Example 20: Verify $(A')'=A$⁠, $(A+B)'=A'+B'$⁠, $(kB)'=kB'$

Basics you need

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

1. Find $A'$⁠, entry by entry

   $A$ has order $2\times3$⁠, so $A'$ has order $3\times2$⁠. Its entries come from $A$ with the two subscripts swapped:

   $$\begin{aligned} &a'_{11} = a_{11} = 3, \quad a'_{12} = a_{21} = 4,\\ &a'_{21} = a_{12} = \sqrt3, \quad a'_{22} = a_{22} = 2,\\ &a'_{31} = a_{13} = 2, \quad a'_{32} = a_{23} = 0. \end{aligned}$$

   Assembling them:

   $$A' = \begin{bmatrix} 3 & 4 \\ \sqrt3 & 2 \\ 2 & 0 \end{bmatrix}.$$

   You can read this off directly: the first row of $A$⁠, which is $3, \sqrt3, 2$⁠, is now the first column; the second row $4, 2, 0$ is now the second column.

   Writing each entry with its address, $a'_{ij} = a_{ji}$⁠, is the definition of transpose applied literally, so no entry can be put in the wrong place. Here $a'_{ij}$ means the $(i,j)$ entry of $A'$⁠.
   Definition 3 (Section 3.5): if $A = [a_{ij}]_{m\times n}$ then $A' = [a_{ji}]_{n\times m}$⁠. Lesson: transpose

2. Transpose again and compare with $A$

   $A'$ has order $3\times2$⁠, so $(A')'$ has order $2\times3$⁠, the same as $A$⁠. Its $(i,j)$ entry is the $(j,i)$ entry of $A'$⁠, which is the $(i,j)$ entry of $A$⁠. Reading it off, the columns of $A'$ become rows:

   $$(A')' = \begin{bmatrix} 3 & \sqrt3 & 2 \\ 4 & 2 & 0 \end{bmatrix}.$$

   Entry by entry this is $3, \sqrt3, 2$ in row 1 and $4, 2, 0$ in row 2, exactly the entries of $A$⁠, and the orders agree. So $(A')' = A$.

   Swapping the two subscripts twice returns them to the original order: $(i,j) \to (j,i) \to (i,j)$⁠. So every entry goes back to its own place. That is why the rule holds for every matrix, not just this one.
   Property (i) of Section 3.5.1: $(A')' = A$ for any matrix $A$⁠.

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

1. Check that $A + B$ is defined, then add entry by entry

   $A$ and $B$ are both $2\times3$⁠, so $A+B$ is defined and is $2\times3$⁠. Add the entries in matching positions:

   $$\begin{aligned} (1,1)\!:&\ 3+2 = 5,\\ (1,2)\!:&\ \sqrt3 + (-1) = \sqrt3 - 1,\\ (1,3)\!:&\ 2+2 = 4,\\ (2,1)\!:&\ 4+1 = 5,\\ (2,2)\!:&\ 2+2 = 4,\\ (2,3)\!:&\ 0+4 = 4. \end{aligned}$$ $$A + B = \begin{bmatrix} 5 & \sqrt3 - 1 & 4 \\ 5 & 4 & 4 \end{bmatrix}.$$
   Matrix addition is only defined for matrices of the same order, because each entry of one must have a partner in the same position of the other. The entry $\sqrt3 - 1$ cannot be simplified further, so we leave it in that exact form rather than rounding.
   Addition of matrices, Section 3.4.1. Basics: adding only like with like

2. Transpose the sum: the left side

   $A+B$ is $2\times3$⁠, so $(A+B)'$ is $3\times2$⁠. Row 1 of $A+B$ becomes column 1 and row 2 becomes column 2:

   $$(A + B)' = \begin{bmatrix} 5 & 5 \\ \sqrt3 - 1 & 4 \\ 4 & 4 \end{bmatrix}.$$
   This is the same "rows become columns" rule as in Part (i), Step 1, now applied to the matrix $A+B$⁠.

3. Transpose first, then add: the right side

   From Part (i), Step 1,

   $$A' = \begin{bmatrix} 3 & 4 \\ \sqrt3 & 2 \\ 2 & 0 \end{bmatrix}.$$

   $B$ is $2\times3$⁠, so $B'$ is $3\times2$⁠: row 1 of $B$⁠, which is $2, -1, 2$⁠, becomes column 1, and row 2, which is $1, 2, 4$⁠, becomes column 2:

   $$B' = \begin{bmatrix} 2 & 1 \\ -1 & 2 \\ 2 & 4 \end{bmatrix}.$$

   $A'$ and $B'$ are both $3\times2$⁠, so they can be added. Entry by entry:

   $$\begin{aligned} &\text{row 1: } 3+2 = 5,\ \ 4+1 = 5,\\ &\text{row 2: } \sqrt3 + (-1) = \sqrt3 - 1,\\ &\phantom{\text{row 2: }} 2+2 = 4,\\ &\text{row 3: } 2+2 = 4,\ \ 0+4 = 4. \end{aligned}$$ $$A' + B' = \begin{bmatrix} 5 & 5 \\ \sqrt3 - 1 & 4 \\ 4 & 4 \end{bmatrix}.$$
   The right side is a different calculation from the left side: here we move the entries first and add afterwards. Doing it independently is what makes the comparison in the next step a genuine check.

4. Compare

   Both sides are $3\times2$⁠, and their six entries agree position by position: $5, 5$⁠; $\sqrt3 - 1, 4$⁠; $4, 4$⁠. So $(A+B)' = A' + B'$.

   In general, the $(i,j)$ entry of $(A+B)'$ is the $(j,i)$ entry of $A+B$⁠, which is $a_{ji} + b_{ji}$⁠. The $(i,j)$ entry of $A'+B'$ is also $a_{ji} + b_{ji}$⁠. Both sides add the same two numbers; they only differ in whether the move happens before or after the addition.
   Property (iii) of Section 3.5.1: $(A+B)' = A' + B'$ whenever $A$ and $B$ have the same order.

Part (iii): $(kB)' = kB'$

1. Multiply $B$ by $k$

   Multiplying a matrix by a number multiplies every entry, and the order stays $2\times3$⁠:

   $$kB = k\begin{bmatrix} 2 & -1 & 2 \\ 1 & 2 & 4 \end{bmatrix} = \begin{bmatrix} 2k & -k & 2k \\ k & 2k & 4k \end{bmatrix}.$$

   For example the $(1,2)$ entry is $k\cdot(-1) = -k$ and the $(2,3)$ entry is $k\cdot 4 = 4k$⁠.

   Scalar multiplication, Section 3.4.2: $kA = [k\,a_{ij}]$⁠.

2. Transpose $kB$ and take out the factor $k$

   $kB$ is $2\times3$⁠, so $(kB)'$ is $3\times2$⁠. Row 1 of $kB$⁠, which is $2k, -k, 2k$⁠, becomes column 1; row 2, which is $k, 2k, 4k$⁠, becomes column 2:

   $$(kB)' = \begin{bmatrix} 2k & k \\ -k & 2k \\ 2k & 4k \end{bmatrix}.$$

   Every entry has the factor $k$⁠, so

   $$(kB)' = k\begin{bmatrix} 2 & 1 \\ -1 & 2 \\ 2 & 4 \end{bmatrix} = kB',$$

   using $B'$ from Part (ii), Step 3. So $(kB)' = kB'$ for every constant $k$⁠.

   Because $k$ is left as a letter, this one calculation covers every constant at once. Taking the common factor $k$ out of every entry is scalar multiplication read backwards. In general the $(i,j)$ entry of $(kB)'$ is $k\,b_{ji}$⁠, which is $k$ times the $(i,j)$ entry of $B'$⁠.
   Property (ii) of Section 3.5.1: $(kA)' = kA'$⁠.

Every fact used, and where it comes from