Exercise 4.5 · Q7: Solve $5x+2y=4$⁠, $7x+3y=5$ by the matrix method

Basics you need

Solution

1. Write the system as $AX = B$

   $$\begin{gathered} A=\begin{bmatrix}5&2\\7&3\end{bmatrix}\\ X=\begin{bmatrix}x\\y\end{bmatrix},\quad B=\begin{bmatrix}4\\5\end{bmatrix} \end{gathered}$$

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

   $$AX=\begin{bmatrix}5\cdot x+2\cdot y\\7\cdot x+3\cdot y\end{bmatrix}=\begin{bmatrix}5x+2y\\7x+3y\end{bmatrix}.$$

   So $AX = B$ is the given system.

   Equal matrices have equal entries, so the one matrix equation holds exactly when both equations hold. Multiplying out $AX$ confirms the coefficients are in the right places.
   Basics: matrix facts the chapter uses

2. Compute $|A|$

   $$\begin{aligned} |A|&=\begin{vmatrix}5&2\\7&3\end{vmatrix}=5\cdot3-7\cdot2\\ &=15-14=1 \end{aligned}$$

   $|A| = 1 \ne 0$⁠, so $A$ is non-singular, $A^{-1}$ exists, and the system has exactly one solution (Section 4.6.1, Case I).

   The matrix method only works when $A^{-1}$ exists, and Theorem 4 says that happens exactly when $|A| \ne 0$⁠. So this check comes before anything else.
   Lesson: determinant of a square matrix

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

   For order 2, each minor is the single entry left after deleting a row and a column; then $A_{ij} = (-1)^{i+j}M_{ij}$⁠:

   $$\begin{aligned} A_{11}&=(-1)^{1+1}(3)=3\\ A_{12}&=(-1)^{1+2}(7)=-7\\ A_{21}&=(-1)^{2+1}(2)=-2\\ A_{22}&=(-1)^{2+2}(5)=5 \end{aligned}$$ $$\operatorname{adj}A=\begin{bmatrix}A_{11}&A_{21}\\A_{12}&A_{22}\end{bmatrix}=\begin{bmatrix}3&-2\\-7&5\end{bmatrix}$$
   The transpose puts $A_{21} = -2$ in row 1, column 2, and $A_{12} = -7$ in row 2, column 1. The shortcut agrees: swap 5 and 3, change the signs of 2 and 7.
   Definition 3 and the Remark after Example 12, Section 4.5.1. Lesson: the adjoint

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

   $$\begin{aligned} A^{-1}&=\frac{1}{|A|}\operatorname{adj}A=\frac{1}{1}\begin{bmatrix}3&-2\\-7&5\end{bmatrix}\\ &=\begin{bmatrix}3&-2\\-7&5\end{bmatrix} \end{aligned}$$

   Row of $A$ times column of $A^{-1}$⁠:

   $$\begin{aligned} (1,1)&:\ 5(3)+2(-7)=15-14=1\\ (1,2)&:\ 5(-2)+2(5)=-10+10=0\\ (2,1)&:\ 7(3)+3(-7)=21-21=0\\ (2,2)&:\ 7(-2)+3(5)=-14+15=1 \end{aligned}$$

   So $AA^{-1} = I$⁠. ✓

   Because $|A| = 1$⁠, dividing by $|A|$ changes nothing, and the inverse has whole-number entries. The product check shows the inverse is right before we use it.
   $A^{-1} = \frac{1}{|A|}\operatorname{adj}A$⁠, Theorem 4. Lesson: inverse of a matrix

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

   $$X=A^{-1}B=\begin{bmatrix}3&-2\\-7&5\end{bmatrix}\begin{bmatrix}4\\5\end{bmatrix}$$ $$\begin{aligned} x&=3(4)+(-2)(5)=12-10=2\\ y&=(-7)(4)+5(5)=-28+25=-3 \end{aligned}$$

   So $x = 2$⁠, $y = -3$⁠.

   $A^{-1}$ is $2\times2$ and $B$ is $2\times1$⁠, so $X$ is $2\times1$⁠: its first entry is $x$⁠, its second $y$⁠. The order $A^{-1}B$ matters; $BA^{-1}$ is not even defined.
   Section 4.6.1, Case I. Lesson: solving systems AX = B

6. Check in both equations

   $$\begin{aligned} 5x+2y&=5(2)+2(-3)=10-6=4\ \checkmark\\ 7x+3y&=7(2)+3(-3)=14-9=5\ \checkmark \end{aligned}$$
   Substituting into the original equations, not into the matrices, catches any mistake made in setting up $A$ and $B$ as well as in the arithmetic.
   Basics: two lines, cross, parallel or the same

Every fact used, and where it comes from