appear on the right hand side of (\ref{pf.thm.det.muli-rows.Atil}) are the
same as the entries $B_{i,\sigma\left(  i\right)  }$ that appear on the right
hand side of (\ref{pf.thm.det.muli-rows.Btil}) and also the same as the
entries $C_{i,\sigma\left(  i\right)  }$ that appear on the right hand side of
(\ref{pf.thm.det.muli-rows.Ctil}) (since no entries from the $k$-th row of any
matrix appear on any of these right hand sides). Therefore, the right hand
sides of the three equalities (\ref{pf.thm.det.muli-rows.Atil}),
(\ref{pf.thm.det.muli-rows.Btil}) and (\ref{pf.thm.det.muli-rows.Ctil}) are
equal. Thus, the left hand sides are equal as well. In other words,%
\begin{equation}
\widetilde{A}_{\sigma}=\widetilde{B}_{\sigma}=\widetilde{C}_{\sigma}.
\label{pf.thm.det.muli-rows.tileqs}%
\end{equation}


Moreover, we assumed that $\operatorname{row}_{k}C=\operatorname{row}%
_{k}A+\operatorname{row}_{k}B$. Thus, $C_{k,\sigma\left(  k\right)
}=A_{k,\sigma\left(  k\right)  }+B_{k,\sigma\left(  k\right)  }$ (since
$A_{k,\sigma\left(  k\right)  }$, $B_{k,\sigma\left(  k\right)  }$ and
$C_{k,\sigma\left(  k\right)  }$ are corresponding entries in
$\operatorname{row}_{k}A$, $\operatorname{row}_{k}B$ and $\operatorname{row}%
_{k}C$). Hence,%
\begin{align*}
C_{\sigma}  &  =\underbrace{C_{k,\sigma\left(  k\right)  }}_{=A_{k,\sigma
\left(  k\right)  }+B_{k,\sigma\left(  k\right)  }}\widetilde{C}_{\sigma
}=\left(  A_{k,\sigma\left(  k\right)  }+B_{k,\sigma\left(  k\right)
}\right)  \widetilde{C}_{\sigma}=A_{k,\sigma\left(  k\right)  }%
\underbrace{\widetilde{C}_{\sigma}}_{\substack{=\widetilde{A}_{\sigma
}\\\text{(by (\ref{pf.thm.det.muli-rows.tileqs}))}}}+B_{k,\sigma\left(
k\right)  }\underbrace{\widetilde{C}_{\sigma}}_{\substack{=\widetilde{B}%
_{\sigma}\\\text{(by (\ref{pf.thm.det.muli-rows.tileqs}))}}}\\
&  =\underbrace{A_{k,\sigma\left(  k\right)  }\widetilde{A}_{\sigma}%
}_{=A_{\sigma}}+\underbrace{B_{k,\sigma\left(  k\right)  }\widetilde{B}%
_{\sigma}}_{=B_{\sigma}}=A_{\sigma}+B_{\sigma}.
\end{align*}


Now, forget that we fixed $\sigma$. We thus have proved that $C_{\sigma
}=A_{\sigma}+B_{\sigma}$ for every permutation $\sigma$. Hence,
(\ref{pf.thm.det.muli-rows.detC=}) becomes%
\begin{align*}
\det C  &  =\sum_{\sigma\text{ is a permutation of }\left[  n\right]
}\operatorname*{sign}\left(  \sigma\right)  \cdot\underbrace{C_{\sigma}%
}_{=A_{\sigma}+B_{\sigma}}\\
&  =\sum_{\sigma\text{ is a permutation of }\left[  n\right]  }%
\operatorname*{sign}\left(  \sigma\right)  \cdot\left(  A_{\sigma}+B_{\sigma
}\right) \\
&  =\underbrace{\sum_{\sigma\text{ is a permutation of }\left[  n\right]
}\operatorname*{sign}\left(  \sigma\right)  \cdot A_{\sigma}}_{\substack{=\det
A\\\text{(by (\ref{pf.thm.det.muli-rows.detA=}))}}}+\underbrace{\sum
_{\sigma\text{ is a permutation of }\left[  n\right]  }\operatorname*{sign}%
\left(  \sigma\right)  \cdot B_{\sigma}}_{\substack{=\det B\\\text{(by
(\ref{pf.thm.det.muli-rows.detB=}))}}}\\
&  =\det A+\det B.
\end{align*}
This proves Theorem \ref{thm.det.muli-rows}.
\end{proof}

\begin{corollary}
\label{cor.det.row-addition}Let $A$ be an $n\times n$-matrix, and let $p$ and
$q$ be two distinct elements of $\left[  n\right]  $. If we add $\lambda
\cdot\operatorname{row}_{p}A$ to the $q$-th row of $A$, then $\det A$ does not change.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.det.row-addition}.]This is another
\textquotedblleft proof by example\textquotedblright: We only consider the
case when $n=3$, $p=3$ and $q=2$, and we write our $3\times3$-matrix $A$ as
$\left(
\begin{array}
[c]{ccc}%
a & b & c\\
d & e & f\\
g & h & i
\end{array}
\right)  $. Then, we need to prove%
\[
\det\left(
\begin{array}
[c]{ccc}%
a & b & c\\
d+\lambda g & e+\lambda h & f+\lambda i\\
g & h & i
\end{array}
\right)  =\det\left(
\begin{array}
[c]{ccc}%
a & b & c\\
d & e & f\\
g & h & i
\end{array}
\right)
\]
(because the matrix on the left hand side is what we obtain if we add
$\lambda\cdot\operatorname{row}_{p}A$ to the $q$-th row of $A$).

By Theorem \ref{thm.det.muli-rows} (applied to $n=3$ and $k=2$), we have%
\begin{align*}
&  \det\left(
\begin{array}
[c]{ccc}%
a & b & c\\
d+\lambda g & e+\lambda h & f+\lambda i\\
g & h & i
\end{array}
\right) \\
&  =\det\left(
\begin{array}
[c]{ccc}%
a & b & c\\
d & e & f\\
g & h & i
\end{array}
\right)  +\underbrace{\det\left(
\begin{array}
[c]{ccc}%
a & b & c\\
\lambda g & \lambda h & \lambda i\\
g & h & i
\end{array}
\right)  }_{\substack{=\lambda\det\left(
\begin{array}
[c]{ccc}%
a & b & c\\
g & h & i\\
g & h & i
\end{array}
\right)  \\\text{(by Theorem \ref{thm.det.scale-row},}\\\text{because the
matrix has had its second row}\\\text{scaled by }\lambda\text{)}}}\\
&  =\det\left(
\begin{array}
[c]{ccc}%
a & b & c\\
d & e & f\\
g & h & i
\end{array}
\right)  +\lambda\underbrace{\det\left(
\begin{array}
[c]{ccc}%
a & b & c\\
g & h & i\\
g & h & i
\end{array}
\right)  }_{\substack{=0\\\text{(by Theorem \ref{thm.det.equal-rows}%
,}\\\text{since the matrix here has two equal rows)}}}\\
&  =\det\left(
\begin{array}
[c]{ccc}%
a & b & c\\
d & e & f\\
g & h & i
\end{array}
\right)  ,
\end{align*}
which is exactly what we need to prove.

The general case (with $n$, $p$ and $q$ arbitrary) is proved in the same way.
\end{proof}

\begin{theorem}
\label{thm.det.row-swap}If we swap two rows of an $n\times n$-matrix, then its
determinant gets multiplied by $-1$ (that is, it flips its sign but preserves
its magnitude).
\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.det.row-swap}.]You can swap rows $i$ and $j$ of a
matrix by performing the following sequence of row operations:\footnote{This
is a version of what is known as the \textquotedblleft XOR swapping
technique\textquotedblright.}

\begin{itemize}
\item add row $i$ to row $j$;

\item then subtract row $j$ from row $i$;

\item then add row $i$ to row $j$;

\item then scale row $i$ by $-1$.
\end{itemize}

\noindent(You can easily convince yourself of this claim by tracking what
happens to the entries in rows $i$ and $j$ when these four row operations are
performed.\footnote{Proof by example: Let us assume that $n=1$ and $i=1$ and
$j=2$. Thus, we want to swap rows $1$ and $2$ of the $2\times2$-matrix
$\left(
\begin{array}
[c]{cc}%
a & b\\
c & d
\end{array}
\right)  $. Our sequence of row operations then does the following:%
\begin{align*}
&  \left(
\begin{array}
[c]{cc}%
a & b\\
c & d
\end{array}
\right)  \overset{\text{add row }1\text{ to row }2}{\longrightarrow}\left(
\begin{array}
[c]{cc}%
a & b\\
a+c & b+d
\end{array}
\right)  \overset{\text{subtract row }2\text{ from row }1}{\longrightarrow
}\left(
\begin{array}
[c]{cc}%
-c & -d\\
a+c & b+d
\end{array}
\right) \\
&  \overset{\text{add row }1\text{ to row }2}{\longrightarrow}\left(
\begin{array}
[c]{cc}%
-c & -d\\
a & b
\end{array}
\right)  \overset{\text{scale row }1\text{ by }-1}{\longrightarrow}\left(
\begin{array}
[c]{cc}%
c & d\\
a & b
\end{array}
\right)  .
\end{align*}
And indeed, the result is what you would get from the original matrix by
swapping rows $1$ and $2$.})

What happens to the determinant when we perform these four row operations? The
first three operations do not change the determinant, because they all have
the form \textquotedblleft add $\lambda\cdot\operatorname{row}_{p}A$ to the
$q$-th row\textquotedblright\ for some values of $p$, $q$ and $\lambda$ (and
as we know from Corollary \ref{cor.det.row-addition}, the determinant does not
change when we perform such operations). The fourth operation multiplies the
determinant by $-1$, because of Theorem \ref{thm.det.scale-row}. Thus,
altogether, the determinant gets multiplied by $-1$.
\end{proof}

\subsection{Computing determinants by row operations}

We recall our old definition of row operations (from
\href{http://www.cip.ifi.lmu.de/~grinberg/t/19fla/2019-09-30.pdf}{the
2019-09-30 classwork}):

\begin{definition}
The following operations on a matrix $A$ are called \textbf{elementary row
operations} (short \textbf{EROs}):

\begin{itemize}
\item \textbf{ERO1:} Exchange two rows.

\item \textbf{ERO2:} Scale a row by a nonzero constant.

\item \textbf{ERO3:} Add a multiple of one row to another row. (That is, add
$\lambda\operatorname{row}_{i}A$ to $\operatorname{row}_{j}A$ for $\lambda
\in\mathbb{R}$ and $i\neq j$.)
\end{itemize}
\end{definition}

Let us give these operations some more descriptive names:

\begin{itemize}
\item ERO1 will be called a \textbf{row swap}.

\item ERO2 will be called a \textbf{row scaling} with scaling factor $\lambda
$, where $\lambda$ is the nonzero constant by which the row is being scaled.

\item ERO3 will be called a \textbf{row addition}.
\end{itemize}

We can now tell what happens to the determinant of an $n\times n$-matrix when
the matrix undergoes any of these row operations:

\begin{itemize}
\item If the matrix undergoes a row swap (i.e., ERO1), then its determinant
gets multiplied by $-1$ (by Theorem \ref{thm.det.row-swap}).).

\item If the matrix undergoes a row scaling (i.e., ERO2) with scaling factor
$\lambda$, then its determinant gets multiplied by $\lambda$ (by Theorem
\ref{thm.det.scale-row}).

\item If the matrix undergoes a row addition (i.e., ERO3), then its
determinant does not change (by Corollary \ref{cor.det.row-addition}).
\end{itemize}

\begin{corollary}
\label{cor.det.rowops}Let $A$ be an $n\times n$-matrix, and let $B$ be a
matrix obtained from $A$ by a sequence of EROs. Assume that this sequence
contains exactly $k$ row swaps and exactly $\ell$ row scalings with scaling
factors $\lambda_{1},\lambda_{2},\ldots,\lambda_{\ell}$. (We don't care how
many row additions it contains.) Then,%
\[
\det B=\left(  -1\right)  ^{k}\underbrace{\lambda_{1}\lambda_{2}\cdots
\lambda_{\ell}}_{\substack{\text{this is understood}\\\text{to mean }1\text{
if }\ell=0}}\det A.
\]

\end{corollary}

\begin{proof}
The previous results show how each row operation affects the determinant:

\begin{itemize}
\item A row addition doesn't change it at all.

\item A row swap multiplies it by $-1$.

\item A row scaling with scaling factor $\lambda$ multiplies it by $\lambda$.
\end{itemize}

Thus, our sequence of EROs that took $A$ to $B$ multiplies the determinant by
$\underbrace{-1,-1,\ldots,-1}_{k\text{ times}},\lambda_{1},\lambda_{2}%
,\ldots,\lambda_{\ell}$. Thus, in total, it multiplies the determinant by
\newline$\left(  -1\right)  ^{k}\lambda_{1}\lambda_{2}\cdots\lambda_{\ell}$.
This proves Corollary \ref{cor.det.rowops}.
\end{proof}

Corollary \ref{cor.det.rowops} gives us a quick way of computing a determinant
of a square matrix $A$: We perform row-reduction to bring $A$ into RREF. The
resulting RREF matrix is upper-triangular (indeed, you can easily check that
any square matrix in RREF is upper-triangular\footnote{Actually, any square
matrix in RREF is diagonal, which is better than upper-triangular.}), so we
can compute its determinant easily (by Theorem \ref{thm.det.triang}). Then, by
Corollary \ref{cor.det.rowops}, we get $\det A$. (We don't even need to go all
the way to an RREF; usually we will obtain an upper-triangular matrix in the
row-reduction process long before we obtain a RREF matrix.)

\begin{example}
Let us compute $\det A$ for $A=\left(
\begin{array}
[c]{cccc}%
1 & 1 & 1 & 1\\
1 & 2 & 2 & 2\\
1 & 2 & 3 & 3\\
1 & 2 & 3 & 4
\end{array}
\right)  $. Indeed, we row-reduce $A$ as follows:%
\begin{align*}
A  &  =\left(
\begin{array}
[c]{cccc}%
1 & 1 & 1 & 1\\
1 & 2 & 2 & 2\\
1 & 2 & 3 & 3\\
1 & 2 & 3 & 4
\end{array}
\right) \\
&  \overset{\text{add }-1\cdot\operatorname{row}1\text{ to }\operatorname{row}%
2}{\longrightarrow}\left(
\begin{array}
[c]{cccc}%
1 & 1 & 1 & 1\\
0 & 1 & 1 & 1\\
1 & 2 & 3 & 3\\
1 & 2 & 3 & 4
\end{array}
\right)  \overset{\text{add }-1\cdot\operatorname{row}1\text{ to
}\operatorname{row}3}{\longrightarrow}\left(
\begin{array}
[c]{cccc}%
1 & 1 & 1 & 1\\
0 & 1 & 1 & 1\\
0 & 1 & 2 & 2\\
1 & 2 & 3 & 4
\end{array}
\right) \\
&  \overset{\text{add }-1\cdot\operatorname{row}1\text{ to }\operatorname{row}%
4}{\longrightarrow}\left(
\begin{array}
[c]{cccc}%
1 & 1 & 1 & 1\\
0 & 1 & 1 & 1\\
0 & 1 & 2 & 2\\
0 & 1 & 2 & 3
\end{array}
\right)  \overset{\text{add }-1\cdot\operatorname{row}2\text{ to
}\operatorname{row}3}{\longrightarrow}\left(
\begin{array}
[c]{cccc}%
1 & 1 & 1 & 1\\
0 & 1 & 1 & 1\\
0 & 0 & 1 & 1\\
0 & 1 & 2 & 3
\end{array}
\right) \\
&  \overset{\text{add }-1\cdot\operatorname{row}2\text{ to }\operatorname{row}%
4}{\longrightarrow}\left(
\begin{array}
[c]{cccc}%
1 & 1 & 1 & 1\\
0 & 1 & 1 & 1\\
0 & 0 & 1 & 1\\
0 & 0 & 1 & 2
\end{array}
\right)  \overset{\text{add }-1\cdot\operatorname{row}3\text{ to
}\operatorname{row}4}{\longrightarrow}\left(
\begin{array}
[c]{cccc}%
1 & 1 & 1 & 1\\
0 & 1 & 1 & 1\\
0 & 0 & 1 & 1\\
0 & 0 & 0 & 1
\end{array}
\right)  .
\end{align*}
We have obtained a triangular matrix with $1,1,1,1$ on the diagonal and thus
determinant $=1$. Since our row operations have included no row scalings and
no row swaps, we thus conclude that $\det A=1$.
\end{example}

\begin{example}
Let us compute $\det A$, where $A=\left(
\begin{array}
[c]{ccc}%
0 & 0 & 1\\
0 & 1 & 2\\
1 & 2 & 3
\end{array}
\right)  $. We row-reduce $A$ as follows:%
\[
A=\left(
\begin{array}
[c]{ccc}%
0 & 0 & 1\\
0 & 1 & 2\\
1 & 2 & 3
\end{array}
\right)  \overset{\text{swap rows }1\text{ and }3}{\longrightarrow}\left(
\begin{array}
[c]{ccc}%
1 & 2 & 3\\
0 & 1 & 2\\
0 & 0 & 1
\end{array}
\right)  .
\]
The matrix we have obtained is upper-triangular with diagonal $1,1,1$, so it
has determinant $1$. Thus,%
\[
1=\left(  -1\right)  ^{1}\cdot\det A,
\]
where the $\left(  -1\right)  ^{1}$ comes from the fact that we have used $1$
row swap operation. Thus, solving this for $\det A$, we obtain $\det A=-1$.
\end{example}

\begin{example}
Let us compute $\det A$, where $A=\left(
\begin{array}
[c]{cc}%
0 & 2\\
3 & 6
\end{array}
\right)  $. Assume that you have forgotten the formula $\det\left(
\begin{array}
[c]{cc}%
a & b\\
c & d
\end{array}
\right)  =ad-bc$. We row-reduce $A$ as follows:%
\begin{align*}
A  &  =\left(
\begin{array}
[c]{cc}%
0 & 2\\
3 & 6
\end{array}
\right)  \overset{\text{swap rows }1\text{ and }2}{\longrightarrow}\left(
\begin{array}
[c]{cc}%
3 & 6\\
0 & 2
\end{array}
\right)  \overset{\text{scale row }1\text{ by }1/3}{\longrightarrow}\left(
\begin{array}
[c]{cc}%
1 & 2\\
0 & 2
\end{array}
\right) \\
&  \overset{\text{scale row }2\text{ by }1/2}{\longrightarrow}\left(
\begin{array}
[c]{cc}%
1 & 2\\
0 & 1
\end{array}
\right)  .
\end{align*}
We have obtained an upper-triangular matrix with diagonal $1,1$, which
therefore has determinant $1$. Thus, Corollary \ref{cor.det.rowops} yields%
\[
1=\left(  -1\right)  ^{1}\cdot\dfrac{1}{3}\cdot\dfrac{1}{2}\cdot\det A.
\]
Solving this for $\det A$, we obtain%
\[
\det A=-6.
\]

\end{example}

\subsection{The determinant determines invertibility}

As a consequence of this method for computing determinants using RREF, we can
get a new criterion for invertibility of a square matrix (which we can add to
our Inverse Matrix Theorem):

\begin{theorem}
\label{thm.det.invertible}Let $A$ be an $n\times n$-matrix. Then, $A$ is
invertible if and only if $\det A\neq0$.
\end{theorem}

\begin{proof}
$\Longrightarrow:$ Assume that $A$ is invertible. We must prove that $\det
A\neq0$.

We have assumed that $A$ is invertible. Hence, the Invertible Matrix Theorem
(more precisely, the implication \textbf{(k)} $\Longrightarrow$ \textbf{(a)}
from Theorem 1.2.1 in
\href{http://www.cip.ifi.lmu.de/~grinberg/t/19fla/2019-10-16.pdf}{the notes
from 2019-10-16}) shows that $A$ can be row-reduced to $I_{n}$. In other
words, we can obtain $I_{n}$ from $A$ by a sequence of row operations. As we
have seen in the previous section, each row operation multiplies the
determinant by a nonzero number (either $1$ in the case of a row addition, or
$-1$ in the case of a row swap, or $\lambda$ if you are scaling a row by
$\lambda$). Hence, $\det\left(  I_{n}\right)  $ is obtained from $\det A$ by
multiplying with a bunch of nonzero numbers. Therefore, $\det\left(
I_{n}\right)  \neq0$ if and only if $\det A\neq0$. But $I_{n}$ is an
upper-triangular matrix with diagonal entries $1,1,\ldots,1$; therefore,
Theorem \ref{thm.det.triang} shows that $\det\left(  I_{n}\right)  =1\neq0$.
Hence, $\det A\neq0$ (since $\det\left(  I_{n}\right)  \neq0$ if and only if
$\det A\neq0$). This completes the proof of the \textquotedblleft%
$\Longrightarrow$\textquotedblright\ direction of Theorem
\ref{thm.det.invertible}.

$\Longleftarrow:$ Assume that $\det A\neq0$. We must prove that $A$ is invertible.

Let $B$ be the RREF of $A$. Then, we can obtain $B$ from $A$ by a sequence of
row operations. As we have seen in the previous section, each row operation
multiplies the determinant by a nonzero number (either $1$ in the case of a
row addition, or $-1$ in the case of a row swap, or $\lambda$ if you are
scaling a row by $\lambda$). Hence, $\det B$ is obtained from $\det A$ by
multiplying with a bunch of nonzero numbers. Therefore, $\det B\neq0$ (since
$\det A\neq0$). From this, we can easily see that the RREF matrix $B$ has a
pivot in each column\footnote{\textit{Proof.} Assume the contrary. Thus, the
matrix $B$ has a column without pivot. Hence, $B$ has $<n$ pivots (since $B$
has $n$ columns, and at most one pivot per column). Therefore, the matrix $B$
has a row without pivot (since $B$ has $n$ rows). This row must be a zero row
(since any nonzero row contains a pivot). But Corollary \ref{cor.det.zero-row}
shows that any matrix with a zero row has determinant $0$. Thus, $\det B=0$.
This contradicts $\det B\neq0$. This contradiction shows that our assumption
was false, qed.}. Hence, $B=I_{n}$ (by Lemma 1.1.14 \textbf{(b)} in
\href{http://www.cip.ifi.lmu.de/~grinberg/t/19fla/2019-10-07.pdf}{the notes
from 2019-10-07}), since $B$ is a square matrix. But $A$ row-reduces to $B$;
in other words, $A$ row-reduces to $I_{n}$ (since $B=I_{n}$). Therefore, the
Invertible Matrix Theorem (more precisely, the implication \textbf{(a)}
$\Longrightarrow$ \textbf{(k)} from Theorem 1.2.1 in
\href{http://www.cip.ifi.lmu.de/~grinberg/t/19fla/2019-10-16.pdf}{the notes
from 2019-10-16}) shows that $A$ is invertible. This proves the
\textquotedblleft$\Longleftarrow$\textquotedblright\ direction of Theorem
\ref{thm.det.invertible}.
\end{proof}

\begin{example}
Given any numbers $a,b,c,d,e,f,g$, we claim that the matrix $\left(
\begin{array}
[c]{cccc}%
0 & 0 & 0 & g\\
0 & 0 & 0 & f\\
0 & 0 & 0 & e\\
a & b & c & d
\end{array}
\right)  $ is not invertible.

To prove this, it suffices (by Theorem \ref{thm.det.invertible}) to show that
its determinant is $0$. Why is its determinant $0$ ?

One way to see this is the following: If we swap row $1$ and row $4$, then we
obtain the matrix $\left(
\begin{array}
[c]{cccc}%
a & b & c & d\\
0 & 0 & 0 & f\\
0 & 0 & 0 & e\\
0 & 0 & 0 & g
\end{array}
\right)  $, which is upper-triangular with diagonal $a,0,0,g$ and thus (by
Theorem \ref{thm.det.triang}) has determinant $a\cdot0\cdot0\cdot g=0$. Thus,
our original matrix must also have determinant $0$ (because the row swap only
multiplied the determinant by $-1$).

Another way to see this is the following: By the definition of the
determinant,
\[
\det\left(
\begin{array}
[c]{cccc}%
0 & 0 & 0 & g\\
0 & 0 & 0 & f\\
0 & 0 & 0 & e\\
a & b & c & d
\end{array}
\right)  =\left(  \text{a sum of }24\text{ products}\right)  .
\]
Each product has $4$ factors, using $1$ entry from each row and $1$ from each
column. Thus, at least two of its $4$ factors must be $0$ (because only $1$ of
the $4$ factors can come from row $4$, and only one can come from column $4$,
but this means that the remaining $2$ or more factors come neither from row
$4$ nor from column $4$ and thus are $0$). Hence, the product is $0$. Thus,%
\[
\det\left(
\begin{array}
[c]{cccc}%
0 & 0 & 0 & g\\
0 & 0 & 0 & f\\
0 & 0 & 0 & e\\
a & b & c & d
\end{array}
\right)  =\left(  \text{a sum of }24\text{ products, each of which is
}0\right)  =0.
\]
This shows that the matrix is not invertible.
\end{example}

\subsection{The determinant of $AB$}

The following theorem is one of the most important properties of determinants:

\begin{theorem}
\label{thm.det.AB}We have $\det\left(  AB\right)  =\det A\cdot\det B$ for any
two $n\times n$-matrices $A$ and $B$.
\end{theorem}

See \cite[Theorem B.17]{Strickland} and \cite[\S 3.2, Theorem 6]{Lay} for two
different proofs of this theorem.

\begin{example}
Let $C=\left(
\begin{array}
[c]{cccc}%
1 & 1 & 1 & 1\\
1 & 2 & 2 & 2\\
1 & 2 & 3 & 3\\
1 & 2 & 3 & 4
\end{array}
\right)  $ again. Here is another way to find $\det C$.

Recall (from midterm 1 Exercise 1) that%
\[
C=\left(
\begin{array}
[c]{cccc}%
1 & 0 & 0 & 0\\
1 & 1 & 0 & 0\\
1 & 1 & 1 & 0\\
1 & 1 & 1 & 1
\end{array}
\right)  \left(
\begin{array}
[c]{cccc}%
1 & 1 & 1 & 1\\
0 & 1 & 1 & 1\\
0 & 0 & 1 & 1\\
0 & 0 & 0 & 1
\end{array}
\right)  .
\]
Hence, by Theorem \ref{thm.det.AB}, we have%
\begin{align*}
\det C  &  =\underbrace{\det\left(
\begin{array}
[c]{cccc}%
1 & 0 & 0 & 0\\
1 & 1 & 0 & 0\\
1 & 1 & 1 & 0\\
1 & 1 & 1 & 1
\end{array}
\right)  }_{\substack{=1\\\text{(since this is a lower-triangular
matrix}\\\text{with diagonal entries }1,1,\ldots,1\text{)}}}\cdot
\underbrace{\det\left(
\begin{array}
[c]{cccc}%
1 & 1 & 1 & 1\\
0 & 1 & 1 & 1\\
0 & 0 & 1 & 1\\
0 & 0 & 0 & 1
\end{array}
\right)  }_{\substack{=1\\\text{(since this is a upper-triangular
matrix}\\\text{with diagonal entries }1,1,\ldots,1\text{)}}}\\
&  =1.
\end{align*}

\end{example}

Another application of Theorem \ref{thm.det.AB} is the following fact:

\begin{corollary}
\label{cor.det.lB}Let $B$ be an $n\times n$-matrix. Let $\lambda$ be a scalar.
Then,%
\[
\det\left(  \lambda B\right)  =\lambda^{n}\det B.
\]

\end{corollary}

\begin{example}
For $n=3$, this is saying%
\[
\det\left(
\begin{array}
[c]{ccc}%
\lambda a & \lambda b & \lambda c\\
\lambda d & \lambda e & \lambda f\\
\lambda g & \lambda h & \lambda i
\end{array}
\right)  =\lambda^{3}\det\left(
\begin{array}
[c]{ccc}%
a & b & c\\
d & e & f\\
g & h & i
\end{array}
\right)  .
\]

\end{example}

\begin{proof}
[First proof of Corollary \ref{cor.det.lB}.]We have%
\begin{align*}
\det\left(  \lambda B\right)    & =\det\left(  \lambda I_{n}\cdot B\right)
\\
& =\det\left(  \lambda I_{n}\right)  \cdot\det B\ \ \ \ \ \ \ \ \ \ \left(
\text{by Theorem \ref{thm.det.AB}, applied to }A=\lambda I_{n}\right)  .
\end{align*}
But $\lambda I_{n}=\left(
\begin{array}
[c]{cccc}%
\lambda & 0 & \cdots & 0\\
0 & \lambda & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & \lambda
\end{array}
\right)  $ is an upper-triangular (and lower-triangular, too) matrix with
diagonal entries $\lambda,\lambda,\ldots,\lambda$. Hence, Theorem
\ref{thm.det.triang} says that $\det\left(  \lambda I_{n}\right)
=\underbrace{\lambda\lambda\cdots\lambda}_{n\text{ times}}=\lambda^{n}$. Thus,
$\det\left(  \lambda B\right)  =\underbrace{\det\left(  \lambda I_{n}\right)
}_{=\lambda^{n}}\cdot\det B=\lambda^{n}\det B$. This proves Corollary
\ref{cor.det.lB}.
\end{proof}

Alternatively, we can prove Corollary \ref{cor.det.lB} without using Theorem
\ref{thm.det.AB}:

\begin{proof}
[Second proof of Corollary \ref{cor.det.lB}.]The matrix $\lambda B$ is
obtained from the $n\times n$-matrix $B$ by successively scaling \textbf{each}
row by $\lambda$. Each time a row is scaled by $\lambda$, the determinant gets
multiplied by $\lambda$ (by Theorem \ref{thm.det.scale-row}). Hence, when
going from $B$ to $\lambda B$, the determinant gets multiplied by $\lambda
^{n}$ (since there are $n$ rows). This proves Corollary \ref{cor.det.lB} as well.
\end{proof}

\subsection{Determinants and columns}

\begin{theorem}
\label{thm.det.cols}Theorem \ref{thm.det.equal-rows}, Theorem
\ref{thm.det.scale-row}, Corollary \ref{cor.det.zero-row}, Theorem
\ref{thm.det.muli-rows}, Corollary \ref{cor.det.row-addition} and Theorem
\ref{thm.det.row-swap} remain valid if we replace \textquotedblleft
row\textquotedblright\ by \textquotedblleft column\textquotedblright%
\ throughout them.
\end{theorem}

Roughly speaking, this means that everything we said above about determinants
and rows holds all the same for determinants and columns. (For example: If you
scale a column of an $n\times n$-matrix by a number $\lambda$, then the
determinant gets multiplied by $\lambda$.)

\begin{proof}
[Proof of Theorem \ref{thm.det.cols} (rough idea).]Let us prove the analogue
of Theorem \ref{thm.det.scale-row} for columns instead of rows:

Let $A$ be an $n\times n$-matrix, and let $\lambda$ be a number. Let $B$ be
the matrix if we scale a column of $A$ (say, the $i$-th column) by $\lambda$.
We must prove that $\det B=\lambda\cdot\det A$.

Theorem \ref{thm.det.transpose} yields $\det A=\det\left(  A^{T}\right)  $ and
$\det B=\det\left(  B^{T}\right)  $. Recall that the matrix $B$ is obtained
from the matrix $A$ by scaling a column by $\lambda$. Hence, its transpose
$B^{T}$ is obtained from $A^{T}$ by scaling a \textbf{row} by $\lambda$
(because the columns of a transpose of a matrix are the transposes of the rows
of the matrix). Therefore, Theorem \ref{thm.det.scale-row} yields $\det\left(
B^{T}\right)  =\lambda\cdot\det\left(  A^{T}\right)  $. In view of $\det
A=\det\left(  A^{T}\right)  $ and $\det B=\det\left(  B^{T}\right)  $, this
rewrites as $\det B=\lambda\cdot\det A$. This proves the analogue of Theorem
\ref{thm.det.scale-row} for columns instead of rows.

Similar arguments can be used to derive the column analogues of all other
theorems from their original (row) versions.
\end{proof}

\begin{example}
Here is an instance of the analogue of Theorem \ref{thm.det.muli-rows} for
columns (with $n=3$ and $k=2$):%
\[
\det\left(
\begin{array}
[c]{ccc}%
a & b+b^{\prime} & c\\
d & e+e^{\prime} & f\\
g & h+h^{\prime} & i
\end{array}
\right)  =\det\left(
\begin{array}
[c]{ccc}%
a & b & c\\
d & e & f\\
g & h & i
\end{array}
\right)  +\det\left(
\begin{array}
[c]{ccc}%
a & b^{\prime} & c\\
d & e^{\prime} & f\\
g & h^{\prime} & i
\end{array}
\right)  .
\]

\end{example}

\subsection{Laplace expansion}

The following fact reduces the computation of the determinant of an $n\times
n$-matrix to the computation of several smaller determinants:

\begin{theorem}
[Laplace expansion along the $p$-th row]\label{thm.det.lap-row}Let $A$ be an
$n\times n$-matrix. For each $p,q\in\left[  n\right]  $, we let $M_{p,q}$ be
the $\left(  n-1\right)  \times\left(  n-1\right)  $-matrix obtained from $A$
by removing row $p$ and column $q$. Then, for each $p\in\left[  n\right]  $,
we have%
\[
\det A=\sum_{q=1}^{n}\left(  -1\right)  ^{p+q}A_{p,q}\det\left(
M_{p,q}\right)  .
\]

\end{theorem}

\begin{example}
Let $n=3$. Then,%
\[
A=\left(
\begin{array}
[c]{ccc}%
A_{1,1} & A_{1,2} & A_{1,3}\\
A_{2,1} & A_{2,2} & A_{2,3}\\
A_{3,1} & A_{3,2} & A_{3,3}%
\end{array}
\right)  .
\]
Thus, the $M_{p,q}$ in Theorem \ref{thm.det.lap-row} are%
\begin{align*}
M_{1,1}  &  =\left(
\begin{array}
[c]{cc}%
A_{2,2} & A_{2,3}\\
A_{3,2} & A_{3,3}%
\end{array}
\right)  ,\ \ \ \ \ \ \ \ \ \ M_{1,2}=\left(
\begin{array}
[c]{cc}%
A_{2,1} & A_{2,3}\\
A_{3,1} & A_{3,3}%
\end{array}
\right)  ,\ \ \ \ \ \ \ \ \ \ M_{1,3}=\left(
\begin{array}
[c]{cc}%
A_{2,1} & A_{2,2}\\
A_{3,1} & A_{3,2}%
\end{array}
\right)  ,\\
& \\
M_{2,1}  &  =\left(
\begin{array}
[c]{cc}%
A_{1,2} & A_{1,3}\\
A_{3,2} & A_{3,3}%
\end{array}
\right)  ,\ \ \ \ \ \ \ \ \ \ M_{2,2}=\left(
\begin{array}
[c]{cc}%
A_{1,1} & A_{1,3}\\
A_{3,1} & A_{3,3}%
\end{array}
\right)  ,\ \ \ \ \ \ \ \ \ \ M_{2,3}=\left(
\begin{array}
[c]{cc}%
A_{1,1} & A_{1,2}\\
A_{3,1} & A_{3,2}%
\end{array}
\right)  ,\\
& \\
M_{3,1}  &  =\left(
\begin{array}
[c]{cc}%
A_{1,2} & A_{1,3}\\
A_{2,2} & A_{2,3}%
\end{array}
\right)  ,\ \ \ \ \ \ \ \ \ \ M_{3,2}=\left(
\begin{array}
[c]{cc}%
A_{1,1} & A_{1,3}\\
A_{2,1} & A_{2,3}%
\end{array}
\right)  ,\ \ \ \ \ \ \ \ \ \ M_{3,3}=\left(
\begin{array}
[c]{cc}%
A_{1,1} & A_{1,2}\\
A_{2,1} & A_{2,2}%
\end{array}
\right)  .
\end{align*}
Thus, Theorem \ref{thm.det.lap-row} (applied to $p=1$) yields%
\begin{align*}
&  \det A\\
&  =\sum_{q=1}^{3}\left(  -1\right)  ^{1+q}A_{1,q}\det\left(  M_{1,q}\right)
\\
&  =\underbrace{\left(  -1\right)  ^{1+1}}_{=1}A_{1,1}\det\left(
M_{1,1}\right)  +\underbrace{\left(  -1\right)  ^{1+2}}_{=-1}A_{1,2}%
\det\left(  M_{1,2}\right)  +\underbrace{\left(  -1\right)  ^{1+3}}%
_{=1}A_{1,3}\det\left(  M_{1,3}\right) \\
&  =A_{1,1}\det\left(  M_{1,1}\right)  -A_{1,2}\det\left(  M_{1,2}\right)
+A_{1,3}\det\left(  M_{1,3}\right) \\
&  =A_{1,1}\det\left(
\begin{array}
[c]{cc}%
A_{2,2} & A_{2,3}\\
A_{3,2} & A_{3,3}%
\end{array}
\right)  -A_{1,2}\det\left(
\begin{array}
[c]{cc}%
A_{2,1} & A_{2,3}\\
A_{3,1} & A_{3,3}%
\end{array}
\right)  +A_{1,3}\det\left(
\begin{array}
[c]{cc}%
A_{2,1} & A_{2,2}\\
A_{3,1} & A_{3,2}%
\end{array}
\right)  .
\end{align*}
Likewise, Theorem \ref{thm.det.lap-row} (applied to $p=2$) yields%
\begin{align*}
&  \det A\\
&  =-A_{2,1}\det\left(
\begin{array}
[c]{cc}%
A_{1,2} & A_{1,3}\\
A_{3,2} & A_{3,3}%
\end{array}
\right)  +A_{2,2}\det\left(
\begin{array}
[c]{cc}%
A_{1,1} & A_{1,3}\\
A_{3,1} & A_{3,3}%
\end{array}
\right)  -A_{2,3}\det\left(
\begin{array}
[c]{cc}%
A_{1,1} & A_{1,2}\\
A_{3,1} & A_{3,2}%
\end{array}
\right)  .
\end{align*}

\end{example}

For a proof of Theorem \ref{thm.det.lap-row}, see \cite[Proposition
B.24]{Strickland}.

\begin{example}
Let's compute $\det A$ for $A=\left(
\begin{array}
[c]{ccc}%
1 & 0 & 2\\
0 & 1 & 0\\
3 & 0 & 4
\end{array}
\right)  $. Theorem \ref{thm.det.lap-row} (for $p=2$) yields%
\begin{align*}
&  \det A\\
&  =-A_{2,1}\det\left(
\begin{array}
[c]{cc}%
A_{1,2} & A_{1,3}\\
A_{3,2} & A_{3,3}%
\end{array}
\right)  +A_{2,2}\det\left(
\begin{array}
[c]{cc}%
A_{1,1} & A_{1,3}\\
A_{3,1} & A_{3,3}%
\end{array}
\right)  -A_{2,3}\det\left(
\begin{array}
[c]{cc}%
A_{1,1} & A_{1,2}\\
A_{3,1} & A_{3,2}%
\end{array}
\right) \\
&  =\underbrace{-0\det\left(
\begin{array}
[c]{cc}%
0 & 2\\
0 & 4
\end{array}
\right)  }_{=0}+1\det\left(
\begin{array}
[c]{cc}%
1 & 2\\
3 & 4
\end{array}
\right)  -\underbrace{0\det\left(
\begin{array}
[c]{cc}%
1 & 0\\
3 & 0
\end{array}
\right)  }_{=0}\\
&  =\det\left(
\begin{array}
[c]{cc}%
1 & 2\\
3 & 4
\end{array}
\right)  =1\cdot4-2\cdot3=-2.
\end{align*}

\end{example}

Applying Theorem \ref{thm.det.lap-row} to some value of $p$ is called
\textquotedblleft Laplace expansion with respect to the $p$-th
row\textquotedblright\ or simply \textquotedblleft expanding (the determinant)
with respect to the $p$-th row\textquotedblright. Generally, a good strategy
when computing determinants is the following: If your matrix $A$ has a row
with many zeroes, you can try computing $\det A$ by Laplace expansion with
respect to this row.

\begin{example}
Let us compute $\det A$, where $A=\left(
\begin{array}
[c]{cccc}%
1 & 0 & 1 & 0\\
0 & 1 & 1 & 1\\
1 & 1 & 1 & 0\\
1 & 1 & 1 & 1
\end{array}
\right)  $. Theorem \ref{thm.det.lap-row} (for $p=1$) yields%
\begin{align*}
\det A  &  =1\det\left(
\begin{array}
[c]{ccc}%
1 & 1 & 1\\
1 & 1 & 0\\
1 & 1 & 1
\end{array}
\right)  -\underbrace{0\det\left(
\begin{array}
[c]{ccc}%
0 & 1 & 1\\
1 & 1 & 0\\
1 & 1 & 1
\end{array}
\right)  }_{=0}\\
&  \ \ \ \ \ \ \ \ \ \ +1\det\left(
\begin{array}
[c]{ccc}%
0 & 1 & 1\\
1 & 1 & 0\\
1 & 1 & 1
\end{array}
\right)  -\underbrace{0\det\left(
\begin{array}
[c]{ccc}%
0 & 1 & 1\\
1 & 1 & 1\\
1 & 1 & 1
\end{array}
\right)  }_{=0}\\
&  =\underbrace{\det\left(
\begin{array}
[c]{ccc}%
1 & 1 & 1\\
1 & 1 & 0\\
1 & 1 & 1
\end{array}
\right)  }_{\substack{=0\\\text{(since this matrix has two equal rows)}}%
}+\det\left(
\begin{array}
[c]{ccc}%
0 & 1 & 1\\
1 & 1 & 0\\
1 & 1 & 1
\end{array}
\right) \\
&  =\det\left(
\begin{array}
[c]{ccc}%
0 & 1 & 1\\
1 & 1 & 0\\
1 & 1 & 1
\end{array}
\right) \\
&  =\underbrace{0\det\left(
\begin{array}
[c]{cc}%
1 & 0\\
1 & 1
\end{array}
\right)  }_{=0}-1\underbrace{\det\left(
\begin{array}
[c]{cc}%
1 & 0\\
1 & 1
\end{array}
\right)  }_{=1}+1\underbrace{\det\left(
\begin{array}
[c]{cc}%
1 & 1\\
1 & 1
\end{array}
\right)  }_{\substack{=0\\\text{(due to two equal rows)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Theorem \ref{thm.det.lap-row} for
}p=1\right) \\
&  =-1.
\end{align*}

\end{example}

\begin{exercise}
Compute%
\begin{align*}
\det\left(  \left(  i+j\right)  _{1\leq i\leq n,\ 1\leq j\leq n}\right)   &
=\det\left(
\begin{array}
[c]{cccc}%
1+1 & 1+2 & \cdots & 1+n\\
2+1 & 2+2 & \cdots & 2+n\\
\vdots & \vdots & \ddots & \vdots\\
n+1 & n+2 & \cdots & n+n
\end{array}
\right) \\
&  =\det\left(
\begin{array}
[c]{cccc}%
2 & 3 & \cdots & n+1\\
3 & 4 & \cdots & n+2\\
\vdots & \vdots & \ddots & \vdots\\
n+1 & n+2 & \cdots & 2n
\end{array}
\right)  .
\end{align*}
You can see that this determinant is $2$ if $n=1$; is $-1$ if $n=2$; is $0$ if
$n=3$; if $0$ if $n=4$; can you spot and prove the general pattern? E.g., for
$n=5$, this is about $\left(
\begin{array}
[c]{ccccc}%
2 & 3 & 4 & 5 & 6\\
3 & 4 & 5 & 6 & 7\\
4 & 5 & 6 & 7 & 8\\
5 & 6 & 7 & 8 & 9\\
6 & 7 & 8 & 9 & 10
\end{array}
\right)  $.
\end{exercise}

\begin{thebibliography}{9999999999}                                                                                       %


\bibitem[Lay]{Lay}David C. Lay, Steven R. Lay, Judi J. McDonald,
\textit{Linear Algebra and its Applications}, 5th edition.

\bibitem[Strickland]{Strickland}Neil Strickland, \textit{MAS201 Linear
Mathematics for Applications}, lecture notes, 28 September 2013.\newline\url{http://neil-strickland.staff.shef.ac.uk/courses/MAS201/}
\end{thebibliography}


\end{document}