- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
3.6. $\quad$ Exercises
$\quad$ In this set of exercises you may assume existence of a determinant function. Determinants of order three may be computed by Equation (3.2).
\begin{align*}
(3.2) \qquad
\det\begin{bmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{bmatrix}
&=
a_{11}\begin{vmatrix}
a_{22} & a_{23} \\ a_{32} & a_{33}
\end{vmatrix}
-
a_{12}\begin{vmatrix}
a_{21} & a_{23} \\ a_{31} & a_{33}
\end{vmatrix}
+
a_{13}\begin{vmatrix}
a_{21} & a_{22} \\ a_{31} & a_{32}
\end{vmatrix}
\end{align*}
3. $\quad$ (a) $\quad$ Prove that $\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} = (b - a)(c - a)(c - b).$
(b) $\quad$ Find corresponding formulas for the determinants
$$\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{vmatrix} \quad \text{and} \quad \begin{vmatrix} 1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix}.$$
Solution. $\quad$
(a) $\quad$ Prove that $\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} = (b - a)(c - a)(c - b).$ $\quad$
$\quad$ Proof. \begin{align*} \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} &= \begin{vmatrix} b & c \\ b^2 & c^2 \end{vmatrix} - \begin{vmatrix} a & c \\ a^2 & c^2 \end{vmatrix} + \begin{vmatrix} a & b \\ a^2 & b^2 \end{vmatrix} \\ &= (bc^2 - cb^2) - (ac^2 - ca^2) + (ab^2 - ba^2) \\ &= bc(c - b) - a^2(c - b) + a(b^2 - c^2) \\ &= (c - b)(bc - a^2) + a(b + c)(b - c) \\ &= (c - b)[bc - a^2 - a(b + c)] \\ &= (c - b)[b(c - a) - a(c - a)] \\ &= (c - a)(b - a)(c - b). \quad \blacksquare \end{align*}
(b) $\quad$ Find corresponding formulas for the determinants $$\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{vmatrix} \quad \text{and} \quad \begin{vmatrix} 1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix}.$$
By the Gauss-Jordan elimination process, We have \begin{align*} \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{vmatrix} &= \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ 0 & b^3 - a^2b & c^3 - a^2c \end{vmatrix} \\ &= \begin{vmatrix} 1 & 1 & 1 \\ 0 & b - a & c - a \\ 0 & b^3 - a^2b & c^3 - a^2c \end{vmatrix} \\ &= \begin{vmatrix} 1 & 1 & 1 \\ 0 & b - a & c - a \\ 0 & 0 & (c-a)[c(c+a) - b(b+a)] \end{vmatrix} \end{align*} Then, as we proved in Section 4, Example 3, the determinant of an upper-triangular matrix is the product of its diagonal elements. That is, \begin{align*} \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{vmatrix} &= (b - a)(c - a)[c(c+a) - b(b+a)] \\ &= (b - a)(c - a)[c^2 + ac - b^2 - ab] \\ &= (b - a)(c - a)[(c^2 - b^2) + ac - ab] \\ &= (b - a)(c - a)(c-b)(a + b + c). \quad \blacksquare \end{align*}
$\quad$ We can do the same to $\begin{vmatrix} 1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix}$ to get \begin{align*} \begin{vmatrix} 1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix} &= \begin{vmatrix} 1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ 0 & b^3 - ab^2 & c^3 - ac^2 \end{vmatrix} \\ &= \begin{vmatrix} 1 & 1 & 1 \\ 0 & b^2 - a^2 & c^2 - a^2 \\ 0 & b^3 - ab^2 & c^3 - ac^2 \end{vmatrix} \\ &= \begin{vmatrix} 1 & 1 & 1 \\ 0 & (b-a)(b+a) & (c-a)(c+a) \\ 0 & b^2(b-a) & c^2(c-a) \end{vmatrix} \\ \end{align*} If $b = 0,$ then the above is an upper-triangular matrix, with determinant $a^3c^2-a^2c^3.$ Otherwise, we have \begin{align*} \begin{vmatrix} 1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix} &= b^2 \begin{vmatrix} 1 & 1 & 1 \\ 0 & (b-a)(b+a) & (c-a)(c+a) \\ 0 & (b-a) & \frac{c^2}{b^2}(c-a) \end{vmatrix} \\ &= -b^2 \begin{vmatrix} 1 & 1 & 1 \\ 0 & (b-a) & \frac{c^2}{b^2}(c-a) \\ 0 & (b-a)(b+a) & (c-a)(c+a) \end{vmatrix} \\ &= -b^2 \begin{vmatrix} 1 & 1 & 1 \\ 0 & (b-a) & \frac{c^2}{b^2}(c-a) \\ 0 & 0 & (c-a)(c+a) - \frac{c^2}{b^2}(c-a)(b+a) \end{vmatrix} \\ &= -b^2(b-a)(c-a)[(c+a) - \frac{c^2}{b^2}(b+a)] \\ &= (b-a)(c-a)[c^2(b+a) - b^2(c+a)] \\ &= (b-a)(c-a)[bc(c - b) + a(c-b)(c+b)] \\ &= (b-a)(c-a)(c-b)(ab + ac + bc). \quad \blacksquare \end{align*}