Example 8: Find $X$ with $2A+3X=5B$

Basics you need

Solution

1. Find the order of $X$

   $A$ and $B$ are both $3\times2$⁠, so $2A$ and $5B$ are $3\times2$⁠. For $2A + 3X$ to be defined, $3X$ must also be $3\times2$⁠, so $X$ is a $3\times2$ matrix.

   We are looking for an unknown matrix, so we should know its shape before hunting for its entries. A sum is defined only for equal orders, and a scalar multiple keeps the order, so $X$ has the same order as $A$⁠.
   Sums need equal orders (Section 3.4.1); $kX$ has the order of $X$ (Section 3.4.2).

2. Isolate $3X$

   Add $-2A$ to both sides of $2A + 3X = 5B$⁠:

   $$\begin{aligned} 2A + 3X - 2A &= 5B - 2A\\ 2A - 2A + 3X &= 5B - 2A\\ O + 3X &= 5B - 2A\\ 3X &= 5B - 2A \end{aligned}$$

   The reasons, line by line: adding the same matrix $-2A$ to equal matrices gives equal matrices; then the commutative law $3X + (-2A) = (-2A) + 3X$⁠, with the associative law to regroup, lets us bring $2A$ and $-2A$ together; then $2A + (-2A) = O$ because $-2A$ is the additive inverse of $2A$⁠; finally $O + 3X = 3X$ because $O$ is the additive identity.

   These are the matrix versions of what we do without thinking for numbers. Matrix multiplication is not commutative (Section 3.4.5, after Example 14), so in matrix algebra we should never assume an order can be swapped unless a property says so. Here only addition is being reordered, and addition of matrices is commutative.
   Commutative law, additive inverse, additive identity: Section 3.4.3 (i), (iv), (iii). Lesson: properties of the operations

3. Isolate $X$

   Multiply both sides of $3X = 5B - 2A$ by the scalar $\frac13$⁠:

   $$X = \tfrac13\,(5B - 2A).$$
   Every entry of $3X$ is $3x_{ij}$⁠, and $\frac13\cdot3x_{ij} = x_{ij}$⁠, so $\frac13(3X) = X$⁠. Multiplying equal matrices by the same scalar keeps them equal. This is the matrix version of "divide both sides by 3".
   $(kl)A = k(lA)$ holds entry by entry because multiplication of numbers is associative: $\frac13(3x_{ij}) = (\frac13\cdot3)x_{ij} = 1\cdot x_{ij}$⁠. Basics: identity and undo

4. Compute $5B$ and $2A$

   Multiply every entry of $B$ by 5, and every entry of $A$ by 2:

   $$5B = \begin{bmatrix} 5\cdot2 & 5\cdot(-2) \\ 5\cdot4 & 5\cdot2 \\ 5\cdot(-5) & 5\cdot1 \end{bmatrix} = \begin{bmatrix} 10 & -10 \\ 20 & 10 \\ -25 & 5 \end{bmatrix},$$ $$2A = \begin{bmatrix} 2\cdot8 & 2\cdot0 \\ 2\cdot4 & 2\cdot(-2) \\ 2\cdot3 & 2\cdot6 \end{bmatrix} = \begin{bmatrix} 16 & 0 \\ 8 & -4 \\ 6 & 12 \end{bmatrix}.$$
   The scalar multiplies every one of the six entries. Writing the products out once makes the sign of each entry clear (for example $5\cdot(-5) = -25$⁠).
   $kA = [k\,a_{ij}]$ (Section 3.4.2). Lesson: addition and scalar multiplication

5. Compute $5B - 2A$

   Subtract corresponding entries, $(5B - 2A)_{ij} = 5b_{ij} - 2a_{ij}$⁠:

   $$\begin{aligned} \text{row 1:}&\quad 10 - 16 = -6, \quad -10 - 0 = -10\\ \text{row 2:}&\quad 20 - 8 = 12, \quad 10 - (-4) = 14\\ \text{row 3:}&\quad -25 - 6 = -31, \quad 5 - 12 = -7 \end{aligned}$$ $$5B - 2A = \begin{bmatrix} -6 & -10 \\ 12 & 14 \\ -31 & -7 \end{bmatrix}.$$
   Both are $3\times2$⁠, so the difference is defined. The entry $10 - (-4) = 14$ is the one to watch: subtracting a negative number adds it.

6. Multiply by $\tfrac13$

   Divide each entry by 3:

   $$\begin{aligned} x_{11} &= \tfrac{-6}{3} = -2, & x_{12} &= -\tfrac{10}{3},\\ x_{21} &= \tfrac{12}{3} = 4, & x_{22} &= \tfrac{14}{3},\\ x_{31} &= -\tfrac{31}{3}, & x_{32} &= -\tfrac{7}{3}. \end{aligned}$$ $$X = \begin{bmatrix} -2 & -\tfrac{10}{3} \\[2pt] 4 & \tfrac{14}{3} \\[2pt] -\tfrac{31}{3} & -\tfrac73 \end{bmatrix}.$$
   Some entries divide exactly by 3 and some do not. Fractions are perfectly good matrix entries, so we leave $-\frac{10}{3}$⁠, $\frac{14}{3}$⁠, $-\frac{31}{3}$⁠, $-\frac73$ as fractions rather than rounding.

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