Exercise 4.5 · Q3: Is $x+3y=5$⁠, $2x+6y=8$ consistent?

Basics you need

Solution

1. Write the system as $AX = B$

   $$\begin{gathered} A=\begin{bmatrix}1&3\\2&6\end{bmatrix}\\ X=\begin{bmatrix}x\\y\end{bmatrix},\quad B=\begin{bmatrix}5\\8\end{bmatrix} \end{gathered}$$

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

   $$AX=\begin{bmatrix}1\cdot x+3\cdot y\\2\cdot x+6\cdot y\end{bmatrix}=\begin{bmatrix}x+3y\\2x+6y\end{bmatrix}.$$

   So $AX = B$ is exactly the given pair of equations.

   Row 1 of $A$ is the coefficients of the first equation, row 2 those of the second. Already the rows look related: row 2, $(2, 6)$⁠, is twice row 1, $(1, 3)$⁠.
   Basics: matrix facts the chapter uses

2. Compute $|A|$

   $$|A|=\begin{vmatrix}1&3\\2&6\end{vmatrix}=1\cdot6-2\cdot3=6-6=0$$

   So $A$ is singular and $A^{-1}$ does not exist (Theorem 4, Section 4.5).

   With $|A| = 0$ we cannot divide by $|A|$⁠, so the route $X = A^{-1}B$ is closed. That does not yet tell us whether solutions exist: a singular system can have none or infinitely many. This is Case II of Section 4.6.1, and it needs one more calculation.
   Lesson: determinant of a square matrix

3. Why $(\operatorname{adj}A)B$ decides it

   Suppose some column $X_0$ did satisfy $AX_0 = B$⁠. Multiply both sides on the left by $\operatorname{adj}A$⁠:

   $$\begin{aligned} (\operatorname{adj}A)B&=(\operatorname{adj}A)(AX_0)\\ &=\big((\operatorname{adj}A)A\big)X_0\\ &=\big(|A|I\big)X_0=|A|\,X_0\\ &=0\cdot X_0=O \end{aligned}$$

   So if a solution exists, then $(\operatorname{adj}A)B = O$. Turned around: if $(\operatorname{adj}A)B \ne O$⁠, no solution can exist, and the system is inconsistent. This is the rule of Section 4.6.1, Case II.

   The second line uses associativity of matrix multiplication, and the third uses Theorem 1, $(\operatorname{adj}A)A = |A|I$⁠, then $IX_0 = X_0$⁠. Theorem 1 holds even when $|A| = 0$⁠: the book's verification of it never divides by $|A|$⁠, it only uses that a row (or column) times its own cofactors gives $|A|$ and times the cofactors of a different row (or column) gives 0. The last line is $0\cdot X_0 = O$ because $|A| = 0$ here. The argument only works in one direction: $(\operatorname{adj}A)B = O$ would not prove that a solution exists. That is why the book says that in that case the system "may be either consistent or inconsistent".
   Theorem 1, Section 4.5.1. Lesson: solving systems AX = B

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

   Each minor of a $2\times2$ matrix is the single entry left after deleting a row and a column, and $A_{ij} = (-1)^{i+j}M_{ij}$⁠:

   $$\begin{aligned} A_{11}&=(-1)^{1+1}(6)=6\\ A_{12}&=(-1)^{1+2}(2)=-2\\ A_{21}&=(-1)^{2+1}(3)=-3\\ A_{22}&=(-1)^{2+2}(1)=1 \end{aligned}$$ $$\operatorname{adj}A=\begin{bmatrix}A_{11}&A_{21}\\A_{12}&A_{22}\end{bmatrix}=\begin{bmatrix}6&-3\\-2&1\end{bmatrix}$$
   The adjoint is the transpose of the matrix of cofactors, so $A_{21} = -3$ goes to row 1, column 2. The shortcut for order 2 agrees: swap 1 and 6 on the diagonal, change the signs of 3 and 2. Cofactors exist for every square matrix; only the division by $|A|$ needs $|A| \ne 0$⁠.
   Definition 3, Section 4.5.1. Lesson: the adjoint

5. Compute $(\operatorname{adj}A)B$ and conclude

   $\operatorname{adj}A$ is $2\times2$ and $B$ is $2\times1$⁠, so the product is $2\times1$⁠:

   $$\begin{aligned} (\operatorname{adj}A)B&=\begin{bmatrix}6&-3\\-2&1\end{bmatrix}\begin{bmatrix}5\\8\end{bmatrix}\\ &=\begin{bmatrix}6\cdot5+(-3)\cdot8\\(-2)\cdot5+1\cdot8\end{bmatrix}\\ &=\begin{bmatrix}30-24\\-10+8\end{bmatrix}=\begin{bmatrix}6\\-2\end{bmatrix} \end{aligned}$$

   Both entries are non-zero, so $(\operatorname{adj}A)B \ne O$⁠. By Step 3, the system is inconsistent: it has no solution.

   One non-zero entry would already be enough. With $|A| = 0$ and $(\operatorname{adj}A)B \ne O$⁠, Step 3 shows that assuming a solution leads to a contradiction, so none exists.
   Row times column. Basics: matrix facts the chapter uses

6. See it directly in the equations

   Multiply the first equation by 2:

   $$2(x+3y)=2(5)\quad\text{gives}\quad 2x+6y=10.$$

   The second equation says $2x + 6y = 8$⁠. The same quantity $2x + 6y$ cannot equal both 10 and 8, so no pair $(x, y)$ satisfies both equations.

   Solving each equation for $y$ gives $y = \frac{5 - x}{3}$ and $y = \frac{4 - x}{3}$⁠: the same slope, different intercepts. This is the picture behind $|A| = 0$ (same slope) and $(\operatorname{adj}A)B \ne O$ (different lines). Had the second equation been $2x + 6y = 10$⁠, the lines would coincide and there would be infinitely many solutions.
   Same slope, different intercepts: parallel lines. Basics: two lines, cross, parallel or the same

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