Example 3: Evaluate a $3\times3$ determinant

Basics you need

Solution

1. Choose the line to expand along

   Count the zeros in each row and column:

   Line	Entries	Zeros
   $R_1$	$1,\ 2,\ 4$	0
   $R_2$	$-1,\ 3,\ 0$	1
   $R_3$	$4,\ 1,\ 0$	1
   $C_1$	$1,\ -1,\ 4$	0
   $C_2$	$2,\ 3,\ 1$	0
   $C_3$	$4,\ 0,\ 0$	2

   The third column $C_3$ has two zeros, so we expand along $C_3$⁠.

   Every line gives the same answer, so we are free to pick the one that needs the least work. With two zeros in $C_3$⁠, two of the three terms vanish and only one $2\times2$ determinant has to be evaluated.
   Remark (i) after Section 4.2.3: expand along the row or column with the maximum number of zeros. Lesson: expanding a 3 × 3 determinant

2. Write the signs for column 3

   The entries of $C_3$ are $a_{13} = 4$⁠, $a_{23} = 0$⁠, $a_{33} = 0$⁠. Their signs are

   $$\begin{aligned} (-1)^{1+3} &= (-1)^4 = +1,\\ (-1)^{2+3} &= (-1)^5 = -1,\\ (-1)^{3+3} &= (-1)^6 = +1. \end{aligned}$$

   So the pattern down the third column is $+,\ -,\ +$⁠.

   The sign belongs to the position, not to the entry. An even exponent gives $+1$ and an odd one gives $-1$⁠, which is the chessboard of Basic 2 read down its last column.
   $(-1)^{i+j}$ is $+1$ when $i+j$ is even and $-1$ when it is odd. Basics: the + − + sign pattern

3. Form the three $2\times2$ determinants

   For each entry of $C_3$⁠, delete its row and column 3, and keep what is left in the same arrangement.

   For $a_{13} = 4$⁠, delete $R_1$ and $C_3$⁠:

   $$\begin{aligned} \begin{vmatrix} -1 & 3 \\ 4 & 1 \end{vmatrix} &= (-1)(1) - (4)(3)\\ &= -1 - 12 = -13. \end{aligned}$$

   For $a_{23} = 0$⁠, delete $R_2$ and $C_3$⁠:

   $$\begin{aligned} \begin{vmatrix} 1 & 2 \\ 4 & 1 \end{vmatrix} &= (1)(1) - (4)(2)\\ &= 1 - 8 = -7. \end{aligned}$$

   For $a_{33} = 0$⁠, delete $R_3$ and $C_3$⁠:

   $$\begin{aligned} \begin{vmatrix} 1 & 2 \\ -1 & 3 \end{vmatrix} &= (1)(3) - (-1)(2)\\ &= 3 + 2 = 5. \end{aligned}$$
   Only the first of these values is actually needed, because the other two will be multiplied by 0. They are worked out here so that every number in the book's line is accounted for, and because Step 5 uses the full picture as a check.
   Each smaller determinant comes from deleting the row and column of the entry it belongs to (Section 4.2.3, Steps 1 to 3). Section 4.4 later names it the minor of that entry. Lesson: minors and cofactors

4. Add the signed terms

   $$\begin{aligned} \Delta &= +4\begin{vmatrix} -1 & 3 \\ 4 & 1 \end{vmatrix} - 0\begin{vmatrix} 1 & 2 \\ 4 & 1 \end{vmatrix}\\ &\quad + 0\begin{vmatrix} 1 & 2 \\ -1 & 3 \end{vmatrix}\\ &= 4(-13) - 0(-7) + 0(5)\\ &= -52 - 0 + 0\\ &= -52. \end{aligned}$$
   Each term is sign $\times$ entry $\times$ its $2\times2$ determinant. The signs $+, -, +$ come from Step 2 and the values $-13, -7, 5$ from Step 3. The two zero entries kill their terms, so $\Delta = 4\times(-13) = -52$⁠.
   Expansion along $C_3$⁠: each of $a_{13}$⁠, $a_{23}$⁠, $a_{33}$ times its sign (Step 2) times its $2\times2$ determinant (Step 3), added. Lesson: expanding a 3 × 3 determinant

5. Check by expanding along the first row

   The first row $1, 2, 4$ has signs $+, -, +$⁠. Its three $2\times2$ determinants are:

   $$\begin{aligned} \begin{vmatrix} 3 & 0 \\ 1 & 0 \end{vmatrix} &= 3(0) - 1(0) = 0,\\ \begin{vmatrix} -1 & 0 \\ 4 & 0 \end{vmatrix} &= (-1)(0) - 4(0) = 0,\\ \begin{vmatrix} -1 & 3 \\ 4 & 1 \end{vmatrix} &= -1 - 12 = -13. \end{aligned}$$

   So

   $$\Delta = 1(0) - 2(0) + 4(-13) = -52.$$
   Section 4.2.3 works out the expansions along $R_1$⁠, $R_2$ and $C_1$ in general, finds the same six terms each time, and states that every row and column gives this value. Expanding a second way is therefore a genuine check: a slip in a sign or a minor would almost certainly make the two answers differ. Here both give $-52$⁠.
   "Expanding a determinant along any row or column gives same value", Section 4.2.3. Lesson: expanding a 3 × 3 determinant

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