Exercise 3.2 · Q1: Find $A+B$⁠, $A-B$⁠, $3A-C$⁠, $AB$⁠, $BA$

Basics you need

Solution

1. Orders: check every operation is defined

   $A$⁠, $B$ and $C$ each have 2 rows and 2 columns, so each has order $2\times2$⁠.

   • $A+B$⁠, $A-B$⁠: same order, so defined, and the result is $2\times2$⁠.
   • $3A$ is $2\times2$⁠, the same as $C$⁠, so $3A-C$ is defined and $2\times2$⁠.
   • $AB$⁠: $A$ has 2 columns and $B$ has 2 rows, so $AB$ is defined, of order $2\times2$⁠. The same count works for $BA$⁠.
   Each operation has a size rule, and checking it first is not a formality: with different orders there are entries with no partner, and the operation simply does not exist. Square matrices of the same order pass every rule, which is why all five parts make sense here.
   Sum: same order. Product $PQ$⁠: columns of $P$ = rows of $Q$⁠; then $(m\times n)(n\times p)$ is $m\times p$⁠. Lesson: matrix multiplication

2. (i) $A+B$⁠: add matching entries

   Each entry of the answer is the sum of the two entries in the same position:

   $$A+B=\begin{bmatrix}2+1&4+3\\3+(-2)&2+5\end{bmatrix}=\begin{bmatrix}3&7\\1&7\end{bmatrix}.$$
   The definition of the sum (Section 3.4.1) says $c_{ij}=a_{ij}+b_{ij}$⁠. The only place to slip is the $(2,1)$ entry, where $b_{21}=-2$⁠: $3+(-2)=1$⁠.
   Addition is position by position. Basics: adding like with like

3. (ii) $A-B$⁠: subtract matching entries

   $$\begin{aligned}A-B&=\begin{bmatrix}2-1&4-3\\3-(-2)&2-5\end{bmatrix}\\&=\begin{bmatrix}1&1\\5&-3\end{bmatrix}.\end{aligned}$$
   The difference is defined by $d_{ij}=a_{ij}-b_{ij}$ (Section 3.4.2). Subtracting a negative entry adds: $3-(-2)=3+2=5$⁠. Write the bracket around $-2$ so this sign is not lost.

4. (iii) $3A-C$⁠: first $3A$⁠, then subtract $C$

   Multiply every entry of $A$ by 3:

   $$3A=\begin{bmatrix}3\cdot2&3\cdot4\\3\cdot3&3\cdot2\end{bmatrix}=\begin{bmatrix}6&12\\9&6\end{bmatrix}.$$

   Now subtract $C$ entry by entry:

   $$\begin{aligned}3A-C&=\begin{bmatrix}6-(-2)&12-5\\9-3&6-4\end{bmatrix}\\&=\begin{bmatrix}8&7\\6&2\end{bmatrix}.\end{aligned}$$
   In $3A-C$ the 3 multiplies only $A$⁠, just as in ordinary algebra, so the scalar multiple is done first. Again the negative entry needs care: $c_{11}=-2$⁠, and $6-(-2)=8$⁠.
   $kA=[k\,a_{ij}]$⁠: the number multiplies every entry. Lesson: addition and scalar multiplication

5. (iv) $AB$⁠: row of $A$ times column of $B$⁠, four times

   Write the rows of $A$ and the columns of $B$⁠:

   • rows of $A$⁠: row 1 is $(2,\ 4)$⁠, row 2 is $(3,\ 2)$⁠;
   • columns of $B$⁠: column 1 is $(1,\ -2)$⁠, column 2 is $(3,\ 5)$⁠.

   Entry $(i,k)$ of $AB$ is row $i$ of $A$ times column $k$ of $B$⁠:

   $$\begin{aligned} c_{11}&=2\cdot1+4\cdot(-2)=2-8=-6\\ c_{12}&=2\cdot3+4\cdot5=6+20=26\\ c_{21}&=3\cdot1+2\cdot(-2)=3-4=-1\\ c_{22}&=3\cdot3+2\cdot5=9+10=19 \end{aligned}$$

   Assemble the four entries in their places:

   $$AB=\begin{bmatrix}-6&26\\-1&19\end{bmatrix}.$$
   The definition (Section 3.4.5) is $c_{ik}=a_{i1}b_{1k}+a_{i2}b_{2k}$⁠. The row number of the entry tells you which row of the first matrix to use, and the column number tells you which column of the second. Writing the rows and columns out first stops you from reading a row of $B$ by mistake, which is the commonest error.
   Pair up, multiply, add: first with first, second with second. Basics: pair up, multiply, add

6. (v) $BA$⁠: now the rows come from $B$

   For $BA$ the first matrix is $B$⁠, so the rows come from $B$ and the columns from $A$⁠:

   • rows of $B$⁠: row 1 is $(1,\ 3)$⁠, row 2 is $(-2,\ 5)$⁠;
   • columns of $A$⁠: column 1 is $(2,\ 3)$⁠, column 2 is $(4,\ 2)$⁠.
   $$\begin{aligned} d_{11}&=1\cdot2+3\cdot3=2+9=11\\ d_{12}&=1\cdot4+3\cdot2=4+6=10\\ d_{21}&=(-2)\cdot2+5\cdot3=-4+15=11\\ d_{22}&=(-2)\cdot4+5\cdot2=-8+10=2 \end{aligned}$$ $$BA=\begin{bmatrix}11&10\\11&2\end{bmatrix}.$$

   Compare with part (iv): not a single entry agrees, so $AB\ne BA$.

   Swapping the order swaps which matrix supplies the rows. $AB$ uses rows of $A$⁠; $BA$ uses rows of $B$⁠. These are genuinely different calculations, so there is no reason for the answers to match. This is the textbook's point in Section 3.4.5 (Examples 13 and 14): matrix multiplication is not commutative, even when both products exist and have the same order.
   For numbers $ab=ba$ always; for matrices, keep the order exactly as written. Lesson: matrix multiplication

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