Exercise 3.2 · Q12: Find $x, y, z, w$ when unknowns appear on both sides

Basics you need

Solution

1. Check the orders

   All three matrices are $2\times2$⁠. The left side $3\times$ (a $2\times2$ matrix) is $2\times2$⁠; the right side is a sum of two $2\times2$ matrices, which is defined and $2\times2$⁠. So the two sides can be compared entry by entry, giving four equations for the four unknowns.

   Matching orders are what make both the addition and the comparison legitimate.

2. Simplify the left side

   $$3\begin{bmatrix} x & y \\ z & w \end{bmatrix} = \begin{bmatrix} 3x & 3y \\ 3z & 3w \end{bmatrix}$$
   The scalar 3 multiplies every entry.
   $kA = [k\,a_{ij}]$⁠. Lesson: scalar multiplication

3. Simplify the right side

   Add entries in the same position:

   $$\begin{aligned} (1,1):&\ \ x + 4\\ (1,2):&\ \ 6 + (x + y) = 6 + x + y\\ (2,1):&\ \ -1 + (z + w) = z + w - 1\\ (2,2):&\ \ 2w + 3 \end{aligned}$$

   so the right side is

   $$\begin{bmatrix} x + 4 & 6 + x + y \\ z + w - 1 & 2w + 3 \end{bmatrix}.$$
   Matrix addition works position by position. Each entry of the second matrix is added to the entry in the same place of the first; brackets can be dropped because addition of numbers is associative.
   $[a_{ij}] + [b_{ij}] = [a_{ij}+b_{ij}]$⁠. Basics: adding like with like

4. Equate corresponding entries

   The two sides are equal $2\times2$ matrices, so:

   $$\begin{aligned} (1,1):&\ \ 3x = x + 4 &(1)\\ (1,2):&\ \ 3y = 6 + x + y &(2)\\ (2,1):&\ \ 3z = z + w - 1 &(3)\\ (2,2):&\ \ 3w = 2w + 3 &(4) \end{aligned}$$
   Definition 2: equal matrices have equal corresponding entries. Notice which unknowns each equation contains: (1) only $x$⁠; (4) only $w$⁠; (2) contains $x$ and $y$⁠; (3) contains $z$ and $w$⁠. That tells us the order to solve in.
   Lesson: equality of matrices

5. Solve (1) for $x$ and (4) for $w$

   (1): subtract $x$ from both sides, then divide by 2:

   $$3x - x = 4 \;\Rightarrow\; 2x = 4 \;\Rightarrow\; x = 2.$$

   (4): subtract $2w$ from both sides:

   $$3w - 2w = 3 \;\Rightarrow\; w = 3.$$
   Each of these equations has a single unknown, appearing on both sides. Subtracting the same term from both sides gathers it on the left.
   Basics: solving equations one unknown at a time

6. Use $x$ in (2) to get $y$

   Subtract $y$ from both sides of (2), then put $x = 2$⁠:

   $$\begin{aligned} 3y - y &= 6 + x\\ 2y &= 6 + 2 = 8\\ y &= 4. \end{aligned}$$
   Equation (2) has two unknowns, but $x$ is now known from Step 5, so only $y$ is left.

7. Use $w$ in (3) to get $z$

   Subtract $z$ from both sides of (3), then put $w = 3$⁠:

   $$\begin{aligned} 3z - z &= w - 1\\ 2z &= 3 - 1 = 2\\ z &= 1. \end{aligned}$$
   Likewise, $w$ from Step 5 leaves $z$ as the only unknown in (3).

8. Substitute back

   With $x = 2$⁠, $y = 4$⁠, $z = 1$⁠, $w = 3$⁠, the left side is

   $$3\begin{bmatrix} 2 & 4 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 6 & 12 \\ 3 & 9 \end{bmatrix},$$

   and the right side is

   $$\begin{bmatrix} 2 & 6 \\ -1 & 6 \end{bmatrix} + \begin{bmatrix} 4 & 6 \\ 4 & 3 \end{bmatrix} = \begin{bmatrix} 6 & 12 \\ 3 & 9 \end{bmatrix}.$$

   The two sides match. ✓

   Here $2w = 6$⁠, $x + y = 6$ and $z + w = 4$⁠. Every position agrees, so the values are correct. Since every solving step was reversible, they are the only values.

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