Exercise 4.5 · Q1: Is $x+2y=2$⁠, $2x+3y=3$ consistent?

Basics you need

Solution

1. Write the system as $AX = B$

   Take the coefficients of $x$ and $y$ row by row, the unknowns as a column, and the right-hand sides as a column:

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

   Orders: $A$ is $2\times2$ and $X$ is $2\times1$⁠, so $AX$ is defined and is $2\times1$⁠, the same order as $B$⁠. Row times column:

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

   So $AX = B$ says $x + 2y = 2$ and $2x + 3y = 3$⁠: exactly the given system.

   Two matrices are equal when their corresponding entries are equal, so the single equation $AX = B$ holds precisely when both original equations hold. Multiplying $AX$ out is the check that the coefficients sit in the right places (row 1 of $A$ belongs to the first equation, column 1 of $A$ to $x$⁠).
   Row $i$ of $A$ times the column $X$ gives the left side of equation $i$⁠. Basics: matrix facts the chapter uses

2. Compute $|A|$

   $$\begin{aligned} |A|&=\begin{vmatrix}1&2\\2&3\end{vmatrix}=1\cdot3-2\cdot2\\ &=3-4=-1 \end{aligned}$$

   So $|A| = -1 \ne 0$⁠: $A$ is non-singular.

   The determinant decides which case of Section 4.6.1 we are in. A non-zero value, positive or negative, is all that matters; the minus sign does not make it "fail". It only matters later, because we divide by $|A|$⁠.
   Main diagonal product minus the other diagonal product. Lesson: determinant of a square matrix

3. Conclude: the system is consistent

   Because $|A| \ne 0$⁠, $A$ is invertible (Theorem 4, Section 4.5). Multiply $AX = B$ on the left by $A^{-1}$⁠:

   $$\begin{aligned} A^{-1}(AX)&=A^{-1}B\\ (A^{-1}A)X&=A^{-1}B\\ X&=A^{-1}B \end{aligned}$$

   So any solution must equal $A^{-1}B$⁠, and $A^{-1}B$ really is a solution, since $A(A^{-1}B) = (AA^{-1})B = B$⁠. A solution exists and is unique, so the system is consistent.

   "Consistent" only asks whether a solution exists. This is Case I of Section 4.6.1: a non-singular $A$ always gives exactly one solution, so no further test is needed. We still find the solution below, because it confirms the conclusion and is what the matrix method is for.
   $|A| \ne 0$ $\Rightarrow$ unique solution $X = A^{-1}B$⁠. Lesson: solving systems AX = B

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

   Cofactor $A_{ij} = (-1)^{i+j}M_{ij}$⁠, where the minor $M_{ij}$ is what is left after deleting row $i$ and column $j$⁠. For a $2\times2$ matrix each minor is a single entry, and the signs $(-1)^{i+j}$ follow the pattern

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

   $\operatorname{adj}A$ is the transpose of the matrix of cofactors, so the cofactors of row $i$ go down column $i$⁠:

   $$\operatorname{adj}A=\begin{bmatrix}A_{11}&A_{21}\\A_{12}&A_{22}\end{bmatrix}=\begin{bmatrix}3&-2\\-2&1\end{bmatrix}$$
   Deleting row 1 and column 1 leaves the entry 3; deleting row 1 and column 2 leaves 2 (the bottom-left entry); and so on. Here the matrix of cofactors happens to be symmetric, so transposing it changes nothing, but the transpose is still part of the definition. The quick rule agrees: swap the diagonal entries 1 and 3, change the signs of the two 2s.
   Definition 3, Section 4.5.1, and the Remark after Example 12 for order 2. Lesson: the adjoint

5. $A^{-1}$⁠, and a check that it is right

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

   Check $AA^{-1} = I$⁠, entry by entry (row of $A$ times column of $A^{-1}$⁠):

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

   So $AA^{-1} = I$⁠. ✓

   Dividing by $|A| = -1$ flips every sign of $\operatorname{adj}A$⁠. Forgetting that sign is the most common slip, and the product check catches it at once: with the wrong sign the diagonal entries would come out $-1$⁠.
   $A^{-1} = \frac{1}{|A|}\operatorname{adj}A$ (Theorem 4). Basics: why the inverse divides by |A|

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

   $$X=A^{-1}B=\begin{bmatrix}-3&2\\2&-1\end{bmatrix}\begin{bmatrix}2\\3\end{bmatrix}$$ $$\begin{aligned} x&=(-3)(2)+2(3)=-6+6=0\\ y&=2(2)+(-1)(3)=4-3=1 \end{aligned}$$

   So $x = 0$⁠, $y = 1$⁠.

   $A^{-1}$ is $2\times2$ and $B$ is $2\times1$⁠, so the product is $2\times1$⁠: one entry for $x$⁠, one for $y$⁠. Each entry is a row of $A^{-1}$ times the column $B$⁠.
   Row times column. Basics: matrix facts the chapter uses

7. Check in both original equations

   $$\begin{aligned} x+2y&=0+2(1)=2\ \checkmark\\ 2x+3y&=2(0)+3(1)=3\ \checkmark \end{aligned}$$
   Substituting back into the equations as printed in the question (not into the matrices) catches any slip made while building $A$ and $B$⁠. The picture shows the meaning: each equation is a line, and the lines have different slopes, $-\tfrac12$ and $-\tfrac23$⁠, so they cross at exactly one point.
   One crossing point means exactly one solution. Basics: two lines, cross, parallel or the same

Every fact used, and where it comes from