Exercise 4.5 · Q16: Onion, wheat and rice prices by the matrix method

Basics you need

Solution

1. Name the unknowns and turn each bill into an equation

   Let the cost per kg be ₹ $x$ for onion, ₹ $y$ for wheat and ₹ $z$ for rice. Put the data in a table, one row per bill (quantities in kg, totals in ₹):

   Bill	Onion	Wheat	Rice	Total
   1	4	3	2	60
   2	2	4	6	90
   3	6	2	3	70

   Each bill: quantity times price for each item, added, equals the total:

   $$\begin{aligned} 4x + 3y + 2z &= 60\\ 2x + 4y + 6z &= 90\\ 6x + 2y + 3z &= 70 \end{aligned}$$
   The unknowns must be the things asked for, the prices per kg, and each must keep the same meaning in every equation: $x$ is always the onion price. The third bill in the book reads "6 kg onion 2 kg wheat" without a comma; the numbers are 6 kg onion and 2 kg wheat.
   Forming linear equations from words, as in Example 18 (Section 4.6.1). Basics: algebra the chapter leans on

2. Write the system as $AX = B$

   The table's quantity columns become $A$ (rows: bills; columns: onion, wheat, rice), the prices become $X$⁠, the totals become $B$⁠:

   $$A = \begin{bmatrix} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{bmatrix},$$ $$X = \begin{bmatrix} x \\ y \\ z \end{bmatrix},\quad B = \begin{bmatrix} 60 \\ 90 \\ 70 \end{bmatrix}.$$

   $A$ is $3\times3$ and $X$ is $3\times1$⁠, so $AX$ is defined and $3\times1$⁠, like $B$⁠:

   $$AX = \begin{bmatrix} 4x + 3y + 2z \\ 2x + 4y + 6z \\ 6x + 2y + 3z \end{bmatrix}.$$
   Why $AX$ and not $XA$⁠? Row $i$ of $A$ lists what was bought on bill $i$⁠, and the column $X$ lists the price of each item in the same order (onion, wheat, rice). Row times column pairs each quantity with its own price and adds, which is precisely the bill. $XA$ (a $3\times1$ times a $3\times3$⁠) is not even defined.
   Section 4.6.1 (matrix form); Section 3.4.5 (row times column). Lesson: solving systems AX = B

3. Compute $|A|$ by expanding along row 1

   $$\begin{aligned} |A| &= 4\begin{vmatrix} 4 & 6 \\ 2 & 3 \end{vmatrix} - 3\begin{vmatrix} 2 & 6 \\ 6 & 3 \end{vmatrix}\\ &\qquad + 2\begin{vmatrix} 2 & 4 \\ 6 & 2 \end{vmatrix}\\ &= 4(0) - 3(-30) + 2(-20)\\ &= 0 + 90 - 40 = 50. \end{aligned}$$

   (The minors $0$⁠, $-30$⁠, $-20$ are worked out in Step 4.) $|A| = 50 \ne 0$⁠, so $A$ is nonsingular: there is exactly one set of prices that fits all three bills.

   The first minor is 0 because $(4, 6)$ is twice $(2, 3)$⁠, so that term drops out. In real terms, $|A| \ne 0$ guarantees that the three bills pin down one set of prices (Case I of Section 4.6.1). (Later courses show that $|A| = 0$ happens exactly when one bill's quantities are a combination of the other two; this reprint does not prove that, and we do not need it here.)
   Section 4.2.3 (expansion along R1); Section 4.6.1, Case I. Lesson: expanding a 3 × 3 determinant

4. Find all nine minors and cofactors

   Row 1 (entries $4, 3, 2$⁠):

   $$\begin{aligned} M_{11} &= \begin{vmatrix} 4 & 6 \\ 2 & 3 \end{vmatrix}\\ &= (4)(3) - (6)(2) = 0,\\ A_{11} &= (-1)^{1+1}(0) = 0,\\ M_{12} &= \begin{vmatrix} 2 & 6 \\ 6 & 3 \end{vmatrix}\\ &= (2)(3) - (6)(6) = -30,\\ A_{12} &= (-1)^{1+2}(-30) = 30,\\ M_{13} &= \begin{vmatrix} 2 & 4 \\ 6 & 2 \end{vmatrix}\\ &= (2)(2) - (4)(6) = -20,\\ A_{13} &= (-1)^{1+3}(-20) = -20. \end{aligned}$$

   Row 2 (entries $2, 4, 6$⁠):

   $$\begin{aligned} M_{21} &= \begin{vmatrix} 3 & 2 \\ 2 & 3 \end{vmatrix}\\ &= (3)(3) - (2)(2) = 5,\\ A_{21} &= (-1)^{2+1}(5) = -5,\\ M_{22} &= \begin{vmatrix} 4 & 2 \\ 6 & 3 \end{vmatrix}\\ &= (4)(3) - (2)(6) = 0,\\ A_{22} &= (-1)^{2+2}(0) = 0,\\ M_{23} &= \begin{vmatrix} 4 & 3 \\ 6 & 2 \end{vmatrix}\\ &= (4)(2) - (3)(6) = -10,\\ A_{23} &= (-1)^{2+3}(-10) = 10. \end{aligned}$$

   Row 3 (entries $6, 2, 3$⁠):

   $$\begin{aligned} M_{31} &= \begin{vmatrix} 3 & 2 \\ 4 & 6 \end{vmatrix}\\ &= (3)(6) - (2)(4) = 10,\\ A_{31} &= (-1)^{3+1}(10) = 10,\\ M_{32} &= \begin{vmatrix} 4 & 2 \\ 2 & 6 \end{vmatrix}\\ &= (4)(6) - (2)(2) = 20,\\ A_{32} &= (-1)^{3+2}(20) = -20,\\ M_{33} &= \begin{vmatrix} 4 & 3 \\ 2 & 4 \end{vmatrix}\\ &= (4)(4) - (3)(2) = 10,\\ A_{33} &= (-1)^{3+3}(10) = 10. \end{aligned}$$
   Each minor is the $2\times2$ determinant left after deleting the row and column of that entry; the sign $(-1)^{i+j}$ is then applied. Two cofactors are 0 ($A_{11}$ and $A_{22}$⁠), which is allowed and makes later sums shorter.
   Definitions 1 and 2, Section 4.4. Lesson: minors and cofactors

5. Build $\operatorname{adj} A$ and $A^{-1}$⁠, and check

   Matrix of cofactors:

   $$\begin{bmatrix} 0 & 30 & -20 \\ -5 & 0 & 10 \\ 10 & -20 & 10 \end{bmatrix}$$

   Transpose, then divide by $|A| = 50$⁠:

   $$\operatorname{adj} A = \begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix},$$ $$A^{-1} = \frac{1}{50}\begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix}.$$

   Check $A(\operatorname{adj} A)$⁠, row of $A$ times column of $\operatorname{adj} A$⁠:

   $$\begin{aligned} (1,1)&:\ 4(0) + 3(30) + 2(-20)\\ &= 0 + 90 - 40 = 50,\\ (1,2)&:\ 4(-5) + 3(0) + 2(10)\\ &= -20 + 0 + 20 = 0,\\ (1,3)&:\ 4(10) + 3(-20) + 2(10)\\ &= 40 - 60 + 20 = 0,\\ (2,1)&:\ 2(0) + 4(30) + 6(-20)\\ &= 0 + 120 - 120 = 0,\\ (2,2)&:\ 2(-5) + 4(0) + 6(10)\\ &= -10 + 0 + 60 = 50,\\ (2,3)&:\ 2(10) + 4(-20) + 6(10)\\ &= 20 - 80 + 60 = 0,\\ (3,1)&:\ 6(0) + 2(30) + 3(-20)\\ &= 0 + 60 - 60 = 0,\\ (3,2)&:\ 6(-5) + 2(0) + 3(10)\\ &= -30 + 0 + 30 = 0,\\ (3,3)&:\ 6(10) + 2(-20) + 3(10)\\ &= 60 - 40 + 30 = 50. \end{aligned}$$ $$A(\operatorname{adj} A) = \begin{bmatrix} 50 & 0 & 0 \\ 0 & 50 & 0 \\ 0 & 0 & 50 \end{bmatrix} = 50\,I,$$

   so $AA^{-1} = \frac{1}{50}(50\,I) = I$⁠.

   The adjoint is the transpose of the cofactor matrix (Definition 3). The product check is Theorem 1: $|A|$ on the diagonal, 0 elsewhere. It confirms all nine cofactors before we use them on the money.
   Definition 3, Section 4.5.1; Theorems 1 and 4, Section 4.5. Lesson: the adjoint

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

   $A^{-1}$ is $3\times3$⁠, $B$ is $3\times1$⁠, so the product is $3\times1$⁠. First $(\operatorname{adj} A)B$⁠:

   $$\begin{aligned} \text{row 1}&:\ 0(60) + (-5)(90) + 10(70)\\ &= 0 - 450 + 700 = 250,\\ \text{row 2}&:\ 30(60) + 0(90) + (-20)(70)\\ &= 1800 + 0 - 1400 = 400,\\ \text{row 3}&:\ (-20)(60) + 10(90) + 10(70)\\ &= -1200 + 900 + 700 = 400. \end{aligned}$$ $$X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{50}\begin{bmatrix} 250 \\ 400 \\ 400 \end{bmatrix} = \begin{bmatrix} 5 \\ 8 \\ 8 \end{bmatrix}.$$

   So $x = 5$⁠, $y = 8$⁠, $z = 8$.

   Multiplying $AX = B$ on the left by $A^{-1}$ gives $X = A^{-1}B$⁠, so these are the only prices that fit. Units: $A^{-1}$ has units of "1 per kg" and $B$ is in ₹, so $X$ is in ₹ per kg, as a price should be.
   Section 4.6.1, Case I. Lesson: inverse of a matrix

7. Check against every bill, and say what it means

   $$\begin{aligned} \text{Bill 1}&:\ 4(5) + 3(8) + 2(8)\\ &= 20 + 24 + 16 = 60,\\ \text{Bill 2}&:\ 2(5) + 4(8) + 6(8)\\ &= 10 + 32 + 48 = 90,\\ \text{Bill 3}&:\ 6(5) + 2(8) + 3(8)\\ &= 30 + 16 + 24 = 70. \end{aligned}$$

   All three totals (₹ 60, ₹ 90, ₹ 70) are reproduced. So onion costs ₹ 5 per kg, wheat ₹ 8 per kg and rice ₹ 8 per kg.

   The check is done on the bills themselves, in rupees, which tests the whole chain from the words to the answer. All three prices are positive, as prices must be; a negative value would have signalled an error in setting up the equations (or data that no real prices could fit).

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