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\textbf{Exercises on properties of determinants}
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\begin{problem} (5.1 \#10.  {\em Introduction to Linear Algebra:} Strang) If the entries in every row of a square matrix $A$ add to zero, solve $A \vb x = \vb 0$ to prove that $\det A = 0.$ If those entries add to one, show that $\det(A-I)=0.$ Does this mean that $\det A = 1?$
\end{problem}

\solution{ If the entries of every row of $A$ sum to zero, then $A \vb x = 0$ when $\vb x = (1,\dots,1)$ since each component of $A \vb x$ is the sum of the entries in a row of $A$.  Since $A$ has a non-zero nullspace, it is not invertible and $\det A = 0.$ 

If the entries of every row of $A$ sum to one, then the entries in every row of $A - I$ sum to zero. Hence $A-I$ has a non-zero nullspace and $\det(A-I)=0.$ 

If $\det(\mb A - \mb I) = 0$ it is \textbf{not} necessarily true that
$\det A = 1.$ For example, the rows of 
$A = \small \twobytwo 0 1 1 0$ sum to one but $\det A = -1.$
}

\begin{problem}(5.1 \#18.) Use row operations and the properties of the determinant to calculate the three by three ``Vandermonde determinant":

$$\det \threebythree 1 a {a^2} 1 b {b^2} 1 c {c^2} = (b-a) (c-a) (c-b).$$
\end{problem}

\solution{ Using row operations and properties of the determinant, we have:

\begin{align*}
\det \threebythree 1 a {a^2} 1 b {b^2} 1 c {c^2} &= \det \threebythreec 1 a {a^2} 0 {b-a} {b^2-a^2} 1 c {c^2} \\ \\
&= \det \threebythreec 1 a {a^2} 0 {b-a} {b^2-a^2} 0 {c-a} {c^2 -a^2}\\ \\
&= (b-a) \det \threebythreec 1 a {a^2} 0 1 {b+a} 1 {c-a} {c^2 -a^2} \\ \\
&= (b-a) \det \threebythreec 1 a {a^2} 0 1 {b+a} 0 0 {(c-a)(c-b)}\\ \\
&= (b-a)(c-a)(c-b) \det \threebythreec 1 a {a^2} 0 1 {b+a} 0 0 1\\ \\
&= (b-a)(c-a)(c-b) \det \threebythree 1 0 0 0 1 0 0 0 1 \\ \\
&=(b-a)(c-a)(c-b). \checkmark 
\end{align*}
}

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