Exercise 3.3 · Q3: Verify $(A+B)'=A'+B'$ and $(A-B)'=A'-B'$ given $A'$

Basics you need

Getting ready: find $A$ and $B'$

1. Recover $A$ by transposing $A'$

   The rows of $A'$ are $3, 4$⁠; $-1, 2$⁠; $0, 1$⁠. Since $A = (A')'$⁠, we write these rows as columns:

   $$A = (A')' = \begin{bmatrix} 3 & -1 & 0 \\ 4 & 2 & 1 \end{bmatrix}_{2\times3}.$$
   Transposing is its own undo (Basic 2). The order confirms it: $A'$ is $3\times2$⁠, so $A$ must be $2\times3$⁠, the same order as $B$⁠. That is exactly what is needed for $A+B$ and $A-B$ to exist.
   $(A')' = A$⁠, Section 3.5.1, property (i). Lesson: transpose

2. Find $B'$ and list all orders

   The rows of $B$ are $-1, 2, 1$ and $1, 2, 3$⁠. As columns:

   $$B' = \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix}_{3\times2}.$$

   Orders: $A$ and $B$ are $2\times3$⁠, so $A\pm B$ is $2\times3$ and $(A\pm B)'$ is $3\times2$⁠. $A'$ and $B'$ are $3\times2$⁠, so $A'\pm B'$ is $3\times2$⁠. Every operation in the question is defined, and both sides of each equation are $3\times2$⁠.

   Order checks first: this question is a trap for anyone who tries to add the given $A'$ to the given $B$⁠. Those have orders $3\times2$ and $2\times3$⁠, so $A'+B$ is not defined at all.
   Sums need equal orders (Basic 3). Basics: adding like with like

Part (i): $(A+B)' = A'+B'$

1. (i) Left side: add, then transpose

   $$\begin{aligned} A + B &= \begin{bmatrix} 3+(-1) & -1+2 & 0+1 \\ 4+1 & 2+2 & 1+3 \end{bmatrix} \\[1mm] &= \begin{bmatrix} 2 & 1 & 1 \\ 5 & 4 & 4 \end{bmatrix}. \end{aligned}$$

   The rows $2, 1, 1$ and $5, 4, 4$ become columns:

   $$(A + B)' = \begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{bmatrix}.$$
   Sum entry by entry, Section 3.4.1; transpose, Definition 3.

2. (i) Right side: add the transposes

   $A'$ is given and $B'$ was found above:

   $$\begin{aligned} A' + B' &= \begin{bmatrix} 3+(-1) & 4+1 \\ -1+2 & 2+2 \\ 0+1 & 1+3 \end{bmatrix} \\[1mm] &= \begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{bmatrix}. \end{aligned}$$
   For the right side we can use the given $A'$ directly; recovering $A$ was only needed for the left side.

3. (i) Compare

   Both sides are $3\times2$ with entries $2, 5$⁠; $1, 4$⁠; $1, 4$ in the same positions, so $(A+B)' = A'+B'$.

   Equal matrices: same order, equal corresponding entries (Definition 2).

Part (ii): $(A-B)' = A'-B'$

1. (ii) Left side: subtract, then transpose

   $$\begin{aligned} A - B &= \begin{bmatrix} 3-(-1) & -1-2 & 0-1 \\ 4-1 & 2-2 & 1-3 \end{bmatrix} \\[1mm] &= \begin{bmatrix} 4 & -3 & -1 \\ 3 & 0 & -2 \end{bmatrix}. \end{aligned}$$

   Rows become columns:

   $$(A - B)' = \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix}.$$
   $3-(-1) = 3+1 = 4$⁠: subtracting a negative adds. Every other entry is a plain subtraction.
   Difference $A-B = [a_{ij} - b_{ij}]$⁠, Section 3.4.2. Basics: adding like with like

2. (ii) Right side: subtract the transposes

   $$\begin{aligned} A' - B' &= \begin{bmatrix} 3-(-1) & 4-1 \\ -1-2 & 2-2 \\ 0-1 & 1-3 \end{bmatrix} \\[1mm] &= \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix}. \end{aligned}$$

3. (ii) Compare

   Both sides are $3\times2$ with the same six entries, so $(A-B)' = A'-B'$.

   This is what the general rule predicts: the $(i,j)$ entry of $(A-B)'$ is $a_{ji} - b_{ji}$⁠, and so is the $(i,j)$ entry of $A'-B'$⁠. The computation confirms it for this particular pair.
   $(A\pm B)' = A' \pm B'$⁠, Section 3.5.1, property (iii) (with $(kA)' = kA'$ for $k = -1$ giving the minus case). Lesson: transpose

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