Exercise 3.2 · Q7: Find $X$ and $Y$

Basics you need

Part (i): $X+Y$ and $X-Y$ given

1. The order of $X$ and $Y$

   $X+Y$ is defined only when $X$ and $Y$ have the same order, and then the sum has that order too. Since $X+Y$ is $2\times2$⁠, both $X$ and $Y$ are $2\times2$⁠.

   Knowing the order in advance tells us what kind of object we are solving for, and guarantees that all the sums below are defined.

2. Add the two equations to remove $Y$

   Adding the left sides and the right sides:

   $$\begin{aligned}&(X+Y)+(X-Y)\\&=\begin{bmatrix}7&0\\2&5\end{bmatrix}+\begin{bmatrix}3&0\\0&3\end{bmatrix}.\end{aligned}$$

   On the left, regroup using the commutative and associative laws: $(X+X)+(Y-Y)=2X+O=2X$⁠. On the right, add position by position:

   $$2X=\begin{bmatrix}7+3&0+0\\2+0&5+3\end{bmatrix}=\begin{bmatrix}10&0\\2&8\end{bmatrix}.$$
   $Y$ and $-Y$ cancel because $-Y$ is the additive inverse of $Y$⁠. This is the matrix version of adding $x+y=7$ and $x-y=3$ to get $2x=10$⁠. The regrouping is legal because matrix addition is commutative and associative (Section 3.4.3).
   Lesson: properties of addition

3. Divide each entry by 2

   $2X$ has entries twice those of $X$⁠, so each entry of $X$ is half the matching entry:

   $$X=\frac12\begin{bmatrix}10&0\\2&8\end{bmatrix}=\begin{bmatrix}5&0\\1&4\end{bmatrix}.$$
   "Dividing by 2" means multiplying by the scalar $\frac12$⁠: $\frac12(2X)=(\frac12\cdot2)X=X$⁠. This is scalar multiplication, which is always allowed.

4. Subtract the equations to remove $X$

   $$\begin{aligned}&(X+Y)-(X-Y)\\&=\begin{bmatrix}7&0\\2&5\end{bmatrix}-\begin{bmatrix}3&0\\0&3\end{bmatrix}\end{aligned}$$

   On the left, $X+Y-X+Y=(X-X)+(Y+Y)=2Y$⁠. On the right:

   $$2Y=\begin{bmatrix}7-3&0-0\\2-0&5-3\end{bmatrix}=\begin{bmatrix}4&0\\2&2\end{bmatrix},$$

   and halving each entry,

   $$Y=\begin{bmatrix}2&0\\1&1\end{bmatrix}.$$
   Subtracting $X-Y$ subtracts both of its parts, so $-(X-Y)=-X+Y$⁠. This is the same sign rule as with numbers: $(x+y)-(x-y)=2y$⁠.

5. Check in both original equations

   $$\begin{aligned} X+Y&=\begin{bmatrix}5+2&0+0\\1+1&4+1\end{bmatrix}=\begin{bmatrix}7&0\\2&5\end{bmatrix}\\ X-Y&=\begin{bmatrix}5-2&0-0\\1-1&4-1\end{bmatrix}=\begin{bmatrix}3&0\\0&3\end{bmatrix} \end{aligned}$$

   Both match the given matrices. ✓

   Substituting back catches any arithmetic slip, and confirms that the pair we found really satisfies both conditions.

Part (ii): $2X+3Y$ and $3X+2Y$ given

1. Names and orders

   Write $P$ and $Q$ for the two right-hand sides:

   $$P=\begin{bmatrix}2&3\\4&0\end{bmatrix},\qquad Q=\begin{bmatrix}2&-2\\-1&5\end{bmatrix}.$$

   So $2X+3Y=P$ and $3X+2Y=Q$⁠. As in Part (i), Step 1, $X$ and $Y$ must be $2\times2$⁠.

   Short names keep the elimination readable. $2X$ and $3Y$ have the orders of $X$ and $Y$⁠, and their sum is $2\times2$⁠, so $X$ and $Y$ are $2\times2$⁠.

2. Eliminate $Y$ to find $X$

   Multiply the first equation by 2 and the second by 3, so that $Y$ has coefficient 6 in both:

   $$4X+6Y=2P,\qquad 9X+6Y=3Q.$$

   Subtract the first from the second: $(9X+6Y)-(4X+6Y)=5X$⁠, so $5X=3Q-2P$⁠. Compute the two scalar multiples:

   $$3Q=\begin{bmatrix}6&-6\\-3&15\end{bmatrix},\qquad 2P=\begin{bmatrix}4&6\\8&0\end{bmatrix}.$$ $$\begin{aligned}5X&=3Q-2P\\&=\begin{bmatrix}6-4&-6-6\\-3-8&15-0\end{bmatrix}\\&=\begin{bmatrix}2&-12\\-11&15\end{bmatrix}\end{aligned}$$

   Multiply by $\frac15$⁠:

   $$X=\begin{bmatrix}\frac25&-\frac{12}{5}\\[1mm]-\frac{11}{5}&3\end{bmatrix}.$$
   Multiplying an equation of matrices by a number multiplies both sides, and $2(2X+3Y)=4X+6Y$ by the distributive law $k(A+B)=kA+kB$⁠. After subtracting, $9X-4X=5X$ by $(k+l)A=kA+lA$⁠, and $6Y-6Y=O$⁠.
   Elimination, as for numbers. Basics: solving equations

3. Eliminate $X$ to find $Y$

   Multiply the first equation by 3 and the second by 2, so that $X$ has coefficient 6 in both:

   $$6X+9Y=3P,\qquad 6X+4Y=2Q.$$

   Subtract the second from the first: $5Y=3P-2Q$⁠.

   $$3P=\begin{bmatrix}6&9\\12&0\end{bmatrix},\qquad 2Q=\begin{bmatrix}4&-4\\-2&10\end{bmatrix}.$$ $$\begin{aligned}5Y&=3P-2Q\\&=\begin{bmatrix}6-4&9-(-4)\\12-(-2)&0-10\end{bmatrix}\\&=\begin{bmatrix}2&13\\14&-10\end{bmatrix}\end{aligned}$$ $$Y=\begin{bmatrix}\frac25&\frac{13}{5}\\[1mm]\frac{14}{5}&-2\end{bmatrix}.$$
   Two entries involve subtracting a negative: $9-(-4)=13$ and $12-(-2)=14$⁠. Finding $Y$ by its own elimination (rather than substituting $X$⁠) avoids fractions until the very end.

4. Check in both original equations

   $2X$ has rows $(\frac45,\ -\frac{24}{5})$⁠, $(-\frac{22}{5},\ 6)$⁠, and $3Y$ has rows $(\frac65,\ \frac{39}{5})$⁠, $(\frac{42}{5},\ -6)$⁠. So

   $$\begin{aligned}2X+3Y&=\begin{bmatrix}\frac{4+6}{5}&\frac{-24+39}{5}\\[1mm]\frac{-22+42}{5}&6-6\end{bmatrix}\\&=\begin{bmatrix}2&3\\4&0\end{bmatrix}=P.\end{aligned}$$

   This is the given first right-hand side. ✓

   $3X$ has rows $(\frac65,\ -\frac{36}{5})$⁠, $(-\frac{33}{5},\ 9)$⁠, and $2Y$ has rows $(\frac45,\ \frac{26}{5})$⁠, $(\frac{28}{5},\ -4)$⁠. So

   $$\begin{aligned}3X+2Y&=\begin{bmatrix}\frac{6+4}{5}&\frac{-36+26}{5}\\[1mm]\frac{-33+28}{5}&9-4\end{bmatrix}\\&=\begin{bmatrix}2&-2\\-1&5\end{bmatrix}=Q.\end{aligned}$$

   This is the given second right-hand side. ✓

   Both equations are satisfied, so the pair is a solution. It is the only one: Part (ii), Steps 2 and 3 show that any pair satisfying both equations must have $5X=3Q-2P$ and $5Y=3P-2Q$⁠, so $X$ and $Y$ are forced to be the matrices found.

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