Example 16: Solve two equations by the matrix method

Basics you need

Solution

1. Write the system as $AX = B$

   $$A = \begin{bmatrix} 2 & 5 \\ 3 & 2 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \end{bmatrix}, \quad B = \begin{bmatrix} 1 \\ 7 \end{bmatrix}.$$

   $A$ is $2\times2$ and $X$ is $2\times1$⁠, so $AX$ is defined and is $2\times1$⁠, the same order as $B$⁠. Its entries are

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

   so $AX = B$ says exactly $2x + 5y = 1$ and $3x + 2y = 7$⁠.

   Row 1 of $A$ holds the coefficients of the first equation, $2$ and $5$⁠, in the same order as the unknowns in $X$⁠; row 2 holds $3$ and $2$⁠. Multiplying out $AX$ confirms that nothing was misplaced.
   Section 4.6.1: a system is written as $AX = B$⁠. Lesson: solving systems AX = B

2. Compute $|A|$

   $$|A| = \begin{vmatrix} 2 & 5 \\ 3 & 2 \end{vmatrix} = 2\cdot2 - 5\cdot3 = 4 - 15 = -11.$$

   Since $|A| = -11 \ne 0$⁠, $A$ is non-singular, so $A^{-1}$ exists and the system has a unique solution.

   This is Case I of Section 4.6.1. The determinant decides whether the matrix method can be used at all: with $|A| = 0$ there would be no inverse, and the system would have either no solution or infinitely many.
   Case I, Section 4.6.1; Theorem 4, Section 4.5. Basics: two lines: cross, parallel or the same

3. Find $A^{-1}$

   The cofactors are $A_{11} = 2$⁠, $A_{12} = -3$⁠, $A_{21} = -5$⁠, $A_{22} = 2$⁠. Transposing their matrix (equivalently: swap the diagonal 2s, change the signs of 5 and 3):

   $$\operatorname{adj}A = \begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix},$$ $$A^{-1} = -\frac{1}{11}\begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix}.$$
   $A^{-1} = \frac{1}{|A|}\operatorname{adj}A$ and $\frac{1}{|A|} = \frac{1}{-11} = -\frac{1}{11}$⁠. Keeping the fraction outside the matrix avoids working with elevenths until the very end.
   Theorem 4, Section 4.5; Remark after Example 12. Lesson: inverse of a matrix

4. Why $X = A^{-1}B$

   Multiply both sides of $AX = B$ on the left by $A^{-1}$⁠:

   $$\begin{aligned} A^{-1}(AX) &= A^{-1}B\\ (A^{-1}A)X &= A^{-1}B\\ IX &= A^{-1}B\\ X &= A^{-1}B. \end{aligned}$$
   Each line uses one rule: pre-multiplying both sides by the same matrix keeps them equal; the associative law regroups; $A^{-1}A = I$⁠; and $IX = X$⁠. It must be the left side, because $X$ sits to the right of $A$⁠; $BA^{-1}$ would not even be defined ($2\times1$ times $2\times2$⁠). Since $A^{-1}$ is unique, so is $X$⁠.
   Case I, Section 4.6.1. Lesson: solving systems AX = B

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

   $$X = -\frac{1}{11}\begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix}\begin{bmatrix} 1 \\ 7 \end{bmatrix}.$$

   Row times column:

   $$\begin{aligned} \text{row 1}&: \ 2\cdot1 + (-5)\cdot7 = -33,\\ \text{row 2}&: \ (-3)\cdot1 + 2\cdot7 = 11. \end{aligned}$$ $$\begin{bmatrix} x \\ y \end{bmatrix} = -\frac{1}{11}\begin{bmatrix} -33 \\ 11 \end{bmatrix} = \begin{bmatrix} 3 \\ -1 \end{bmatrix}.$$

   Equal matrices have equal entries, so $x = 3$⁠, $y = -1$.

   $-\frac{1}{11}\cdot(-33) = 3$ and $-\frac{1}{11}\cdot 11 = -1$⁠. Comparing the entries of two equal $2\times1$ matrices gives one equation per entry.
   Equality of matrices, Chapter 3, Section 3.3.1. Basics: rows, columns and positions

6. Check in both equations, and $AA^{-1} = I$

   With $x = 3$⁠, $y = -1$⁠:

   $$\begin{aligned} 2x + 5y &= 6 - 5 = 1,\\ 3x + 2y &= 9 - 2 = 7. \end{aligned}$$

   Both equations hold. Also $A\,\operatorname{adj}A$ should be $|A|\,I = -11I$⁠:

   $$\begin{aligned} (1,1)&: \ 2\cdot2 + 5(-3) = -11,\\ (1,2)&: \ 2(-5) + 5\cdot2 = 0,\\ (2,1)&: \ 3\cdot2 + 2(-3) = 0,\\ (2,2)&: \ 3(-5) + 2\cdot2 = -11, \end{aligned}$$

   so $AA^{-1} = -\frac{1}{11}(-11I) = I$⁠.

   Substituting into the original equations is the real test of a solution. Checking $AA^{-1} = I$ also confirms the inverse used in Step 5.

7. The picture: two lines crossing once

   Each equation is a line. Their slopes are $-\frac{2}{5}$ and $-\frac{3}{2}$⁠, which differ, so the lines are not parallel and cross at exactly one point, $(3, -1)$⁠.

   For two lines $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$⁠, the determinant $a_1b_2 - a_2b_1$ is 0 exactly when the two lines have the same direction (parallel or identical lines). Here it is $-11 \ne 0$⁠, which is the algebraic way of saying "the lines cross once".
   $|A| \ne 0$ means a unique solution. Basics: two lines: cross, parallel or the same

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