Exercise 4.3 · Q4: Evaluate using cofactors of the third column

Basics you need

Solution

1. The elements of the third column

   $a_{13} = yz$⁠, $a_{23} = zx$⁠, $a_{33} = xy$⁠.

   Each element of $C_3$ is the product of the two letters other than the one in the middle of its row: row 1 has $x$ and $yz$⁠, row 2 has $y$ and $zx$⁠, row 3 has $z$ and $xy$⁠. This symmetry is a hint that the answer will be symmetric too.

2. The three minors

   Delete $C_3$⁠; each minor uses columns 1 and 2, with one row removed:

   $$\begin{aligned} M_{13} &= \begin{vmatrix} 1 & y \\ 1 & z \end{vmatrix} = 1\cdot z - y\cdot1 = z - y,\\ M_{23} &= \begin{vmatrix} 1 & x \\ 1 & z \end{vmatrix} = 1\cdot z - x\cdot1 = z - x,\\ M_{33} &= \begin{vmatrix} 1 & x \\ 1 & y \end{vmatrix} = 1\cdot y - x\cdot1 = y - x. \end{aligned}$$
   $M_{13}$ deletes $R_1$⁠, leaving rows 2 and 3; $M_{23}$ deletes $R_2$⁠, leaving rows 1 and 3; $M_{33}$ deletes $R_3$⁠, leaving rows 1 and 2. With a column of 1s, each minor is "lower letter minus upper letter".
   $2\times2$ rule, Section 4.2.2. Basics: where the determinant comes from

3. The three cofactors

   $$\begin{aligned} A_{13} &= (-1)^{1+3}M_{13} = z - y,\\ A_{23} &= (-1)^{2+3}M_{23}\\ &= -(z - x) = x - z,\\ A_{33} &= (-1)^{3+3}M_{33} = y - x. \end{aligned}$$
   Only the middle position (2, 3) has an odd $i + j$⁠, so only $A_{23}$ changes sign, and $-(z - x)$ is $x - z$⁠.
   Definition 2, Section 4.4. Basics: the + − + sign pattern

4. Multiply and add

   $$\begin{aligned} \Delta &= a_{13}A_{13} + a_{23}A_{23} + a_{33}A_{33}\\ &= yz(z - y) + zx(x - z)\\ &\quad + xy(y - x)\\ &= yz^2 - y^2z + x^2z - xz^2\\ &\quad + xy^2 - x^2y. \end{aligned}$$
   This is already a correct value of $\Delta$⁠. The six terms are all of degree 3, and the next step shows they combine into a neat product.
   Section 4.4, Remark after Example 10. Lesson: minors and cofactors

5. Collect as a quadratic in $x$

   Group the terms by powers of $x$⁠:

   $$\begin{aligned} \Delta &= x^2(z - y) + x(y^2 - z^2)\\ &\quad + (yz^2 - y^2z). \end{aligned}$$

   Each bracket has a factor $z - y$⁠:

   $$\begin{aligned} y^2 - z^2 &= -(z - y)(z + y),\\ yz^2 - y^2z &= yz(z - y). \end{aligned}$$

   So

   $$\Delta = (z - y)\big[x^2 - (y + z)x + yz\big].$$
   The $x^2$ terms are $x^2z - x^2y$⁠, the $x$ terms are $xy^2 - xz^2$⁠, and the rest have no $x$⁠. Why look for the factor $z - y$⁠? If $y = z$⁠, rows 2 and 3 become identical, and the Note in Section 4.4 uses (without proof in this reprint) the fact that such a determinant is 0; so $z - y$ is a good guess. We do not rely on that fact: putting $y = z$ in the grouped form makes all three brackets 0, and the two lines just above show the factor $z - y$ directly.
   $y^2 - z^2 = (y - z)(y + z)$⁠. Basics: algebra the chapter leans on

6. Factorise completely

   The bracket is $x^2 - (y + z)x + yz = (x - y)(x - z)$⁠, because the two numbers with sum $y + z$ and product $yz$ are $y$ and $z$⁠. Hence

   $$\Delta = (z - y)(x - y)(x - z).$$

   Rewrite $(z - y)(x - z)$ as $(y - z)(z - x)$ (two sign changes cancel):

   $$\Delta = (x - y)(y - z)(z - x).$$
   The final form is the cyclic one, $x \to y \to z \to x$⁠, which is how the book's answer is written. Changing the sign of two factors at once multiplies by $(-1)(-1) = 1$⁠, so the value is unchanged.
   $(x-p)(x-q) = x^2 - (p+q)x + pq$ (Basic 4).

Every fact used, and where it comes from