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\textbf{Exercises on Cramer's rule, inverse matrix, and volume}
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\begin{problem} (5.3 \#8.  {\em Introduction to Linear Algebra:} Strang) Suppose
$$A = \threebythree 1 1 4 1 2 2 1 2 5.$$  
Find its cofactor matrix $\mb C$ and multiply $AC^T$ to find $\det(A)$.
$$C = \threebythree 6 {-3} 0 \cdot \cdot \cdot \cdot
  \cdot \cdot \hspace{4pt}\text{ and }AC^T =
  \rule{10mm}{0.4pt}.$$
If you change $a_{1,3} = 4$ to $100$, why is $\det(A)$ unchanged?
\end{problem}

\solution{ We fill in the cofactor matrix $C$ and then multiply to obtain $AC^T$:
$$ C = \threebythree 6 {-3} 0 3 1 {-1} {-6} 2 1 $$ 
and
$$A C^T = \threebythree 1 1 4 1 2 2 1 2 5 \threebythree 6 3 {-6} {-3} 1 2 0 {-1} 1 = \threebythree 3 0 0 0 3 0 0 0 3 = 3I.$$ 
Since $AC^T = \det (A) I,$ we have $\det(A) = 3.$ If 4 is changed to 100, $\det(A)$ is unchanged because the cofactor of that entry is 0, and thus its value does not contribute to the determinant.
}

\begin{problem}(5.3 \#28.) Spherical coordinates $\rho, \phi, \theta$ satisfy 
$$x = \rho \sin \phi \cos \theta,\; y = \rho \sin\phi\sin\theta \text{
    and } z = \rho\cos\phi.$$ Find the three by three matrix of
  partial derivatives: 
$$\threebythree 
{\partial x / \partial \rho} {\partial x / \partial \phi} {\partial x / \partial \theta}
{\partial y / \partial \rho} {\partial y / \partial \phi} {\partial y / \partial \theta}
{\partial z / \partial \rho} {\partial z / \partial \phi} {\partial z / \partial \theta}.
$$ Simplify its determinant to $J = \rho^2 \sin \phi.$ In spherical
coordinates, $$dV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta$$ is the volume
of an infinitesimal ``coordinate box."
\end{problem}

\solution{ The rows are formed by the partials of $x,y,$ and $z$ with respect to $\rho, \phi,$ and $\theta$:
$$ \threebythreec {\sin \phi \cos \theta} {\rho \cos\phi\cos\theta} {-\rho\sin\phi\sin\theta} {\sin\phi\sin\theta} {\rho\cos\phi\sin\theta} {\rho\sin\phi\cos\theta} {\cos\phi} {-\rho\sin\phi} {0}.$$
Expanding its determinant $J$ along the bottom row, we get:
\begin{eqnarray*} 
J &= &\cos\phi \twobytwo {\rho \cos\phi\cos\theta} {-\rho\sin\phi\sin\theta} {\rho\cos\phi\sin\theta} {\rho\sin\phi\cos\theta} \\&&- (-\rho\sin\phi) \twobytwo {\sin \phi \cos \theta}  {-\rho\sin\phi\sin\theta} {\sin\phi\sin\theta} {\rho\sin\phi\cos\theta} + 0  \\
\\
&=& \cos\phi(\rho^2 \cos\phi\sin\phi \cos^2 \theta + \rho^2  \cos\phi \sin\phi\sin^2 \theta)\\&& + \rho\sin\phi(\rho \sin^2 \phi \cos^2 \theta + \rho \sin^2 \phi \sin^2 \theta) 
\\
\\
&=& \cos\phi(\rho^2 \cos\phi\sin\phi (\cos^2 \theta + \sin^2 \theta)) + \rho\sin\phi(\rho \sin^2 \phi (\cos^2 \theta + \sin^2 \theta)) \\
&=& \cos\phi(\rho^2 \cos\phi\sin\phi) + \rho^2 \sin^3\phi\\ 
&=& \rho^2 \sin\phi(\cos^2 \phi + \sin^2 \phi) \\
J &=& \rho^2 \sin \phi.
\end{eqnarray*}
}

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