Commit a39cf176 authored by Jim Hefferon's avatar Jim Hefferon

all cramers rule edits

parent 31863f3c
......@@ -509,6 +509,9 @@
Patrick J.~Ryan,
\emph{Euclidean and Non-Euclidean Geometry: an Analytic Approach},
Cambridge University Press, 1986.
\bibitem[Schmidt]{SchmidtSO}
Jack Schmidt (\url{http://math.stackexchange.com/users/583/jack-schmidt}),
\url{http://math.stackexchange.com/a/98558/12012} (version: 2012-01-12).
\bibitem[Shepelev]{Shepelev}
Anton Shepelev,
private communication,
......
This diff is collapsed.
......@@ -22327,18 +22327,43 @@ octave:6> gplot z
\end{ans}
\begin{ans}{Four.I.3.27}
\cite{SchmidtSO}
We will show that $P\trans{P}=I$; the $\trans{P}P=I$ argument is similar.
The $i,j$ entry of $P\trans{P}$ is the sum of terms of the form
$p_{i,k}q_{k,j}$ where the entries of $\trans{P}$ are denoted with $q$'s,
that is, $q_{k,j}=p_{j,k}$.
Thus the $i,j$ entry of $P\trans{P}$ is the sum
\(
\sum_{k=1}^n p_{i,k}p_{j,k}
\).
But $p_{i,k}$ is usually $0$, and so $P_{i,k}P_{j,k}$ is usually $0$.
The only time $P_{i,k}$ is nonzero is when it is $1$,
but then there are no other $i^\prime\neq i$ such that $P_{i^\prime,k}$
is nonzero ($i$ is the only row with a $1$ in column~$k$).
In other words,
\begin{equation*}
\sum_{k=1}^n p_{i,k}p_{j,k}=
\begin{cases}
1 &i=j \\
0 &\text{otherwise}
\end{cases}
\end{equation*}
and this is exactly the formula for the entries of the identity matrix.
\end{ans}
\begin{ans}{Four.I.3.28}
In
\( \deter{A}=\deter{\trans{A}}=\deter{-A}=(-1)^n\deter{A} \)
the exponent $n$ must be even.
\end{ans}
\begin{ans}{Four.I.3.28}
\begin{ans}{Four.I.3.29}
Showing that no placement of three zeros suffices is routine.
Four zeroes does suffice; put them all in the same
row or column.
\end{ans}
\begin{ans}{Four.I.3.29}
\begin{ans}{Four.I.3.30}
The $n=3$ case shows what to do.
The row combination operations of
$-x_1\rho_2+\rho_3$ and $-x_1\rho_1+\rho_2$
......@@ -22375,7 +22400,7 @@ octave:6> gplot z
\end{equation*}
\end{ans}
\begin{ans}{Four.I.3.30}
\begin{ans}{Four.I.3.31}
Let \( T \) be \( \nbyn{n} \),
let \( J \) be \( \nbyn{p} \),
and let \( K \) be \( \nbyn{q} \).
......@@ -22411,7 +22436,7 @@ octave:6> gplot z
\end{equation*}
\end{ans}
\begin{ans}{Four.I.3.31}
\begin{ans}{Four.I.3.32}
The $n=3$ case shows what happens.
\begin{equation*}
\deter{T-rI}
......@@ -22434,7 +22459,7 @@ octave:6> gplot z
A polynomial of degree \( n \) has at most \( n \) roots.
\end{ans}
\begin{ans}{Four.I.3.32}
\begin{ans}{Four.I.3.33}
\answerasgiven
When two rows of a determinant are interchanged, the sign of the
determinant is changed.
......@@ -22445,7 +22470,7 @@ octave:6> gplot z
which sums to zero.
\end{ans}
\begin{ans}{Four.I.3.33}
\begin{ans}{Four.I.3.34}
\answerasgiven
When the elements of any column are subtracted from the elements of
each of the other two, the elements in two of the columns of the derived
......@@ -22470,7 +22495,7 @@ octave:6> gplot z
\end{equation*}
\end{ans}
\begin{ans}{Four.I.3.34}
\begin{ans}{Four.I.3.35}
\answerasgiven
Let
\begin{equation*}
......@@ -22514,7 +22539,7 @@ octave:6> gplot z
\end{equation*}
\end{ans}
\begin{ans}{Four.I.3.35}
\begin{ans}{Four.I.3.36}
\answerasgiven
Denote by \( D_n \) the determinant in question and by \( a_{i,j} \)
the element in the \( i \)-th row and \( j \)-th column.
......@@ -23601,7 +23626,7 @@ octave:6> gplot z
0 &0 &1
\end{vmat}=1
\end{equation*}
dosn't equal the result of
doesn't equal the result of
expanding down the diagonal.
\begin{equation*}
1\cdot (+1)\begin{vmat}[r]
......@@ -23874,25 +23899,25 @@ octave:6> gplot z
\partsitem
\begin{equation*}
x=
\frac{ \begin{vmatrix}
\frac{ \begin{vmat}[r]
4 &-1 \\
-7 &2
\end{vmatrix} }{
\begin{vmatrix}
\end{vmat} }{
\begin{vmat}[r]
1 &-1 \\
-1 &2
\end{vmatrix} }
\end{vmat} }
=\frac{1}{1}=1
\qquad
y=
\frac{ \begin{vmatrix}
\frac{ \begin{vmat}[r]
1 &4 \\
-1 &-7
\end{vmatrix} }{
\begin{vmatrix}
\end{vmat} }{
\begin{vmat}[r]
1 &-1 \\
-1 &2
\end{vmatrix} }
\end{vmat} }
=\frac{-3}{1}=-3
\end{equation*}
\partsitem $x=2$, $y=2$
......@@ -23943,7 +23968,8 @@ octave:6> gplot z
\end{ans}
\begin{ans}{6}
Of course, singular systems have \( \deter{A} \) equal to zero, but
the infinitely many solutions case is characterized by the fact that
we can characterize
the infinitely many solutions case is by the fact that
all of the \( \deter{B_i} \) are zero as well.
\end{ans}
......@@ -23960,7 +23986,7 @@ octave:6> gplot z
value for $c$ yields no solutions.
The corresponding vector equation
\begin{equation*}
x_1\cdot\colvec{1 \\ 1}+x_2\cdot\colvec{2 \\ 2}=\colvec{6 \\ c}
x_1\cdot\colvec[r]{1 \\ 1}+x_2\cdot\colvec[r]{2 \\ 2}=\colvec[r]{6 \\ c}
\end{equation*}
gives a picture of two overlapping vectors.
Both lie on the line $y=x$.
......@@ -3,14 +3,14 @@
% 2001-Jun-12
\topic{Cramer's Rule}
\index{Cramer's rule|(}
We have introduced determinant functions algebraically by looking
for a formula to decide whether a matrix is nonsingular.
After that introduction we saw a geometric interpretation,
that the determinant function
gives the size of the box with sides formed by the columns of the matrix.
This Topic makes a connection between the two views.
% We have introduced determinant functions algebraically by looking
% for a formula to decide whether a matrix is nonsingular.
% After that introduction we saw a geometric interpretation,
% that the determinant function
% gives the size of the box with sides formed by the columns of the matrix.
% Here we make a connection between the two views.
First, a linear system
We have seen that a linear system
\begin{equation*}
\begin{linsys}{2}
x_1 &+ &2x_2 &= &6 \\
......@@ -21,23 +21,22 @@ is equivalent to a linear relationship among vectors.
\begin{equation*}
x_1\cdot\colvec{1 \\ 3}+x_2\cdot\colvec{2 \\ 1}=\colvec{6 \\ 8}
\end{equation*}
The picture below shows a parallelogram with sides formed from
$\binom{1}{3}$ and $\binom{2}{1}$ nested inside a parallelogram
This pictures that vector equation.
A parallelogram with sides formed from
$\binom{1}{3}$ and $\binom{2}{1}$ is nested inside a parallelogram
with sides formed from $x_1\binom{1}{3}$ and $x_2\binom{2}{1}$.
\begin{center}
\includegraphics{ch4.1}
\end{center}
So even without determinants
we can state the algebraic issue that opened this book,
finding the solution of a linear system,
That is,
we can restate the algebraic question of
finding the solution of a linear system
in geometric terms:~by
what factors $x_1$ and $x_2$ must we dilate the vectors to expand the small
parallegram to fill the larger one?
parallelogram so that it will fill the larger one?
However, by employing the geometric significance of determinants
we can get something that is not just a restatement, but also
gives us a new insight and sometimes allows us to compute
answers quickly.
We can apply the geometric significance of determinants
to that picture to get a new formula.
Compare the sizes of these shaded boxes.
\begin{center}
\includegraphics{ch4.2}
......@@ -46,88 +45,89 @@ Compare the sizes of these shaded boxes.
\hfil
\includegraphics{ch4.4}
\end{center}
The second is formed from $x_1\binom{1}{3}$ and $\binom{2}{1}$, and
The second is defined by the vectors $x_1\binom{1}{3}$ and $\binom{2}{1}$, and
one of the properties of the size function\Dash the determinant\Dash is
that its size is therefore \( x_1 \) times the size of the
that therefore the size of the second box is \( x_1 \) times the size of the
first box.
Since the third box is formed from
Since the third box is defined by the vector
$x_1\binom{1}{3}+x_2\binom{2}{1}=\binom{6}{8}$
and $\binom{2}{1}$,
and the determinant is unchanged by adding $x_2$
and the vector $\binom{2}{1}$,
and since the determinant does not change when we add $x_2$
times the second column to the first column,
the size of the third box equals that of the second.
We have this.
\begin{equation*}
\begin{vmatrix}
\begin{vmat}[r]
6 &2 \\
8 &1
\end{vmatrix}
\end{vmat}
=
\begin{vmatrix}
\begin{vmat}
x_1\cdot 1 &2 \\
x_1\cdot 3 &1
\end{vmatrix}
\end{vmat}
=
x_1\cdot \begin{vmatrix}
x_1\cdot \begin{vmat}[r]
1 &2 \\
3 &1
\end{vmatrix}
\end{vmat}
\end{equation*}
Solving gives the value of one of the variables.
\begin{equation*}
x_1=
\frac{\begin{vmatrix}
\frac{\begin{vmat}[r]
6 &2 \\
8 &1
\end{vmatrix} }{
\begin{vmatrix}
\end{vmat} }{
\begin{vmat}[r]
1 &2 \\
3 &1
\end{vmatrix} }
\end{vmat} }
=\frac{-10}{-5}=2
\end{equation*}
The theorem that generalizes this example, \definend{Cramer's Rule}%
The generalization of this example is \definend{Cramer's Rule}:%
\index{determinant!Cramer's rule}%
\index{linear equation!solutions of!Cramer's rule},
is:~if \( \deter{A}\neq 0 \) then the system \( A\vec{x}=\vec{b} \) has the
\index{linear equation!solutions of!Cramer's rule}
if \( \deter{A}\neq 0 \) then the system \( A\vec{x}=\vec{b} \) has the
unique solution
$
x_i=\deter{B_i}/\deter{A}
$
where the matrix $B_i$ is formed from $A$ by replacing column~$i$
with the vector \( \vec{b} \).
\nearbyexercise{ex:CramerRule} asks for a proof.
The proof is \nearbyexercise{ex:CramerRule}.
For instance, to solve this system for \( x_2 \)
\begin{equation*}
\begin{pmatrix}
\begin{mat}[r]
1 &0 &4 \\
2 &1 &-1 \\
1 &0 &1
\end{pmatrix}
\end{mat}
\colvec{x_1 \\ x_2 \\ x_3}
=\colvec{2 \\ 1 \\ -1}
=\colvec[r]{2 \\ 1 \\ -1}
\end{equation*}
we do this computation.
\begin{equation*}
x_2=
\frac{ \begin{vmatrix}
\frac{ \begin{vmat}[r]
1 &2 &4 \\
2 &1 &-1 \\
1 &-1 &1
\end{vmatrix} }{
\begin{vmatrix}
\end{vmat} }{
\begin{vmat}[r]
1 &0 &4 \\
2 &1 &-1 \\
1 &0 &1
\end{vmatrix} }
\end{vmat} }
=\frac{-18}{-3}
\end{equation*}
Cramer's Rule allows us to solve
many two equations/two unknowns systems by eye.
It is also sometimes used for three equations/three unknowns systems.
simple two equations/two unknowns systems by eye
(they must be simple in that we can mentally compute with the numbers
in the system).
With practice a person can also do simple equations/three unknowns systems.
But computing large determinants takes a long time, so solving
large systems by Cramer's Rule is not practical.
......@@ -149,25 +149,25 @@ large systems by Cramer's Rule is not practical.
\partsitem
\begin{equation*}
x=
\frac{ \begin{vmatrix}
\frac{ \begin{vmat}[r]
4 &-1 \\
-7 &2
\end{vmatrix} }{
\begin{vmatrix}
\end{vmat} }{
\begin{vmat}[r]
1 &-1 \\
-1 &2
\end{vmatrix} }
\end{vmat} }
=\frac{1}{1}=1
\qquad
y=
\frac{ \begin{vmatrix}
\frac{ \begin{vmat}[r]
1 &4 \\
-1 &-7
\end{vmatrix} }{
\begin{vmatrix}
\end{vmat} }{
\begin{vmat}[r]
1 &-1 \\
-1 &2
\end{vmatrix} }
\end{vmat} }
=\frac{-3}{1}=-3
\end{equation*}
\partsitem $x=2$, $y=2$
......@@ -240,13 +240,14 @@ large systems by Cramer's Rule is not practical.
solutions and one with infinitely many?
\begin{answer}
Of course, singular systems have \( \deter{A} \) equal to zero, but
the infinitely many solutions case is characterized by the fact that
we can characterize
the infinitely many solutions case is by the fact that
all of the \( \deter{B_i} \) are zero as well.
\end{answer}
\item
The first picture in this Topic (the one that doesn't use determinants)
shows a unique solution case.
Produce a similar picture for the case of infintely many solutions,
Produce a similar picture for the case of infinitely many solutions,
and the case of no solutions.
\begin{answer}
We can consider the two nonsingular cases together with this
......@@ -261,7 +262,7 @@ large systems by Cramer's Rule is not practical.
value for $c$ yields no solutions.
The corresponding vector equation
\begin{equation*}
x_1\cdot\colvec{1 \\ 1}+x_2\cdot\colvec{2 \\ 2}=\colvec{6 \\ c}
x_1\cdot\colvec[r]{1 \\ 1}+x_2\cdot\colvec[r]{2 \\ 2}=\colvec[r]{6 \\ c}
\end{equation*}
gives a picture of two overlapping vectors.
Both lie on the line $y=x$.
......
......@@ -2979,6 +2979,31 @@ Determinant some, \\*
\begin{answer}
\( n\cdot(n-1)\cdots 2\cdot 1=n! \)
\end{answer}
\item Show that the inverse of a permutation matrix is its transpose.
\begin{answer}
\cite{SchmidtSO}
We will show that $P\trans{P}=I$; the $\trans{P}P=I$ argument is similar.
The $i,j$ entry of $P\trans{P}$ is the sum of terms of the form
$p_{i,k}q_{k,j}$ where the entries of $\trans{P}$ are denoted with $q$'s,
that is, $q_{k,j}=p_{j,k}$.
Thus the $i,j$ entry of $P\trans{P}$ is the sum
\(
\sum_{k=1}^n p_{i,k}p_{j,k}
\).
But $p_{i,k}$ is usually $0$, and so $P_{i,k}P_{j,k}$ is usually $0$.
The only time $P_{i,k}$ is nonzero is when it is $1$,
but then there are no other $i^\prime\neq i$ such that $P_{i^\prime,k}$
is nonzero ($i$ is the only row with a $1$ in column~$k$).
In other words,
\begin{equation*}
\sum_{k=1}^n p_{i,k}p_{j,k}=
\begin{cases}
1 &i=j \\
0 &\text{otherwise}
\end{cases}
\end{equation*}
and this is exactly the formula for the entries of the identity matrix.
\end{answer}
\item
A matrix \( A \) is
\definend{skew-symmetric}\index{matrix!skew-symmetric}%
......
......@@ -924,7 +924,7 @@ Gauss-Jordan method.
0 &0 &1
\end{vmat}=1
\end{equation*}
dosn't equal the result of
doesn't equal the result of
expanding down the diagonal.
\begin{equation*}
1\cdot (+1)\begin{vmat}[r]
......
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