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\textbf{Exercises on determinant formulas and cofactors}
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\begin{problem} Compute the determinant of: $$A = \left[\begin{array}{cccc} 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \end{array}\right].$$ Which method of computing the determinant do you prefer for this problem, and why?
\end{problem}

\solution{ The preferred method is that of using \textbf{cofactors.} We apply the Big Formula:
$$ \det A = \displaystyle\sum_{P = (\alpha, \beta, \dots, \omega)} (\det P) a_{1\alpha} a_{2\beta} \cdots a_{n\omega}$$
to $A$:
\begin{align*}
\det A &=  0 \threebythreedet 0 0 0 1 0 0 0 1 0  - 0 \threebythreedet 1 0 0 0 0 0 0 1 0 + 0 \threebythreedet 1 0 0 0 1 0 0 0 0  - 1 \threebythreedet 1 0 0 0 1 0 0 0 1\\ \\ &= -1 \threebythreedet 1 0 0 0 1 0 0 0 1 = -1. 
\end{align*}
This is quicker than row exchange:
\begin{align*}
 \det A &= \det \left [ \begin{array}{rrrr} 
0 & 0 & 0 & 1 \\
1 & 0 & 0 &0  \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0  \end{array} \right ]  = -\det \left [ \begin{array}{rrrr} 
1 & 0 & 0 &0  \\
0 & 0 & 0 & 1 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0  \end{array} \right ] \\ \\
&= \det \left [ \begin{array}{rrrr} 
1 & 0 & 0 &0  \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0  \end{array} \right ] = -\det\left [ \begin{array}{rrrr} 
1 & 0 & 0 &0  \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \end{array} \right ] \\ \\&= -1. 
\end{align*}
}

\begin{problem} (5.2 \#33.  {\em Introduction to Linear Algebra:} Strang) The symmetric Pascal matrices have determinant 1. If I subtract 1 from the $n,n$ entry, why does the determinant become zero? (Use rule 3 or cofactors.)
$$\det \left [ \small \begin{array}{rrrr} 
1 & 1 & 1 & 1 \\
1 & 2 & 3 & 4 \\
1 & 3 & 6 & 10 \\
1 & 4 & 10 & 20 \\  \end{array} \right ] = 1 \text{ (known)} \hspace{20pt} \det 
 \left [ \small \begin{array}{rrrr} 
1 & 1 & 1 & 1 \\
1 & 2 & 3 & 4 \\
1 & 3 & 6 & 10 \\
1 & 4 & 10 & \mathbf{19} \\  \end{array} \right ] =\vb 0 \text{ (to explain).} $$
\end{problem}

\solution{ The difference in the $n,n$ entry (in the example, the difference between 19 and 20) multiplies its cofactor, the determinant of the $n-1$ by $n-1$ symmetric Pascal matrix.  In our example this matrix is
$$\threebythree 1 1 1 1 2 3 1 3 6. $$ We're told that this matrix has
  determinant 1. Since the $n,n$ entry multiplies its cofactor
  positively, the overall determinant drops by 1 to become 0.  
}

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