CBSE Sample Paper 2026-27 · Section E · 4 marks
Q38: Case study: fruit baskets, prices by the matrix inverse
With chapter links Answers only
The question
To celebrate their child's birthday, a family decided to distribute fruits among children in three orphanages. The fruit vendor packed three different types of baskets, according to the quantity required at each orphanage, details are given as follows:
Basket A contains 5 kg Apples, 10 kg Oranges and 5 kg Pomegranate
Basket B contains 10 kg Apples, 10 kg Oranges and 5kg Pomegranate
Basket C contains 8 kg Apples, 15 kg Oranges and 7kg Pomegranate
Cost of basket A, B and C are respectively ₹2750, ₹3500, and ₹4100.
Based on the given information, answer the following questions:
(i) Write the matrix form of the equations involved in this situation and find the adjoint of the coefficient matrix involved. (2 marks)
(ii) Using inverse of the matrix, calculate the price of each fruit per kg. (2 marks)
Exam answer Show all explanations
(i) Let the prices of 1 kg of apples, oranges and pomegranate be ₹ x x x y y y z z z
5 x + 10 y + 5 z = 2750 10 x + 10 y + 5 z = 3500 8 x + 15 y + 7 z = 4100 \begin{aligned} 5x + 10y + 5z &= 2750 \\ 10x + 10y + 5z &= 3500 \\ 8x + 15y + 7z &= 4100 \end{aligned} 5 x + 10 y + 5 z 10 x + 10 y + 5 z 8 x + 15 y + 7 z = 2750 = 3500 = 4100 Why?
The cost of a basket is the sum of (kilograms of each fruit) × \times × 5 x + 10 y + 5 z 5x + 10y + 5z 5 x + 10 y + 5 z
In the matrix form below, each row of A holds the coefficients of one equation, in the order x , y , z x, y, z x , y , z
The matrix form is A X = B AX = B A X = B
A = [ 5 10 5 10 10 5 8 15 7 ] , A = \begin{bmatrix} 5 & 10 & 5 \\ 10 & 10 & 5 \\ 8 & 15 & 7 \end{bmatrix}, A = 5 10 8 10 10 15 5 5 7 , X = [ x y z ] , B = [ 2750 3500 4100 ] X = \begin{bmatrix} x \\ y \\ z \end{bmatrix},\ B = \begin{bmatrix} 2750 \\ 3500 \\ 4100 \end{bmatrix} X = x y z , B = 2750 3500 4100 ∣ A ∣ = 5 ( 70 − 75 ) − 10 ( 70 − 40 ) + 5 ( 150 − 80 ) = − 25 − 300 + 350 = 25 ≠ 0 \begin{aligned} |A| &= 5(70 - 75) - 10(70 - 40) \\ &\quad + 5(150 - 80) \\ &= -25 - 300 + 350 = 25 \ne 0 \end{aligned} ∣ A ∣ = 5 ( 70 − 75 ) − 10 ( 70 − 40 ) + 5 ( 150 − 80 ) = − 25 − 300 + 350 = 25 = 0 Why?
Expand along the first row, with the signs + , − , + +, -, + + , − , +
∣ A ∣ = 5 ∣ 10 5 15 7 ∣ − 10 ∣ 10 5 8 7 ∣ + 5 ∣ 10 10 8 15 ∣ \begin{aligned} \lvert A\rvert &= 5\begin{vmatrix} 10 & 5 \\ 15 & 7 \end{vmatrix} - 10\begin{vmatrix} 10 & 5 \\ 8 & 7 \end{vmatrix} \\ &\quad + 5\begin{vmatrix} 10 & 10 \\ 8 & 15 \end{vmatrix} \end{aligned} ∣ A ∣ = 5 10 15 5 7 − 10 10 8 5 7 + 5 10 8 10 15 Each 2 × 2 2 \times 2 2 × 2
∣ 10 5 15 7 ∣ = ( 10 ) ( 7 ) − ( 5 ) ( 15 ) = 70 − 75 = − 5 ∣ 10 5 8 7 ∣ = ( 10 ) ( 7 ) − ( 5 ) ( 8 ) = 70 − 40 = 30 ∣ 10 10 8 15 ∣ = ( 10 ) ( 15 ) − ( 10 ) ( 8 ) = 150 − 80 = 70 \begin{aligned} \begin{vmatrix} 10 & 5 \\ 15 & 7 \end{vmatrix} &= (10)(7) - (5)(15) \\ &= 70 - 75 = -5 \\[4pt] \begin{vmatrix} 10 & 5 \\ 8 & 7 \end{vmatrix} &= (10)(7) - (5)(8) \\ &= 70 - 40 = 30 \\[4pt] \begin{vmatrix} 10 & 10 \\ 8 & 15 \end{vmatrix} &= (10)(15) - (10)(8) \\ &= 150 - 80 = 70 \end{aligned} 10 15 5 7 10 8 5 7 10 8 10 15 = ( 10 ) ( 7 ) − ( 5 ) ( 15 ) = 70 − 75 = − 5 = ( 10 ) ( 7 ) − ( 5 ) ( 8 ) = 70 − 40 = 30 = ( 10 ) ( 15 ) − ( 10 ) ( 8 ) = 150 − 80 = 70 Multiply by the entries and their signs, then add:
5 ( − 5 ) = − 25 − 10 ( 30 ) = − 300 5 ( 70 ) = + 350 − 25 − 300 = − 325 − 325 + 350 = 25 \begin{aligned} 5(-5) &= -25 \\ -10(30) &= -300 \\ 5(70) &= +350 \\[4pt] -25 - 300 &= -325 \\ -325 + 350 &= 25 \end{aligned} 5 ( − 5 ) − 10 ( 30 ) 5 ( 70 ) − 25 − 300 − 325 + 350 = − 25 = − 300 = + 350 = − 325 = 25 So the system is consistent and has a unique solution. The cofactors of A are
Why?
∣ A ∣ ≠ 0 |A| \ne 0 ∣ A ∣ = 0 A − 1 A^{-1} A − 1 X = A − 1 B X = A^{-1}B X = A − 1 B
A 11 = 70 − 75 = − 5 A 12 = − ( 70 − 40 ) = − 30 A 13 = 150 − 80 = 70 \begin{aligned} A_{11} &= 70 - 75 = -5 \\ A_{12} &= -(70 - 40) = -30 \\ A_{13} &= 150 - 80 = 70 \end{aligned} A 11 A 12 A 13 = 70 − 75 = − 5 = − ( 70 − 40 ) = − 30 = 150 − 80 = 70 A 21 = − ( 70 − 75 ) = 5 A 22 = 35 − 40 = − 5 A 23 = − ( 75 − 80 ) = 5 \begin{aligned} A_{21} &= -(70 - 75) = 5 \\ A_{22} &= 35 - 40 = -5 \\ A_{23} &= -(75 - 80) = 5 \end{aligned} A 21 A 22 A 23 = − ( 70 − 75 ) = 5 = 35 − 40 = − 5 = − ( 75 − 80 ) = 5 A 31 = 50 − 50 = 0 A 32 = − ( 25 − 50 ) = 25 A 33 = 50 − 100 = − 50 \begin{aligned} A_{31} &= 50 - 50 = 0 \\ A_{32} &= -(25 - 50) = 25 \\ A_{33} &= 50 - 100 = -50 \end{aligned} A 31 A 32 A 33 = 50 − 50 = 0 = − ( 25 − 50 ) = 25 = 50 − 100 = − 50 Why?
The cofactor of the entry in row i i i j j j A i j = ( − 1 ) i + j M i j A_{ij} = (-1)^{i + j}M_{ij} A ij = ( − 1 ) i + j M ij M i j M_{ij} M ij − - − i + j i + j i + j
For example, A 23 A_{23} A 23
∣ 5 10 8 15 ∣ = 75 − 80 = − 5 \begin{vmatrix} 5 & 10 \\ 8 & 15 \end{vmatrix} = 75 - 80 = -5 5 8 10 15 = 75 − 80 = − 5 and 2 + 3 2 + 3 2 + 3 A 23 = − ( − 5 ) = 5 A_{23} = -(-5) = 5 A 23 = − ( − 5 ) = 5
The signs ( − 1 ) i + j (-1)^{i + j} ( − 1 ) i + j
[ + − + − + − + − + ] on A = [ 5 10 5 10 10 5 8 15 7 ] \begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}\ \text{ on }\ A = \begin{bmatrix} 5 & 10 & 5 \\ 10 & 10 & 5 \\ 8 & 15 & 7 \end{bmatrix} + − + − + − + − + on A = 5 10 8 10 10 15 5 5 7 Row 1 (cover row 1 and one column each time):
A 11 = + ∣ 10 5 15 7 ∣ = + ( ( 10 ) ( 7 ) − ( 5 ) ( 15 ) ) = + ( 70 − 75 ) = − 5 A 12 = − ∣ 10 5 8 7 ∣ = − ( ( 10 ) ( 7 ) − ( 5 ) ( 8 ) ) = − ( 70 − 40 ) = − 30 A 13 = + ∣ 10 10 8 15 ∣ = + ( ( 10 ) ( 15 ) − ( 10 ) ( 8 ) ) = + ( 150 − 80 ) = 70 \begin{aligned} A_{11} &= +\begin{vmatrix} 10 & 5 \\ 15 & 7 \end{vmatrix} \\ &= +\big((10)(7) - (5)(15)\big) \\ &= +(70 - 75) = -5 \\[4pt] A_{12} &= -\begin{vmatrix} 10 & 5 \\ 8 & 7 \end{vmatrix} \\ &= -\big((10)(7) - (5)(8)\big) \\ &= -(70 - 40) = -30 \\[4pt] A_{13} &= +\begin{vmatrix} 10 & 10 \\ 8 & 15 \end{vmatrix} \\ &= +\big((10)(15) - (10)(8)\big) \\ &= +(150 - 80) = 70 \end{aligned} A 11 A 12 A 13 = + 10 15 5 7 = + ( ( 10 ) ( 7 ) − ( 5 ) ( 15 ) ) = + ( 70 − 75 ) = − 5 = − 10 8 5 7 = − ( ( 10 ) ( 7 ) − ( 5 ) ( 8 ) ) = − ( 70 − 40 ) = − 30 = + 10 8 10 15 = + ( ( 10 ) ( 15 ) − ( 10 ) ( 8 ) ) = + ( 150 − 80 ) = 70 Row 2 (cover row 2):
A 21 = − ∣ 10 5 15 7 ∣ = − ( ( 10 ) ( 7 ) − ( 5 ) ( 15 ) ) = − ( 70 − 75 ) = − ( − 5 ) = 5 A 22 = + ∣ 5 5 8 7 ∣ = + ( ( 5 ) ( 7 ) − ( 5 ) ( 8 ) ) = + ( 35 − 40 ) = − 5 A 23 = − ∣ 5 10 8 15 ∣ = − ( ( 5 ) ( 15 ) − ( 10 ) ( 8 ) ) = − ( 75 − 80 ) = − ( − 5 ) = 5 \begin{aligned} A_{21} &= -\begin{vmatrix} 10 & 5 \\ 15 & 7 \end{vmatrix} \\ &= -\big((10)(7) - (5)(15)\big) \\ &= -(70 - 75) = -(-5) = 5 \\[4pt] A_{22} &= +\begin{vmatrix} 5 & 5 \\ 8 & 7 \end{vmatrix} \\ &= +\big((5)(7) - (5)(8)\big) \\ &= +(35 - 40) = -5 \\[4pt] A_{23} &= -\begin{vmatrix} 5 & 10 \\ 8 & 15 \end{vmatrix} \\ &= -\big((5)(15) - (10)(8)\big) \\ &= -(75 - 80) = -(-5) = 5 \end{aligned} A 21 A 22 A 23 = − 10 15 5 7 = − ( ( 10 ) ( 7 ) − ( 5 ) ( 15 ) ) = − ( 70 − 75 ) = − ( − 5 ) = 5 = + 5 8 5 7 = + ( ( 5 ) ( 7 ) − ( 5 ) ( 8 ) ) = + ( 35 − 40 ) = − 5 = − 5 8 10 15 = − ( ( 5 ) ( 15 ) − ( 10 ) ( 8 ) ) = − ( 75 − 80 ) = − ( − 5 ) = 5 Row 3 (cover row 3):
A 31 = + ∣ 10 5 10 5 ∣ = + ( ( 10 ) ( 5 ) − ( 5 ) ( 10 ) ) = + ( 50 − 50 ) = 0 A 32 = − ∣ 5 5 10 5 ∣ = − ( ( 5 ) ( 5 ) − ( 5 ) ( 10 ) ) = − ( 25 − 50 ) = − ( − 25 ) = 25 A 33 = + ∣ 5 10 10 10 ∣ = + ( ( 5 ) ( 10 ) − ( 10 ) ( 10 ) ) = + ( 50 − 100 ) = − 50 \begin{aligned} A_{31} &= +\begin{vmatrix} 10 & 5 \\ 10 & 5 \end{vmatrix} \\ &= +\big((10)(5) - (5)(10)\big) \\ &= +(50 - 50) = 0 \\[4pt] A_{32} &= -\begin{vmatrix} 5 & 5 \\ 10 & 5 \end{vmatrix} \\ &= -\big((5)(5) - (5)(10)\big) \\ &= -(25 - 50) = -(-25) = 25 \\[4pt] A_{33} &= +\begin{vmatrix} 5 & 10 \\ 10 & 10 \end{vmatrix} \\ &= +\big((5)(10) - (10)(10)\big) \\ &= +(50 - 100) = -50 \end{aligned} A 31 A 32 A 33 = + 10 10 5 5 = + ( ( 10 ) ( 5 ) − ( 5 ) ( 10 ) ) = + ( 50 − 50 ) = 0 = − 5 10 5 5 = − ( ( 5 ) ( 5 ) − ( 5 ) ( 10 ) ) = − ( 25 − 50 ) = − ( − 25 ) = 25 = + 5 10 10 10 = + ( ( 5 ) ( 10 ) − ( 10 ) ( 10 ) ) = + ( 50 − 100 ) = − 50 adj A = [ − 5 5 0 − 30 − 5 25 70 5 − 50 ] \operatorname{adj}A = \begin{bmatrix} -5 & 5 & 0 \\ -30 & -5 & 25 \\ 70 & 5 & -50 \end{bmatrix} adj A = − 5 − 30 70 5 − 5 5 0 25 − 50 Why?
adj A is the transpose of the matrix of cofactors: the cofactors of row 1 of A, − 5 , − 30 , 70 -5, -30, 70 − 5 , − 30 , 70
In the same way, the row 2 cofactors 5 , − 5 , 5 5, -5, 5 5 , − 5 , 5 0 , 25 , − 50 0, 25, -50 0 , 25 , − 50
Watch out: forgetting to transpose is the most common slip here.
(ii) A − 1 = 1 ∣ A ∣ adj A = 1 25 adj A A^{-1} = \frac{1}{|A|}\operatorname{adj}A = \frac{1}{25}\operatorname{adj}A A − 1 = ∣ A ∣ 1 adj A = 25 1 adj A
Why?
From A X = B AX = B A X = B A − 1 A^{-1} A − 1 A − 1 A X = A − 1 B A^{-1}AX = A^{-1}B A − 1 A X = A − 1 B A − 1 A = I A^{-1}A = I A − 1 A = I X = A − 1 B X = A^{-1}B X = A − 1 B
X = A − 1 B = 1 25 [ − 5 5 0 − 30 − 5 25 70 5 − 50 ] × [ 2750 3500 4100 ] \begin{aligned} X &= A^{-1}B \\ &= \frac{1}{25}\begin{bmatrix} -5 & 5 & 0 \\ -30 & -5 & 25 \\ 70 & 5 & -50 \end{bmatrix} \\ &\qquad \times \begin{bmatrix} 2750 \\ 3500 \\ 4100 \end{bmatrix} \end{aligned} X = A − 1 B = 25 1 − 5 − 30 70 5 − 5 5 0 25 − 50 × 2750 3500 4100 Here ( adj A ) B (\operatorname{adj}A)B ( adj A ) B
[ − 13750 + 17500 + 0 − 82500 − 17500 + 102500 192500 + 17500 − 205000 ] \begin{bmatrix} -13750 + 17500 + 0 \\ -82500 - 17500 + 102500 \\ 192500 + 17500 - 205000 \end{bmatrix} − 13750 + 17500 + 0 − 82500 − 17500 + 102500 192500 + 17500 − 205000 = [ 3750 2500 5000 ] = \begin{bmatrix} 3750 \\ 2500 \\ 5000 \end{bmatrix} = 3750 2500 5000 ∴ X = 1 25 [ 3750 2500 5000 ] = [ 150 100 200 ] \therefore\ X = \frac{1}{25}\begin{bmatrix} 3750 \\ 2500 \\ 5000 \end{bmatrix} = \begin{bmatrix} 150 \\ 100 \\ 200 \end{bmatrix} ∴ X = 25 1 3750 2500 5000 = 150 100 200 Why?
Each entry is a row of adj A times the column B. For example, row 1:
( − 5 ) ( 2750 ) + 5 ( 3500 ) + 0 ( 4100 ) = − 13750 + 17500 = 3750 \begin{aligned} &(-5)(2750) + 5(3500) \\ &\quad + 0(4100) \\ &= -13750 + 17500 \\ &= 3750 \end{aligned} ( − 5 ) ( 2750 ) + 5 ( 3500 ) + 0 ( 4100 ) = − 13750 + 17500 = 3750 Row 2, products first, then add two at a time:
( − 30 ) ( 2750 ) = − 82500 ( − 5 ) ( 3500 ) = − 17500 ( 25 ) ( 4100 ) = 102500 − 82500 − 17500 = − 100000 − 100000 + 102500 = 2500 \begin{aligned} &(-30)(2750) = -82500 \\ &(-5)(3500) = -17500 \\ &(25)(4100) = 102500 \\[4pt] &-82500 - 17500 = -100000 \\ &-100000 + 102500 = 2500 \end{aligned} ( − 30 ) ( 2750 ) = − 82500 ( − 5 ) ( 3500 ) = − 17500 ( 25 ) ( 4100 ) = 102500 − 82500 − 17500 = − 100000 − 100000 + 102500 = 2500 Row 3:
( 70 ) ( 2750 ) = 192500 ( 5 ) ( 3500 ) = 17500 ( − 50 ) ( 4100 ) = − 205000 192500 + 17500 = 210000 210000 − 205000 = 5000 \begin{aligned} &(70)(2750) = 192500 \\ &(5)(3500) = 17500 \\ &(-50)(4100) = -205000 \\[4pt] &192500 + 17500 = 210000 \\ &210000 - 205000 = 5000 \end{aligned} ( 70 ) ( 2750 ) = 192500 ( 5 ) ( 3500 ) = 17500 ( − 50 ) ( 4100 ) = − 205000 192500 + 17500 = 210000 210000 − 205000 = 5000 Then divide each entry by 25: 3750 ÷ 25 = 150 3750 \div 25 = 150 3750 ÷ 25 = 150 2500 ÷ 25 = 100 2500 \div 25 = 100 2500 ÷ 25 = 100 5000 ÷ 25 = 200 5000 \div 25 = 200 5000 ÷ 25 = 200
3750 25 = 150 ( 25 × 150 = 3750 ) 2500 25 = 100 ( 25 × 100 = 2500 ) 5000 25 = 200 ( 25 × 200 = 5000 ) \begin{aligned} \frac{3750}{25} &= 150 \quad (25 \times 150 = 3750) \\ \frac{2500}{25} &= 100 \quad (25 \times 100 = 2500) \\ \frac{5000}{25} &= 200 \quad (25 \times 200 = 5000) \end{aligned} 25 3750 25 2500 25 5000 = 150 ( 25 × 150 = 3750 ) = 100 ( 25 × 100 = 2500 ) = 200 ( 25 × 200 = 5000 ) ⇒ x = 150 \Rightarrow x = 150 ⇒ x = 150 y = 100 y = 100 y = 100 z = 200 z = 200 z = 200
Why?
Check in the first equation:
5 ( 150 ) + 10 ( 100 ) + 5 ( 200 ) = 750 + 1000 + 1000 = 2750 \begin{aligned} &5(150) + 10(100) + 5(200) \\ &= 750 + 1000 + 1000 \\ &= 2750 \end{aligned} 5 ( 150 ) + 10 ( 100 ) + 5 ( 200 ) = 750 + 1000 + 1000 = 2750 It matches the cost of basket A.
The other two baskets check out too:
10 ( 150 ) + 10 ( 100 ) + 5 ( 200 ) = 1500 + 1000 + 1000 = 3500 8 ( 150 ) + 15 ( 100 ) + 7 ( 200 ) = 1200 + 1500 + 1400 = 4100 \begin{aligned} &10(150) + 10(100) + 5(200) \\ &= 1500 + 1000 + 1000 = 3500 \\[4pt] &8(150) + 15(100) + 7(200) \\ &= 1200 + 1500 + 1400 = 4100 \end{aligned} 10 ( 150 ) + 10 ( 100 ) + 5 ( 200 ) = 1500 + 1000 + 1000 = 3500 8 ( 150 ) + 15 ( 100 ) + 7 ( 200 ) = 1200 + 1500 + 1400 = 4100 ∴ \therefore ∴
Where this comes from Chapter 4, Determinants : System of equations by the matrix inverse.
Question type: Solve equations by the matrix method
How to spot it: click to see Type 9. Solve equations by the matrix method
Spot it: "solve, using the matrix method", with two or three equations. A word problem (prices, three numbers) is the same: write the equations first.
Here: (i) Each basket gives one equation in the prices x , y , z x, y, z x , y , z A X = B AX = B A X = B ∣ A ∣ = 25 |A| = 25 ∣ A ∣ = 25 adj A \operatorname{adj}A adj A X = 1 25 ( adj A ) B X = \frac{1}{25}(\operatorname{adj}A)B X = 25 1 ( adj A ) B
Quick lesson · Worked NCERT solutions · Interactive lesson
Also uses Chapter 3, Matrices .
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