Miscellaneous · Q8: Find $X$ from $XA=B$

Basics you need

Solution

1. The orders force X to be $2\times2$

   Let X be $m\times n$⁠. A is $2\times3$ and B is $2\times3$⁠.

   • For XA to be defined, the columns of X must equal the rows of A: $n = 2$⁠.
   • Then XA is $m\times3$⁠. For it to equal B ($2\times3$⁠), we need $m = 2$⁠.

   So X is a $2\times2$ matrix.

   Finding the order first turns "find a matrix" into "find four numbers". Without it we would not know how many unknowns to write down.
   Section 3.4.5 (order of a product) and Definition 2 (equal matrices have the same order). Lesson: matrix multiplication

2. Write X with unknown entries and compute XA

   $$X = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$

   The columns of A are $(1, 4)$⁠, $(2, 5)$⁠, $(3, 6)$⁠. Row 1 of X, $(a, b)$⁠, times each column, and then row 2, $(c, d)$⁠, times each column:

   $$\begin{aligned} (XA)_{11} &= a\cdot1 + b\cdot4 = a + 4b \\ (XA)_{12} &= a\cdot2 + b\cdot5 = 2a + 5b \\ (XA)_{13} &= a\cdot3 + b\cdot6 = 3a + 6b \\ (XA)_{21} &= c\cdot1 + d\cdot4 = c + 4d \\ (XA)_{22} &= c\cdot2 + d\cdot5 = 2c + 5d \\ (XA)_{23} &= c\cdot3 + d\cdot6 = 3c + 6d \end{aligned}$$
   Notice that row 1 of XA involves only $a$ and $b$ (row 1 of X), and row 2 involves only $c$ and $d$⁠. That splits the problem into two independent small systems.
   Entry formula of Section 3.4.5. Basics: pair up, multiply, add

3. Equate with B, entry by entry

   XA and B are both $2\times3$⁠, so $XA = B$ means these six equations all hold:

   $$\begin{aligned} a + 4b &= -7 & c + 4d &= 2 \\ 2a + 5b &= -8 & 2c + 5d &= 4 \\ 3a + 6b &= -9 & 3c + 6d &= 6 \end{aligned}$$

   The left column comes from row 1 of B, the right column from row 2.

   Six equations but only four unknowns: each system has one equation more than it needs. We solve with two of them and use the third as a test. If the third failed, there would be no X at all.
   Definition 2 (Section 3.3.1). Lesson: equality of matrices

4. Solve for $a$ and $b$

   From the first equation, $a = -7 - 4b$⁠. Substitute into the second:

   $$\begin{aligned} 2(-7 - 4b) + 5b &= -8 \\ -14 - 8b + 5b &= -8 \\ -3b &= 6 \\ b &= -2 \end{aligned}$$

   and then $a = -7 - 4(-2) = -7 + 8 = 1$⁠.

   Third equation: $3a + 6b = 3(1) + 6(-2) = 3 - 12 = -9$ ✓. So $a = 1$⁠, $b = -2$⁠.

   Substitution reduces two unknowns to one. Dividing by $-3$ is allowed and reversible, so $b = -2$ is the only value; hence $a = 1$ is forced too. The third equation is not used to find anything, only to confirm that this pair satisfies all three.
   Substitution method. Basics: solving equations, one unknown at a time

5. Solve for $c$ and $d$

   From the first equation, $c = 2 - 4d$⁠. Substitute into the second:

   $$\begin{aligned} 2(2 - 4d) + 5d &= 4 \\ 4 - 8d + 5d &= 4 \\ -3d &= 0 \\ d &= 0 \end{aligned}$$

   and then $c = 2 - 4(0) = 2$⁠.

   Third equation: $3c + 6d = 3(2) + 6(0) = 6$ ✓. So $c = 2$⁠, $d = 0$⁠.

   The same method as Step 4. A zero entry is a perfectly good answer: $d = 0$ is the unique solution of $-3d = 0$⁠.
   Substitution method. Basics: solving equations, one unknown at a time

6. Assemble X and check

   $$X = \begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix}$$

   Check every entry of XA against B:

   $$\begin{aligned} (1,1) &: 1\cdot1 + (-2)\cdot4 = -7 \\ (1,2) &: 1\cdot2 + (-2)\cdot5 = -8 \\ (1,3) &: 1\cdot3 + (-2)\cdot6 = -9 \\ (2,1) &: 2\cdot1 + 0\cdot4 = 2 \\ (2,2) &: 2\cdot2 + 0\cdot5 = 4 \\ (2,3) &: 2\cdot3 + 0\cdot6 = 6 \end{aligned}$$

   These are exactly the entries $-7, -8, -9$ and $2, 4, 6$ of B ✓. Because each pair $(a, b)$ and $(c, d)$ was forced, this X is the only solution.

   The final check multiplies the actual numbers, independently of the algebra, so it would catch any slip in Steps 2 to 5.
   Row times column. Basics: pair up, multiply, add

Every fact used, and where it comes from