Exercise 3.2 · Q9: Find $x$ and $y$ from $2A+B=C$

Basics you need

Solution

1. Check the orders

   Each of the three matrices has 2 rows and 2 columns, so each is of order $2\times2$⁠. Multiplying by the number 2 does not change the order, so the first term (2 times the first matrix) is also $2\times2$⁠. Two $2\times2$ matrices can be added, and the sum is $2\times2$⁠. The right side is $2\times2$ too, so comparing the two sides makes sense.

   Addition is only defined for matrices of the same order, and two matrices can only be equal if they have the same order. Checking this first tells us the question is meaningful and that we will get exactly $2\times2 = 4$ entry equations.
   Sum and scalar multiple keep the order. Lesson: addition and scalar multiplication

2. Multiply the first matrix by 2

   Multiply each of the four entries by 2:

   $$2\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} = \begin{bmatrix} 2\cdot 1 & 2\cdot 3 \\ 2\cdot 0 & 2\cdot x \end{bmatrix} = \begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix}$$
   The scalar 2 multiplies the whole matrix, so it reaches every entry, including the 0 (which stays 0) and the unknown $x$ (which becomes $2x$⁠).
   $kA = [k\,a_{ij}]$⁠: every entry is multiplied by $k$⁠. Lesson: scalar multiplication

3. Add the second matrix, entry by entry

   Add the entries that sit in the same position:

   $$\begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 2+y & 6+0 \\ 0+1 & 2x+2 \end{bmatrix}$$

   which simplifies to

   $$\begin{bmatrix} 2+y & 6 \\ 1 & 2x+2 \end{bmatrix}.$$
   Matrix addition is defined position by position: the $(1,1)$ entries are added together, the $(1,2)$ entries together, and so on. We cannot simplify $2+y$ or $2x+2$ further, because $x$ and $y$ are still unknown.
   $[a_{ij}] + [b_{ij}] = [a_{ij} + b_{ij}]$⁠. Basics: adding like with like

4. Equate corresponding entries

   The equation now reads

   $$\begin{bmatrix} 2+y & 6 \\ 1 & 2x+2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}.$$

   Both sides are $2\times2$⁠, so they are equal exactly when all four pairs of corresponding entries are equal:

   • position $(1,1)$⁠: $2 + y = 5$⁠;
   • position $(1,2)$⁠: $6 = 6$⁠, true whatever $x, y$ are;
   • position $(2,1)$⁠: $1 = 1$⁠, true whatever $x, y$ are;
   • position $(2,2)$⁠: $2x + 2 = 8$⁠.
   This is the definition of equal matrices put to work. The two entries without unknowns must still be checked: if one of them had said something false, such as $6 = 7$⁠, no choice of $x$ and $y$ could make the matrices equal and the equation would have no solution. Here both are true, so only the two equations with unknowns remain.
   Equal matrices: same order and $a_{ij} = b_{ij}$ for all $i, j$ (Definition 2). Lesson: equality of matrices

5. Solve the two equations

   From position $(1,1)$⁠: subtract 2 from both sides of $2 + y = 5$⁠:

   $$y = 5 - 2 = 3.$$

   From position $(2,2)$⁠: subtract 2 from both sides of $2x + 2 = 8$⁠, then divide both sides by 2:

   $$2x = 8 - 2 = 6, \qquad x = \frac{6}{2} = 3.$$
   Each equation has only one unknown, so each can be solved on its own. Doing the same operation to both sides keeps the equation true, and each operation is reversible, so $x = 3$ and $y = 3$ are the only values that work.
   Undo "$+2$" by subtracting 2; undo "$\times 2$" by dividing by 2. Basics: solving equations

6. Substitute back

   Put $x = 3$ and $y = 3$ into the original left side:

   $$2\begin{bmatrix} 1 & 3 \\ 0 & 3 \end{bmatrix} + \begin{bmatrix} 3 & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 6 \\ 0 & 6 \end{bmatrix} + \begin{bmatrix} 3 & 0 \\ 1 & 2 \end{bmatrix}$$ $$= \begin{bmatrix} 2+3 & 6+0 \\ 0+1 & 6+2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix},$$

   which is the right side. ✓

   Substituting into the original equation, not into our simplified one, also catches any slip made in Steps 2 and 3.

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