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\textbf{Mathematics 262} \\
\textbf{Homework 3 solutions}
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Assignment: Section 5.4, problems 1, 2, 14

\S 5.4, \#1:
You just use the formula.  The inverses of the two given matrices are:
\[
\frac{1}{72}
\begin{bmatrix}
-3 & 5 & 9 \\ 18 & -6 & 18 \\ 6 & 14 & -18 
\end{bmatrix}, 
\begin{bmatrix}
\cos \theta & 0 & \sin \theta \\
0 & 1 & 0 \\ 
-\sin \theta & 0 & \cos \theta
\end{bmatrix}.
\]
There's no real excuse for getting these wrong, since you can find the
inverse other ways, and since it's easy to check your work (just
multiply the original matrix by the supposed inverse).  Of course, I
probably screwed up and gave you the wrong answers, but I'm not
getting graded on this\dots

The second matrix in the problem has a geometric interpretation: it
corresponds to rotating $\mathbf{R}^{3}$ around the $y$-axis by the
angle $-\theta$.  The inverse matrix should precisely undo this, so
it's not surprising that the inverse matrix rotates $\mathbf{R}^{3}$
around the $y$-axis by the angle $\theta$.

\S 5.4, \#2:
Again, you just use the formula.  For part (a), the system of
equations is 
\[
\begin{pmatrix}
1 & 1 & 1 \\ 2 & -6 & -1 \\ 3 & 4 & 2
\end{pmatrix}
\begin{pmatrix}
x \\ y \\ z
\end{pmatrix}
= \begin{pmatrix}
11 \\ 0 \\ 0
\end{pmatrix}.
\]
The determinant of the coefficient matrix is $11$, so 
according to Cramer's rule, 
\[
x = \frac{\det \begin{pmatrix}
11 & 1 & 1 \\ 0 & -6 & -1 \\ 0 & 4 & 2
\end{pmatrix}}{11} = -8.
\]
Similarly, $y=-7$ and $z=26$.  As with the previous problem, since you
can check your work by plugging in the numbers, you really should have
the right answer.

For part (b), the system is 
\[
\begin{pmatrix}
3 & -2 & 0 \\ 0 & 3 & -2 \\ -2 & 0 & 3
\end{pmatrix}
\begin{pmatrix}
x \\ y \\ z
\end{pmatrix}
= \begin{pmatrix}
7 \\ 6 \\ -1
\end{pmatrix}
\]
Since the determinant of the coefficient matrix is 19, then
\[
x = \frac{\det \begin{pmatrix}
7 & -2 & 0 \\ 6 & 3 & -2 \\ -1 & 0 & 3
\end{pmatrix}}{19} = 5.
\]
Similarly, $y=4$ and $z=3$.

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\pagebreak

\S 5.4, \#14:
(I should apologize for assigning this problem; it was harder than I
expected.  On the other hand, challenges are good for you.)

Let $M$ denote the $2n \times 2n$ matrix 
\[
M = \begin{bmatrix}
A & B \\ C & D
\end{bmatrix}
\]
where $A$, $B$, $C$, and $D$ are commuting $n \times n$ matrices.  I
want to show that $\det M = \det (AD-BC)$.  If $C$ is invertible, this
is easy: the matrix
\[
\begin{bmatrix}
D & C^{-1} \\ -C & 0
\end{bmatrix}
\]
has determinant one, so 
\begin{align*}
\det M 
& = \det 
\begin{bmatrix}
A & B \\ C & D
\end{bmatrix}
= \det \left( 
\begin{bmatrix}
A & B \\ C & D
\end{bmatrix}
\begin{bmatrix}
D & C^{-1} \\ -C & 0
\end{bmatrix}
\right) \\ 
& = \det
\begin{bmatrix}
AD-BC & AC^{-1} \\ CD-DC & I
\end{bmatrix}
= \det
\begin{bmatrix}
AD-BC & AC^{-1} \\ 0 & I
\end{bmatrix} \\
& = \det (AD-BC).
\end{align*}

If $C$ is not invertible, then either $C = 0$, in which case I'm done
by a formula from class, or else $C \neq 0$.  In this case, I'll add
things to $C$ until it becomes invertible: notice that for any scalar
$r$, 
\begin{equation}\label{}
\begin{bmatrix}
A & B \\ C & D 
\end{bmatrix}
\begin{bmatrix}
I & 0 \\ rI & I
\end{bmatrix} = 
\begin{bmatrix}
A + rB & B \\ C + rD & D
\end{bmatrix}.
\end{equation}
The determinant of this matrix is the same as the determinant of $M$.
The entries all commute with each other, and the supposed formula for
the determinant doesn't change, either:
\[
(A+rB)D - B(C+rD) = AD + rBD - BC -rBD = AD - BC.
\]
So I may as well work with the matrix 
\[
\begin{bmatrix}
A + rB & B \\ C + rD & D
\end{bmatrix}.
\]
I claim that for some scalar $r$, $C+rD$ is invertible, and so I can
use the previous argument.  As long as the field $F$ has at least $n$
elements, I can appeal to problem 11 in Section 5.4 (which has the
same solution as problem 10, which we discussed in class).  If the
field has fewer than $n$ elements, things are a bit trickier; I'll
skip that case.

\end{document}


  I
claim that the formula $\det(AD - BC)$ is $2n$-linear and alternating
\emph{as a function of $M$}.  If $M$ is the $2n \times 2n$ identity
matrix, then this formula gives $\det(I_{n} I_{n} - 0_{n} 0_{n}) =
\det(I_{n}) = 1$, where $I_{n}$ denotes the $n \times n$ identity
matrix, and $0_{n}$ denotes the $n\times n$ zero matrix.  Therefore if
I can verify the claim, I'm done by our theorem about the uniqueness
of the determinant.

\emph{$2n$-linearity}: look at the $i$th row of $M$.  There are two
cases: either $1 \leq i \leq n$ or $n+1 \leq i \leq 2n$.  In the first
case, if $1 \leq i \leq n$ and the $i$th row of $M$ is a sum of two
vectors, then this is true for the $i$th row of $A$ and of $B$.  Then
when you compute the product $AD-BC$, the same will be true of the
matrices $AD$ and $BC$, and therefore of their difference.  (You
should really include some details here; I'm just hitting the high
points.)  A similar thing happens when the $i$th row of $M$ is
multiplied by a scalar.

In the second case, if $n+1\leq i \leq 2n$ and the $i$th row of $M$ is
the sum of two vectors, then you use the same reasoning, more or less:
Since the matrices commute, then $AD-BC = DA - CB$.  In $DA - CB$,
since $C$ and $D$ are on the left, then the products $DA$ and $CB$ are
linear in the rows of $C$ and $D$, and the argument goes through as
before.

\emph{Alternating}: if two of the first $n$ rows of $M$ are equal,
then the same will be true of each product $AD$ and $BC$, and hence of
$AD-BC$; therefore $\det(AD-BC)$ will be zero.  The same works if two
of the last $n$ rows of $M$ are equal, again using commutativity.  The
remaining case is that one of the first $n$ rows equals one of the
last $n$ rows, say row $i$ equals row $n+k$, where both $i$ and $k$
are numbers between $1$ and $n$.  Unfortunately, I'm not quite sure
how to deal with this case.  Let me know if you figure it out.

