Exercise 4.5 · Q10: Solve $5x+2y=3$⁠, $3x+2y=5$ by the matrix method

Basics you need

Solution

1. Write the system as $AX = B$

   $$A = \begin{bmatrix} 5 & 2 \\ 3 & 2 \end{bmatrix},\quad X = \begin{bmatrix} x \\ y \end{bmatrix},\quad B = \begin{bmatrix} 3 \\ 5 \end{bmatrix}.$$

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

   $$AX = \begin{bmatrix} 5x + 2y \\ 3x + 2y \end{bmatrix}.$$

   Setting $AX = B$ and comparing entries gives back the two equations, so $A$⁠, $X$ and $B$ are right.

   Row 1 of $A$ holds the coefficients of the first equation, row 2 those of the second, and the columns follow the order $x$⁠, then $y$⁠. Any other arrangement would multiply out to different equations.
   Section 4.6.1 (matrix form of a system); Section 3.4.5 (row times column). Lesson: solving systems AX = B

2. Compute $|A|$

   $$\begin{aligned} |A| &= \begin{vmatrix} 5 & 2 \\ 3 & 2 \end{vmatrix} = 5(2) - 2(3)\\ &= 10 - 6 = 4. \end{aligned}$$

   $|A| = 4 \ne 0$⁠, so $A$ is nonsingular, $A^{-1}$ exists, and the system has exactly one solution, $X = A^{-1}B$⁠.

   If $AX = B$⁠, multiplying on the left by $A^{-1}$ gives $X = A^{-1}B$⁠, so there is no other solution; and $A(A^{-1}B) = IB = B$⁠, so this one does work. In the picture, the slopes are $-\frac{5}{2}$ and $-\frac{3}{2}$⁠, which differ, so the lines cross once.
   Section 4.6.1, Case I; Definition 5, Section 4.5 (nonsingular). Basics: two lines

3. Find the cofactors and $\operatorname{adj} A$

   With $a_{11} = 5$⁠, $a_{12} = 2$⁠, $a_{21} = 3$⁠, $a_{22} = 2$⁠, each minor is the single entry left after deleting a row and a column:

   $$\begin{aligned} A_{11} &= (-1)^{1+1} M_{11} = +(2) = 2,\\ A_{12} &= (-1)^{1+2} M_{12} = -(3) = -3,\\ A_{21} &= (-1)^{2+1} M_{21} = -(2) = -2,\\ A_{22} &= (-1)^{2+2} M_{22} = +(5) = 5. \end{aligned}$$

   Matrix of cofactors, then its transpose:

   $$\begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix} = \begin{bmatrix} 2 & -3 \\ -2 & 5 \end{bmatrix},$$ $$\operatorname{adj} A = \begin{bmatrix} A_{11} & A_{21} \\ A_{12} & A_{22} \end{bmatrix} = \begin{bmatrix} 2 & -2 \\ -3 & 5 \end{bmatrix}.$$

   Shortcut check: swap 5 and 2 on the diagonal, change the signs of 2 and 3: the same matrix.

   $M_{12}$ is found by deleting row 1 and column 2, which leaves $a_{21} = 3$⁠; its cofactor carries the sign $(-1)^{3} = -1$⁠. Then the transpose sends $A_{12} = -3$ to row 2, column 1 of the adjoint.
   Definitions 1 and 2, Section 4.4; Definition 3 and the Remark after Example 12, Section 4.5.1. Basics: the sign pattern

4. Form $A^{-1}$ and check $AA^{-1} = I$

   $$A^{-1} = \frac{1}{|A|}\operatorname{adj} A = \frac{1}{4}\begin{bmatrix} 2 & -2 \\ -3 & 5 \end{bmatrix}.$$

   Check $A(\operatorname{adj} A)$⁠, entry by entry:

   $$\begin{aligned} (1,1)&:\ 5(2) + 2(-3) = 10 - 6 = 4,\\ (1,2)&:\ 5(-2) + 2(5) = -10 + 10 = 0,\\ (2,1)&:\ 3(2) + 2(-3) = 6 - 6 = 0,\\ (2,2)&:\ 3(-2) + 2(5) = -6 + 10 = 4. \end{aligned}$$ $$A(\operatorname{adj} A) = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} = 4I,$$

   so $AA^{-1} = \frac{1}{4}(4I) = I$⁠.

   Theorem 1 predicts $|A| = 4$ on the diagonal and 0 elsewhere. Getting exactly that confirms the cofactors before we use them.
   Theorem 1 and Theorem 4, Section 4.5. Basics: why the inverse divides by |A|

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

   $A^{-1}$ ($2\times2$⁠) times $B$ ($2\times1$⁠) is $2\times1$⁠. First $(\operatorname{adj} A)B$⁠:

   $$\begin{aligned} \text{row 1}&:\ 2(3) + (-2)(5)\\ &= 6 - 10 = -4,\\ \text{row 2}&:\ (-3)(3) + 5(5)\\ &= -9 + 25 = 16. \end{aligned}$$ $$X = \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{4}\begin{bmatrix} -4 \\ 16 \end{bmatrix} = \begin{bmatrix} -1 \\ 4 \end{bmatrix}.$$

   Comparing entries, $x = -1$⁠, $y = 4$.

   $X = A^{-1}B$ comes from multiplying $AX = B$ on the left by $A^{-1}$⁠, so $A^{-1}$ stays on the left. The $\frac{1}{4}$ is applied last, and here it divides both entries exactly.
   Equality of matrices, Section 3.3.1. Basics: rows, columns and positions

6. Check in both original equations

   $$\begin{aligned} 5x + 2y &= 5(-1) + 2(4) = -5 + 8 = 3,\\ 3x + 2y &= 3(-1) + 2(4) = -3 + 8 = 5. \end{aligned}$$

   Both hold, so $(-1, 4)$ is the solution.

   The point must lie on both lines; checking only one equation would not show that.

Every fact used, and where it comes from