Exercise 3.2 · Q10: Solve for $x, y, z, t$ in $2A+3B=3C$

Basics you need

Solution

1. Check the orders

   All three matrices are $2\times2$⁠. Scalar multiples keep the order, so $2\times$ the first, $3\times$ the second and $3\times$ the third are all $2\times2$⁠. The sum on the left is therefore defined and is $2\times2$⁠, the same order as the right side.

   Matrices of different orders can neither be added nor be equal. With matching orders we expect four entry equations, which is exactly the number of unknowns.

2. Work out the three scalar multiples

   Multiply every entry by the number in front:

   $$2\begin{bmatrix} x & z \\ y & t \end{bmatrix} = \begin{bmatrix} 2x & 2z \\ 2y & 2t \end{bmatrix}$$ $$3\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 3\cdot1 & 3\cdot(-1) \\ 3\cdot0 & 3\cdot2 \end{bmatrix}$$ $$\phantom{3\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix}} = \begin{bmatrix} 3 & -3 \\ 0 & 6 \end{bmatrix}$$ $$3\begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix} = \begin{bmatrix} 3\cdot3 & 3\cdot5 \\ 3\cdot4 & 3\cdot6 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix}$$
   Each scalar belongs to the matrix right after it and reaches every one of its entries. Watch the sign: $3\cdot(-1) = -3$⁠.
   $kA = [k\,a_{ij}]$⁠. Lesson: scalar multiplication

3. Add the two matrices on the left

   $$\begin{bmatrix} 2x & 2z \\ 2y & 2t \end{bmatrix} + \begin{bmatrix} 3 & -3 \\ 0 & 6 \end{bmatrix}$$ $$= \begin{bmatrix} 2x+3 & 2z-3 \\ 2y+0 & 2t+6 \end{bmatrix}$$

   So the equation becomes

   $$\begin{bmatrix} 2x+3 & 2z-3 \\ 2y & 2t+6 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix}.$$
   Matrix addition adds entries in the same position. Adding $-3$ is the same as subtracting 3, which is why the $(1,2)$ entry is $2z - 3$⁠.
   $[a_{ij}] + [b_{ij}] = [a_{ij} + b_{ij}]$⁠. Basics: adding like with like

4. Equate corresponding entries

   Both sides are $2\times2$⁠, so they are equal exactly when

   • position $(1,1)$⁠: $2x + 3 = 9$⁠;
   • position $(1,2)$⁠: $2z - 3 = 15$⁠;
   • position $(2,1)$⁠: $2y = 12$⁠;
   • position $(2,2)$⁠: $2t + 6 = 18$⁠.
   This is Definition 2 of equal matrices: equality of the whole matrices is the same as equality of each pair of entries in the same place.
   Lesson: equality of matrices

5. Solve each equation

   Each equation has one unknown. Subtract the constant, then divide by 2:

   $$\begin{aligned} 2x + 3 = 9 &\;\Rightarrow\; 2x = 6 \;\Rightarrow\; x = 3,\\ 2z - 3 = 15 &\;\Rightarrow\; 2z = 18 \;\Rightarrow\; z = 9,\\ 2y = 12 &\;\Rightarrow\; y = 6,\\ 2t + 6 = 18 &\;\Rightarrow\; 2t = 12 \;\Rightarrow\; t = 6. \end{aligned}$$
   For $z$ we add 3 to both sides (to undo "$-3$"), giving $2z = 18$⁠. Every step is reversible, so these are the only values that satisfy the equation.
   Basics: solving equations one unknown at a time

6. Substitute back

   With $x = 3$⁠, $y = 6$⁠, $z = 9$⁠, $t = 6$ the first matrix, in the question's layout, and its double are

   $$\begin{bmatrix} x & z \\ y & t \end{bmatrix} = \begin{bmatrix} 3 & 9 \\ 6 & 6 \end{bmatrix},$$ $$2\begin{bmatrix} 3 & 9 \\ 6 & 6 \end{bmatrix} = \begin{bmatrix} 6 & 18 \\ 12 & 12 \end{bmatrix},$$

   so the left side is

   $$\begin{bmatrix} 6 & 18 \\ 12 & 12 \end{bmatrix} + \begin{bmatrix} 3 & -3 \\ 0 & 6 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix},$$

   which equals the right side from Step 2. ✓

   Substituting into the original equation confirms both the arithmetic and that each value went into the right position.

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