Exercise 3.2 · Q2: Compute four sums of matrices

Basics you need

Part (i): $a$⁠, $b$ matrices

1. Check the orders

   Both matrices have 2 rows and 2 columns, so both are $2\times2$⁠. The sum is defined and is $2\times2$⁠.

   Addition needs a partner for every entry, which happens exactly when the orders agree.
   Basics: only same-order matrices can be added

2. Add position by position and simplify

   $$\begin{bmatrix}a&b\\-b&a\end{bmatrix}+\begin{bmatrix}a&b\\b&a\end{bmatrix}=\begin{bmatrix}a+a&b+b\\-b+b&a+a\end{bmatrix}$$

   Now simplify each entry: $a+a=2a$⁠, $b+b=2b$⁠, $-b+b=0$⁠, $a+a=2a$⁠. So the sum is

   $$\begin{bmatrix}2a&2b\\0&2a\end{bmatrix}.$$
   Each entry is an ordinary algebraic sum, so like terms collect as usual. The only entry that changes character is the $(2,1)$ entry: $-b$ and $b$ cancel to 0. Notice the two matrices differ only in that position ($-b$ against $b$⁠).
   $(i,j)$ entry of the sum = $a_{ij}+b_{ij}$⁠. Lesson: addition of matrices

Part (ii): sums that become perfect squares

1. Check the orders

   Both matrices are $2\times2$ (each entry, however long, fills one position). The sum is defined and is $2\times2$⁠.

   An entry such as $a^2+b^2$ is a single number in a single position. The order counts positions, not symbols.

2. Add each position and recognise the square

   Take the four positions one at a time. In each, add the two entries and compare with Basic 3.

   $$\begin{aligned} (1,1)&:\ a^2+b^2+2ab=(a+b)^2\\ (1,2)&:\ b^2+c^2+2bc=(b+c)^2\\ (2,1)&:\ a^2+c^2-2ac=(a-c)^2\\ (2,2)&:\ a^2+b^2-2ab=(a-b)^2 \end{aligned}$$

   For example, in the $(2,1)$ position: $a^2+c^2+(-2ac)=a^2-2ac+c^2$⁠, which is $p^2-2pq+q^2$ with $p=a$⁠, $q=c$⁠. Assembling,

   $$\begin{bmatrix}(a+b)^2&(b+c)^2\\ (a-c)^2&(a-b)^2\end{bmatrix}.$$
   The second matrix supplies exactly the middle term $\pm2pq$ that turns each sum of two squares into a perfect square. A plus sign gives $(p+q)^2$⁠; the minus signs in the bottom row ($-2ac$⁠, $-2ab$⁠) give $(p-q)^2$⁠. Writing the answer as squares is the "simplest form" the question is aiming at.
   $p^2\pm2pq+q^2=(p\pm q)^2$ (Basic 3).

Part (iii): a $3\times3$ numerical sum

1. Check the orders

   Both matrices have 3 rows and 3 columns, so both are $3\times3$⁠. The sum is defined and is $3\times3$⁠, with nine entries.

2. Add the nine pairs of entries

   $$\begin{aligned}&\begin{bmatrix}-1+12&4+7&-6+6\\8+8&5+0&16+5\\2+3&8+2&5+4\end{bmatrix}\\&=\begin{bmatrix}11&11&0\\16&5&21\\5&10&9\end{bmatrix}\end{aligned}$$
   Each of the nine positions is a separate small addition. The two places to watch are the negative entries in the first row: $-1+12=11$ and $-6+6=0$⁠.
   Basics: adding like with like

Part (iv): a trigonometric sum

1. Check the orders

   Both matrices are $2\times2$⁠, so the sum is defined and is $2\times2$⁠.

2. Add position by position

   $$\begin{bmatrix}\cos^2x+\sin^2x&\sin^2x+\cos^2x\\\sin^2x+\cos^2x&\cos^2x+\sin^2x\end{bmatrix}$$
   The second matrix has $\sin^2x$ exactly where the first has $\cos^2x$⁠, and the other way round. So every position gets one $\cos^2x$ and one $\sin^2x$⁠.

3. Use $\sin^2x+\cos^2x=1$ in every entry

   Each of the four entries is $\sin^2x+\cos^2x$ (in one order or the other), and that equals 1 for every real $x$⁠. So the sum is

   $$\begin{bmatrix}1&1\\1&1\end{bmatrix}.$$
   The identity holds for every real $x$⁠, so the answer does not depend on $x$ at all: the sum is the same constant matrix whatever angle you choose. Adding numbers is commutative, so $\cos^2x+\sin^2x$ and $\sin^2x+\cos^2x$ are the same thing.
   $\sin^2x+\cos^2x=1$ for all real $x$ (Basic 4).

Every fact used, and where it comes from