Exercise 3.1 · Q6: Find $x, y, z$ from equal matrices

Basics you need

Part (i)

1. Same order, then equate entries

   Both sides are $2\times2$⁠, so equality is possible. Matching the four addresses:

   $$\begin{aligned} (1,1):&\quad 4 = y \\ (1,2):&\quad 3 = z \\ (2,1):&\quad x = 1 \\ (2,2):&\quad 5 = 5 \end{aligned}$$
   Definition 2 requires every corresponding pair of entries to be equal, so all four equations must hold together. The $(2,2)$ equation $5 = 5$ has no unknowns; it is true, so it imposes no restriction. Had it read, say, $5 = 6$⁠, there would be no solution at all.
   Equal matrices: same order and $a_{ij} = b_{ij}$ for every $i, j$⁠. Lesson: equality of matrices

2. Read off the values

   Each equation already isolates one unknown: $x = 1$⁠, $y = 4$⁠, $z = 3$.

   Substituting back, the left side becomes

   $$\begin{bmatrix} 4 & 3 \\ 1 & 5 \end{bmatrix},$$

   which is exactly the right side with $y = 4$⁠, $z = 3$⁠. ✓

   Each unknown appears in exactly one equation, so each has exactly one value and the solution is unique.

Part (ii)

1. Same order, then equate entries

   Both sides are $2\times2$⁠. Matching addresses:

   $$\begin{aligned} (1,1):&\quad x + y = 6 \\ (1,2):&\quad 2 = 2 \\ (2,1):&\quad 5 + z = 5 \\ (2,2):&\quad xy = 8 \end{aligned}$$
   The $(1,2)$ equation is automatically true. The other three must all hold. Note that $x$ and $y$ are tied together by two equations, one linear and one not, so we must solve them as a pair.
   Equate entry by entry. Lesson: equality of matrices

2. Find $z$

   From $5 + z = 5$⁠, subtracting 5 from both sides gives $z = 0$.

   Subtracting the same number from both sides keeps an equation true, and leaves $z$ alone.
   Isolate the unknown by undoing what is done to it. Basics: solving equations

3. Find $x$ and $y$ from their sum and product

   From $x + y = 6$⁠, $y = 6 - x$⁠. Substitute into $xy = 8$⁠:

   $$\begin{aligned} x(6 - x) &= 8 \\ 6x - x^2 &= 8 \\ x^2 - 6x + 8 &= 0 \\ (x - 2)(x - 4) &= 0 \end{aligned}$$

   So $x = 2$ or $x = 4$⁠, and then $y = 6 - x$⁠:

   $$\begin{aligned} x = 2 &\;\Rightarrow\; y = 4, \\ x = 4 &\;\Rightarrow\; y = 2. \end{aligned}$$
   A product is zero only when one of its factors is zero, so $x - 2 = 0$ or $x - 4 = 0$⁠, and there are no other possibilities. The factorisation is right because $(x-2)(x-4) = x^2 - 6x + 8$⁠. Both roots must be kept: nothing in the question says $x > y$ or $x < y$⁠, so the two pairs are both genuine solutions.
   Zero product rule: $ab = 0 \Rightarrow a = 0$ or $b = 0$⁠.

4. Check both solutions

   With $x = 2$⁠, $y = 4$⁠, $z = 0$⁠: $x + y = 6$⁠, $5 + z = 5$⁠, $xy = 8$⁠, so the left side is

   $$\begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix},$$

   the right side ✓. With $x = 4$⁠, $y = 2$⁠, $z = 0$⁠: again $x + y = 6$ and $xy = 8$⁠, giving the same matrix ✓.

   So $x = 4, y = 2, z = 0$ or $x = 2, y = 4, z = 0$.

   Substituting into the original matrix equation confirms that no solution was gained or lost in the algebra. Swapping $x$ and $y$ changes neither $x + y$ nor $xy$⁠, which explains why the two solutions are mirror images of each other.

Part (iii)

1. Same order, then equate entries

   Both sides are $3\times1$ column matrices. Matching the three addresses:

   $$\begin{aligned} (1,1):&\quad x + y + z = 9 \\ (2,1):&\quad x + z = 5 \\ (3,1):&\quad y + z = 7 \end{aligned}$$
   A column matrix has one column, so its addresses are $(1,1)$⁠, $(2,1)$⁠, $(3,1)$⁠. Three equations in three unknowns: we expect a single solution if they are independent.
   Equate entry by entry. Lesson: equality of matrices

2. Subtract to isolate $y$ and $x$

   Subtract the second equation from the first:

   $$(x + y + z) - (x + z) = 9 - 5 \;\Rightarrow\; y = 4.$$

   Subtract the third equation from the first:

   $$(x + y + z) - (y + z) = 9 - 7 \;\Rightarrow\; x = 2.$$
   Each of the shorter equations is the first one with one unknown missing. Subtracting cancels everything except the missing unknown, so we get it directly.
   Subtracting equal quantities from equal quantities gives equal results. Basics: solving equations

3. Find $z$ and check

   From $x + z = 5$ with $x = 2$⁠: $z = 3$⁠. Check in the equation not yet used for $z$⁠: $y + z = 4 + 3 = 7$ ✓, and $x + y + z = 2 + 4 + 3 = 9$ ✓.

   So $x = 2$⁠, $y = 4$⁠, $z = 3$.

   Every one of the three original equations is satisfied, so the values make the two column matrices equal. Each step was forced (no choices were made), so this is the only solution.

Every fact used, and where it comes from