Exercise 4.1 · Q3: Show $|2A|=4|A|$

Basics you need

Solution

1. Compute $|A|$

   $A$ is $2\times2$⁠, so it has a determinant. With $a_{11} = 1$⁠, $a_{12} = 2$⁠, $a_{21} = 4$⁠, $a_{22} = 2$⁠:

   $$\begin{aligned} |A| = \begin{vmatrix} 1 & 2 \\ 4 & 2 \end{vmatrix} &= 1\cdot2 - 4\cdot2\\ &= 2 - 8 = -6. \end{aligned}$$
   Main diagonal $1\cdot2 = 2$⁠; other diagonal $4\cdot2 = 8$⁠; subtract. A negative determinant is perfectly normal: the bars mean "determinant", not "absolute value".
   $|A| = a_{11}a_{22} - a_{21}a_{12}$⁠, Section 4.2.2. Lesson: determinant of a square matrix

2. Form $2A$⁠, entry by entry

   $$\begin{aligned} 2A &= \begin{bmatrix} 2\cdot1 & 2\cdot2 \\ 2\cdot4 & 2\cdot2 \end{bmatrix}\\ &= \begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}. \end{aligned}$$

   $2A$ is again of order $2\times2$⁠.

   Scalar multiplication multiplies all four entries by 2, not just one row. Because every entry changes, the determinant changes by more than a factor of 2, as Step 5 explains.
   $kA = [k\,a_{ij}]$⁠, Chapter 3, Section 3.4.2. Basics: scaling and adding matrices

3. Compute $|2A|$

   $$\begin{aligned} |2A| = \begin{vmatrix} 2 & 4 \\ 8 & 4 \end{vmatrix} &= 2\cdot4 - 8\cdot4\\ &= 8 - 32 = -24. \end{aligned}$$
   Same rule as Step 1, applied to the new entries. Compare the pieces with Step 1: the main diagonal product went from 2 to 8 and the other from 8 to 32. Each is 4 times bigger, because each is a product of two doubled entries.

4. Compare with $4|A|$

   $$4|A| = 4\cdot(-6) = -24 = |2A|.$$

   So $|2A| = 4|A|$, as required. (Note that $2|A| = -12$⁠, which is not $|2A|$⁠.)

   Both sides have been computed independently and they agree, which is exactly what "show that" asks for this particular $A$⁠.

5. Why the factor is 4 for every $2\times2$ matrix

   Take any $2\times2$ matrix with entries $a, b$ in the first row and $c, d$ in the second, and any number $k$⁠. Then $kA$ has entries $ka, kb, kc, kd$⁠, and

   $$\begin{aligned} |kA| &= (ka)(kd) - (kc)(kb)\\ &= k^2 ad - k^2 bc\\ &= k^2(ad - bc) = k^2|A|. \end{aligned}$$

   With $k = 2$ this gives $|2A| = 2^2|A| = 4|A|$ for every $2\times2$ matrix, not only this one.

   Each of the two terms of the determinant contains exactly one entry from each row, so it picks up one factor $k$ from each row: $k\cdot k = k^2$⁠. The book states the general rule in Remark (iii) of Section 4.2.3: if $A = kB$ for square matrices of order $n$⁠, then $|A| = k^n|B|$⁠. Here $n = 2$⁠.
   $|kA| = k^n|A|$ for order $n$⁠. Exercise 4.1 Q4 is the case $n = 3$⁠, where the factor is $3^3 = 27$⁠.

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