Exercise 4.5 · Q9: Solve $4x-3y=3$⁠, $3x-5y=7$ by the matrix method

Basics you need

Solution

1. Write the system as $AX = B$

   Put the coefficients of $x$ and $y$ in the rows of $A$ (one row per equation, $x$-column first), the unknowns in $X$⁠, and the right-hand sides in $B$⁠:

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

   Orders: $A$ is $2\times2$⁠, $X$ is $2\times1$⁠, so $AX$ is defined (2 columns of $A$⁠, 2 rows of $X$⁠) and is $2\times1$⁠, the same order as $B$⁠. Multiplying out,

   $$AX = \begin{bmatrix} 4x - 3y \\ 3x - 5y \end{bmatrix},$$

   so $AX = B$ says exactly $4x - 3y = 3$ and $3x - 5y = 7$⁠.

   Writing the product out is how we know $A$ was built correctly: each entry of $AX$ must reproduce the left side of one equation. Note the coefficient of $y$ in both equations is negative, so the second column of $A$ holds $-3$ and $-5$⁠, signs included.
   Section 4.6.1 writes any system as $AX = B$ in this way. A product is row times column (Section 3.4.5). Lesson: solving systems AX = B

2. Compute $|A|$ and decide whether there is a unique solution

   $$\begin{aligned} |A| &= \begin{vmatrix} 4 & -3 \\ 3 & -5 \end{vmatrix} = 4(-5) - (-3)(3)\\ &= -20 + 9 = -11. \end{aligned}$$

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

   Here is why the solution is unique. If $X$ solves $AX = B$⁠, multiply both sides on the left by $A^{-1}$⁠: $A^{-1}(AX) = A^{-1}B$⁠, so $(A^{-1}A)X = A^{-1}B$⁠, that is $IX = X = A^{-1}B$⁠. So any solution must equal $A^{-1}B$⁠. And $A^{-1}B$ really is a solution, because $A(A^{-1}B) = (AA^{-1})B = IB = B$⁠. Geometrically, the slopes are $\frac{4}{3}$ and $\frac{3}{5}$⁠, which differ, so the two lines cross once.
   Case I of Section 4.6.1: if $A$ is nonsingular, $X = A^{-1}B$ is the unique solution. Definition 5, Section 4.5: nonsingular means $|A| \ne 0$⁠. Basics: two lines

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

   For a $2\times2$ matrix, deleting a row and a column leaves a single entry, and that entry is the minor. With $a_{11} = 4$⁠, $a_{12} = -3$⁠, $a_{21} = 3$⁠, $a_{22} = -5$⁠:

   $$\begin{aligned} A_{11} &= (-1)^{1+1} M_{11} = +(-5) = -5,\\ A_{12} &= (-1)^{1+2} M_{12} = -(3) = -3,\\ A_{21} &= (-1)^{2+1} M_{21} = -(-3) = 3,\\ A_{22} &= (-1)^{2+2} M_{22} = +(4) = 4. \end{aligned}$$

   The matrix of cofactors, and its transpose, the adjoint:

   $$\begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix} = \begin{bmatrix} -5 & -3 \\ 3 & 4 \end{bmatrix},$$ $$\operatorname{adj} A = \begin{bmatrix} A_{11} & A_{21} \\ A_{12} & A_{22} \end{bmatrix} = \begin{bmatrix} -5 & 3 \\ -3 & 4 \end{bmatrix}.$$

   Shortcut check: swapping $a_{11}$ and $a_{22}$ (4 and $-5$⁠) and changing the signs of $a_{12}$ and $a_{21}$ ($-3 \to 3$⁠, $3 \to -3$⁠) gives the same matrix.

   $M_{11}$ is what is left after deleting row 1 and column 1, namely $-5$⁠; $M_{12}$ is left after deleting row 1 and column 2, namely $3$⁠; and so on. The sign $(-1)^{i+j}$ is $+$ when $i+j$ is even and $-$ when it is odd. The transpose matters: cofactor $A_{12}$ goes to position (2, 1) of the adjoint.
   Definitions 1 and 2, Section 4.4 (minor, cofactor); Definition 3, Section 4.5.1 (adjoint); the Remark after Example 12 (the $2\times2$ shortcut). Basics: the sign pattern

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

   $$A^{-1} = \frac{1}{|A|}\operatorname{adj} A = -\frac{1}{11}\begin{bmatrix} -5 & 3 \\ -3 & 4 \end{bmatrix}.$$

   Check: compute $A(\operatorname{adj} A)$ entry by entry (row of $A$ times column of $\operatorname{adj} A$⁠):

   $$\begin{aligned} (1,1)&:\ 4(-5) + (-3)(-3)\\ &= -20 + 9 = -11,\\ (1,2)&:\ 4(3) + (-3)(4)\\ &= 12 - 12 = 0,\\ (2,1)&:\ 3(-5) + (-5)(-3)\\ &= -15 + 15 = 0,\\ (2,2)&:\ 3(3) + (-5)(4)\\ &= 9 - 20 = -11. \end{aligned}$$ $$A(\operatorname{adj} A) = \begin{bmatrix} -11 & 0 \\ 0 & -11 \end{bmatrix} = -11\,I,$$

   so

   $$\begin{aligned} AA^{-1} &= -\frac{1}{11}\,A(\operatorname{adj} A)\\ &= -\frac{1}{11}(-11\,I) = I. \end{aligned}$$
   The check costs four short products and catches almost every slip in the cofactors. The diagonal entries must come out as $|A|$ and the off-diagonal ones as 0, exactly as Theorem 1 says. We keep the factor $-\frac{1}{11}$ outside the matrix to avoid fractions inside it.
   Theorem 1, Section 4.5: $A(\operatorname{adj} A) = |A|\,I$⁠. Theorem 4: $A^{-1} = \frac{1}{|A|}\operatorname{adj} A$⁠. Basics: why the inverse divides by |A|

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

   $A^{-1}$ is $2\times2$ and $B$ is $2\times1$⁠, so $A^{-1}B$ is defined and is $2\times1$⁠, like $X$⁠. First multiply $\operatorname{adj} A$ by $B$⁠:

   $$\begin{aligned} \text{row 1}&:\ (-5)(3) + 3(7)\\ &= -15 + 21 = 6,\\ \text{row 2}&:\ (-3)(3) + 4(7)\\ &= -9 + 28 = 19. \end{aligned}$$

   Then divide by $|A| = -11$⁠:

   $$X = \begin{bmatrix} x \\ y \end{bmatrix} = -\frac{1}{11}\begin{bmatrix} 6 \\ 19 \end{bmatrix} = \begin{bmatrix} -\frac{6}{11} \\ -\frac{19}{11} \end{bmatrix}.$$

   By equality of matrices, $x = -\frac{6}{11}$⁠, $y = -\frac{19}{11}$.

   The order matters: $X = A^{-1}B$⁠, never $BA^{-1}$⁠. We got there by multiplying $AX = B$ on the left by $A^{-1}$⁠, and here $BA^{-1}$ is not even defined (a $2\times1$ times a $2\times2$⁠). Multiplying $\operatorname{adj} A$ by $B$ first keeps the arithmetic in whole numbers until the last line.
   Equality of matrices (Section 3.3.1): two columns are equal when their entries are equal one by one. Basics: rows, columns and positions

6. Check the answer in both original equations

   $$\begin{aligned} 4x - 3y &= 4\left(-\tfrac{6}{11}\right) - 3\left(-\tfrac{19}{11}\right)\\ &= -\tfrac{24}{11} + \tfrac{57}{11} = \tfrac{33}{11} = 3,\\ 3x - 5y &= 3\left(-\tfrac{6}{11}\right) - 5\left(-\tfrac{19}{11}\right)\\ &= -\tfrac{18}{11} + \tfrac{95}{11} = \tfrac{77}{11} = 7. \end{aligned}$$

   Both equations hold, so the solution is correct.

   A check in only one equation is not enough: many wrong pairs satisfy one of the two lines. Checking both confirms the point lies on both lines, which is what "solution of the system" means.

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