Exercise 4.5 · Q8: Solve $2x-y=-2$⁠, $3x+4y=3$ by the matrix method

Basics you need

Solution

1. Write the system as $AX = B$

   $$\begin{gathered} A=\begin{bmatrix}2&-1\\3&4\end{bmatrix}\\ X=\begin{bmatrix}x\\y\end{bmatrix},\quad B=\begin{bmatrix}-2\\3\end{bmatrix} \end{gathered}$$

   $A$ is $2\times2$ and $X$ is $2\times1$⁠, so $AX$ is $2\times1$⁠:

   $$AX=\begin{bmatrix}2\cdot x+(-1)\cdot y\\3\cdot x+4\cdot y\end{bmatrix}=\begin{bmatrix}2x-y\\3x+4y\end{bmatrix}.$$

   So $AX = B$ is the given system.

   The $-y$ is the coefficient $-1$⁠, and the right-hand side $-2$ goes into $B$ with its sign. Both minus signs will matter later.
   Basics: matrix facts the chapter uses

2. Compute $|A|$

   $$\begin{aligned} |A|&=\begin{vmatrix}2&-1\\3&4\end{vmatrix}=2\cdot4-3\cdot(-1)\\ &=8+3=11 \end{aligned}$$

   $|A| = 11 \ne 0$⁠: $A$ is non-singular, so the system has a unique solution and the matrix method applies.

   The other diagonal product is $3\cdot(-1) = -3$⁠; subtracting $-3$ adds 3. A non-zero determinant is exactly the condition for $A^{-1}$ to exist (Theorem 4).
   Lesson: determinant of a square matrix

3. Cofactors and $\operatorname{adj}A$

   $$\begin{aligned} A_{11}&=(-1)^{1+1}(4)=4\\ A_{12}&=(-1)^{1+2}(3)=-3\\ A_{21}&=(-1)^{2+1}(-1)=1\\ A_{22}&=(-1)^{2+2}(2)=2 \end{aligned}$$ $$\operatorname{adj}A=\begin{bmatrix}A_{11}&A_{21}\\A_{12}&A_{22}\end{bmatrix}=\begin{bmatrix}4&1\\-3&2\end{bmatrix}$$
   For $A_{21}$⁠, delete row 2 and column 1: the entry left is $-1$⁠, and the sign $(-1)^3 = -1$ makes the cofactor $+1$⁠. Shortcut check: swap 2 and 4, change $-1$ to $1$ and $3$ to $-3$⁠.
   Definitions 1, 2 (Section 4.4) and Definition 3 (Section 4.5.1). Lesson: the adjoint

4. $A^{-1}$⁠, checked by $A(\operatorname{adj}A) = 11I$

   $$A^{-1}=\frac{1}{|A|}\operatorname{adj}A=\frac{1}{11}\begin{bmatrix}4&1\\-3&2\end{bmatrix}$$

   Row of $A$ times column of $\operatorname{adj}A$⁠:

   $$\begin{aligned} (1,1)&:\ 2(4)+(-1)(-3)=8+3=11\\ (1,2)&:\ 2(1)+(-1)(2)=2-2=0\\ (2,1)&:\ 3(4)+4(-3)=12-12=0\\ (2,2)&:\ 3(1)+4(2)=3+8=11 \end{aligned}$$

   So $A(\operatorname{adj}A) = 11I$⁠, and dividing by 11 gives $AA^{-1} = I$⁠. ✓

   Keeping $\frac{1}{11}$ outside lets the check run in whole numbers. Theorem 1 says the product must be $|A|I = 11I$⁠, which it is.
   Theorem 1, Section 4.5.1; Theorem 4. Lesson: inverse of a matrix

5. The solution $X = A^{-1}B$

   $$X=A^{-1}B=\frac{1}{11}\begin{bmatrix}4&1\\-3&2\end{bmatrix}\begin{bmatrix}-2\\3\end{bmatrix}$$ $$\begin{aligned} x&=\tfrac{1}{11}(4(-2)+1(3))\\ &=\tfrac{1}{11}(-8+3)=-\tfrac{5}{11}\\ y&=\tfrac{1}{11}((-3)(-2)+2(3))\\ &=\tfrac{1}{11}(6+6)=\tfrac{12}{11} \end{aligned}$$

   So $x = -\frac{5}{11}$⁠, $y = \frac{12}{11}$⁠.

   The $2\times2$ times $2\times1$ product is $2\times1$⁠. In the second entry, $(-3)(-2) = +6$⁠: two negatives give a positive. The fractions cannot be simplified because 5 and 12 have no common factor with 11.
   Section 4.6.1, Case I. Lesson: solving systems AX = B

6. Check in both equations

   $$\begin{aligned} 2x-y&=-\tfrac{10}{11}-\tfrac{12}{11}=-\tfrac{22}{11}=-2\ \checkmark\\ 3x+4y&=-\tfrac{15}{11}+\tfrac{48}{11}=\tfrac{33}{11}=3\ \checkmark \end{aligned}$$
   Fractional answers are easy to get slightly wrong, so the exact check matters. Both equations hold, so the solution is right.
   Basics: two lines, cross, parallel or the same

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