Example 9: Find $X$ and $Y$ from $X+Y$ and $X-Y$

Basics you need

Solution

1. Fix the orders

   Call the two given matrices $P$ (the value of $X + Y$⁠) and $Q$ (the value of $X - Y$⁠). Both are $2\times2$⁠. Since $X + Y$ is defined only when $X$ and $Y$ have the same order, and the sum then has that order, $X$ and $Y$ are both $2\times2$.

   Knowing the shape of the unknowns first tells us we are looking for 4 entries in $X$ and 4 in $Y$⁠, and confirms that every sum and difference below is defined.
   Sums and differences need the same order, and keep it (Sections 3.4.1, 3.4.2).

2. Add the two equations to eliminate $Y$

   Adding equal matrices to equal matrices gives equal matrices, so

   $$(X + Y) + (X - Y) = P + Q.$$

   On the left, regroup using the commutative and associative laws, then cancel:

   $$\begin{aligned} (X + Y) + (X - Y) &= (X + X) + (Y - Y)\\ &= 2X + O = 2X. \end{aligned}$$

   On the right, entry by entry:

   $$P + Q = \begin{bmatrix} 5+3 & 2+6 \\ 0+0 & 9+(-1) \end{bmatrix} = \begin{bmatrix} 8 & 8 \\ 0 & 8 \end{bmatrix}.$$

   So

   $$2X = \begin{bmatrix} 8 & 8 \\ 0 & 8 \end{bmatrix}.$$
   $+Y$ and $-Y$ are additive inverses, so they cancel to the zero matrix $O$⁠, and $X + X = 1X + 1X = (1 + 1)X = 2X$⁠. The regrouping is allowed because matrix addition is commutative and associative; this is exactly why the textbook writes $(X + X) + (Y - Y)$⁠.
   $Y + (-Y) = O$⁠, $A + O = A$⁠, and $(k + l)A = kA + lA$ (Sections 3.4.3, 3.4.4). Lesson: properties of the operations

3. Find $X$

   Multiply both sides by $\frac12$⁠:

   $$X = \tfrac12\begin{bmatrix} 8 & 8 \\ 0 & 8 \end{bmatrix} = \begin{bmatrix} 4 & 4 \\ 0 & 4 \end{bmatrix}.$$

   Entry by entry: $\frac12\cdot8 = 4$⁠, $\frac12\cdot8 = 4$⁠, $\frac12\cdot0 = 0$⁠, $\frac12\cdot8 = 4$⁠.

   $\frac12(2X) = X$⁠, because each entry $2x_{ij}$ becomes $x_{ij}$⁠. This is the matrix version of dividing both sides by 2.
   Scalar multiplication acts on every entry (Section 3.4.2). Basics: identity and undo

4. Subtract the equations to eliminate $X$

   Now subtract the second equation from the first:

   $$(X + Y) - (X - Y) = P - Q.$$

   On the left, $-(X - Y) = -X + Y$⁠, so

   $$\begin{aligned} (X + Y) - (X - Y) &= (X - X) + (Y + Y)\\ &= O + 2Y = 2Y. \end{aligned}$$

   On the right, entry by entry:

   $$P - Q = \begin{bmatrix} 5-3 & 2-6 \\ 0-0 & 9-(-1) \end{bmatrix} = \begin{bmatrix} 2 & -4 \\ 0 & 10 \end{bmatrix}.$$
   Subtracting $X - Y$ means adding $(-1)(X - Y) = -X + Y$⁠; the minus sign reaches both terms. Then $X$ and $-X$ cancel and $Y + Y = 2Y$⁠. Watch $9 - (-1) = 10$⁠.
   $k(A + B) = kA + kB$ with $k = -1$ gives $-(X - Y) = -X + Y$ (Section 3.4.4).

5. Find $Y$

   $$Y = \tfrac12\begin{bmatrix} 2 & -4 \\ 0 & 10 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 0 & 5 \end{bmatrix}.$$

   Entry by entry: $\frac12\cdot2 = 1$⁠, $\frac12\cdot(-4) = -2$⁠, $\frac12\cdot0 = 0$⁠, $\frac12\cdot10 = 5$⁠.

   Same reason as Step 3: multiplying by $\frac12$ undoes the factor 2.

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