Exercise 4.1 · Q4: Show $|3A|=27|A|$

Basics you need

Solution

1. Choose where to expand

   $A$ is $3\times3$⁠, so $|A|$ is a determinant of order 3. Count zeros: row 1 has one, row 2 has one, row 3 has two; column 1 has two, column 2 has two, column 3 has none. We expand along column 1 ($C_1$⁠), whose elements are $1, 0, 0$⁠. The signs for $C_1$ are $+, -, +$⁠.

   Row 3, column 1 and column 2 all have two zeros, so any of them is equally quick. Column 1 has the extra convenience that its only nonzero element is 1. Whichever we choose, the value is the same (Section 4.2.3).
   Signs down column 1: $(-1)^{1+1} = +1$⁠, $(-1)^{2+1} = -1$⁠, $(-1)^{3+1} = +1$⁠. Basics: the sign pattern

2. Compute $|A|$ along column 1

   The three $2\times2$ determinants left after deleting the row and column of each element of $C_1$⁠:

   $$\begin{aligned} a_{11}=1:&\ \begin{vmatrix} 1 & 2 \\ 0 & 4 \end{vmatrix} = 1\cdot4 - 0\cdot2 = 4\\ a_{21}=0:&\ \begin{vmatrix} 0 & 1 \\ 0 & 4 \end{vmatrix} = 0\cdot4 - 0\cdot1 = 0\\ a_{31}=0:&\ \begin{vmatrix} 0 & 1 \\ 1 & 2 \end{vmatrix} = 0\cdot2 - 1\cdot1 = -1 \end{aligned}$$

   Signed sum:

   $$\begin{aligned} |A| &= (+1)(1)(4) + (-1)(0)(0)\\ &\quad + (+1)(0)(-1)\\ &= 4 - 0 + 0 = 4. \end{aligned}$$
   For $a_{11}$ we delete row 1 and column 1, leaving rows $1, 2$ and $0, 4$⁠. For $a_{21}$ we delete row 2 and column 1, leaving $0, 1$ and $0, 4$⁠. For $a_{31}$ we delete row 3 and column 1, leaving $0, 1$ and $1, 2$⁠. The last two are multiplied by the element 0, so they contribute nothing whatever their values; we show them only to make the expansion complete.
   Expansion along a column, Section 4.2.3, like the third-column expansion in Example 3. Lesson: expanding a 3 × 3 determinant

3. Form $3A$⁠, entry by entry

   $$\begin{aligned} 3A &= \begin{bmatrix} 3\cdot1 & 3\cdot0 & 3\cdot1 \\ 3\cdot0 & 3\cdot1 & 3\cdot2 \\ 3\cdot0 & 3\cdot0 & 3\cdot4 \end{bmatrix}\\ &= \begin{bmatrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{bmatrix}. \end{aligned}$$

   $3A$ is again $3\times3$⁠, and its zeros are in the same places as those of $A$⁠.

   Scalar multiplication triples all nine entries; zeros stay zero because $3\cdot0 = 0$⁠. So column 1 is still the convenient column to expand along.
   $kA = [k\,a_{ij}]$⁠, Chapter 3, Section 3.4.2. Basics: scaling and adding matrices

4. Compute $|3A|$ along column 1

   Column 1 of $3A$ is $3, 0, 0$⁠, with signs $+, -, +$⁠:

   $$\begin{aligned} 3:&\ \begin{vmatrix} 3 & 6 \\ 0 & 12 \end{vmatrix} = 3\cdot12 - 0\cdot6 = 36\\ 0:&\ \begin{vmatrix} 0 & 3 \\ 0 & 12 \end{vmatrix} = 0\cdot12 - 0\cdot3 = 0\\ 0:&\ \begin{vmatrix} 0 & 3 \\ 3 & 6 \end{vmatrix} = 0\cdot6 - 3\cdot3 = -9 \end{aligned}$$ $$\begin{aligned} |3A| &= (+1)(3)(36) + (-1)(0)(0)\\ &\quad + (+1)(0)(-9)\\ &= 108 - 0 + 0 = 108. \end{aligned}$$
   Compare with Step 2: the element 1 became 3 and the $2\times2$ determinant 4 became 36, which is $3^2\times4$ (by Exercise 4.1 Q3's rule $|kB| = k^2|B|$ for order 2). So the nonzero term grew by $3\times9 = 27$⁠.

5. Compare with $27|A|$

   $$27|A| = 27\cdot4 = 108 = |3A|.$$

   So $|3A| = 27|A|$, as required.

   Both sides were computed independently and they agree. Note that $3|A| = 12$ and $9|A| = 36$ are both different from $|3A|$⁠.

6. Why the factor is $3^3 = 27$ in general

   The full expansion of a $3\times3$ determinant (equation (1) of Section 4.2.3) is a sum of six terms such as $a_{11}a_{22}a_{33}$ and $-a_{12}a_{21}a_{33}$⁠. Each term contains exactly one entry from each of the three rows. In $kA$ each entry is multiplied by $k$⁠, so each term is multiplied by $k\cdot k\cdot k = k^3$⁠:

   $$|kA| = k^3|A| \quad\text{for every } 3\times3 \text{ matrix } A.$$

   With $k = 3$⁠: $|3A| = 27|A|$⁠.

   This is Remark (iii) of Section 4.2.3: if $A = kB$ for square matrices of order $n$⁠, then $|A| = k^n|B|$⁠. The power is the order of the matrix, because that is how many entries each term multiplies together.
   Order 2 gives $k^2$ (Exercise 4.1 Q3); order 3 gives $k^3$⁠.

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