Question

Find the cost (per kg) of each fertilizer A, B and C that a farmer needs to buy, such that 1 kg each of fertilizer A and C added to 2 kg of B costs him ₹ 400. Also, the cost of each kg of fertilizer B and C added together is equal to the cost of 1 kg of fertilizer A. However, the cost of 3 kg of fertilizer B added to ₹ 200 is the same as the cost of 1 kg of fertilizer A and C together. Use the matrix method to find the solution.

Hint:

To verify your solution quickly in a competitive exam, plug the calculated variables back into the simplest equation: $x - y - z = 150 - 50 - 100 = 0$. Since it satisfies the condition, your matrix inverse steps are guaranteed to be correct!

Answer 1

Concept: To solve a system of three linear equations using the matrix inversion method, we express the system in the matrix form: \[ AX = B \implies X = A^{-1}B \] where $A$ is the $3 \times 3$ coefficient matrix, $X$ is the column matrix of unknown variables, and $B$ is the column matrix of constants. The inverse matrix is computed using the formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$, where $|A|$ is the matrix determinant and $\text{adj}(A)$ is the transpose of the cofactor matrix.

Step 1: Translating the word problem into a system of linear equations.
Let the cost per kg of fertilizer $A$, $B$, and $C$ be represented by the variables $x$, $y$, and $z$ respectively (in ₹).
• From the first condition: "1 kg each of fertilizer A and C added to 2 kg of B costs ₹ 400": \[ 1x + 2y + 1z = 400 \implies x + 2y + z = 400 \]
• From the second condition: "the cost of each kg of fertilizer B and C added together is equal to the cost of 1 kg of fertilizer A": \[ y + z = x \implies x - y - z = 0 \]
• From the third condition: "the cost of 3 kg of fertilizer B added to ₹ 200 is the same as the cost of 1 kg of fertilizer A and C together": \[ 3y + 200 = x + z \implies x - 3y + z = 200 \] Our system of linear equations is: \[ \begin{aligned} x + 2y + z &= 400 x - y - z &= 0 x - 3y + z &= 200 \end{aligned} \]

Step 2: Writing the system in matrix form $AX = B$ and evaluating determinant $|A|$.
Define matrices $A$, $X$, and $B$: \[ A = \begin{bmatrix} 1 & 2 & 1 1 & -1 & -1 1 & -3 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x y z \end{bmatrix}, \quad B = \begin{bmatrix} 400 0 200 \end{bmatrix} \] Let us compute the determinant of matrix $A$ by expanding along the first row: \[ |A| = 1\begin{bmatrix} -1 & -1 -3 & 1 \end{bmatrix} - 2\begin{bmatrix} 1 & -1 1 & 1 \end{bmatrix} + 1\begin{bmatrix} 1 & -1 1 & -3 \end{bmatrix} \] \[ |A| = 1((-1)(1) - (-1)(-3)) - 2((1)(1) - (-1)(1)) + 1((1)(-3) - (-1)(1)) \] \[ |A| = 1(-1 - 3) - 2(1 + 1) + 1(-3 + 1) = 1(-4) - 2(2) + 1(-2) \] \[ |A| = -4 - 4 - 2 = -10 \] Since $|A| = -10 \neq 0$, the matrix inverse $A^{-1}$ exists, and the system has a unique solution.

Step 3: Finding the cofactor matrix and the adjugate matrix $\text{adj}(A)$.
Let us calculate the cofactors $C_{ij}$ for all nine elements of matrix $A$: \[ \begin{aligned} C_{11} &= +(-1 - 3) = -4, & C_{12} &= -(1 - (-1)) = -2, & C_{13} &= +(-3 - (-1)) = -2 C_{21} &= -(2 - (-3)) = -5, & C_{22} &= +(1 - 1) = 0, & C_{23} &= -(-3 - 2) = 5 C_{31} &= +(-2 - (-1)) = -1, & C_{32} &= -(-1 - 1) = 2, & C_{33} &= +(-1 - 2) = -3 \end{aligned} \] Form the cofactor matrix and transpose it to obtain the adjugate matrix $\text{adj}(A)$: \[ \text{Cofactor Matrix} = \begin{bmatrix} -4 & -2 & -2 -5 & 0 & 5 -1 & 2 & -3 \end{bmatrix} \implies \text{adj}(A) = \begin{bmatrix} -4 & -5 & -1 -2 & 0 & 2 -2 & 5 & -3 \end{bmatrix} \]

Step 4: Computing the inverse $A^{-1}$ and solving for $X$.
Using the matrix equation $X = A^{-1}B = \frac{1}{|A|} \text{adj}(A)B$: \[ \begin{bmatrix} x y z \end{bmatrix} = \frac{1}{-10} \begin{bmatrix} -4 & -5 & -1 -2 & 0 & 2 -2 & 5 & -3 \end{bmatrix} \begin{bmatrix} 400 0 200 \end{bmatrix} \] Perform row-by-column multiplication: \[ \begin{bmatrix} -4(400) + (-5)(0) + (-1)(200) -2(400) + (0)(0) + (2)(200) -2(400) + (5)(0) + (-3)(200) \end{bmatrix} = \begin{bmatrix} -1600 + 0 - 200 -800 + 0 + 400 -800 + 0 - 600 \end{bmatrix} = \begin{bmatrix} -1800 -400 -1400 \end{bmatrix} \] Now divide each component by the determinant value $-10$: \[ x = \frac{-1800}{-10} = 150, \quad y = \frac{-400}{-10} = 50, \quad z = \frac{-1400}{-10} = 100 \] Thus, the individual prices are: Fertilizer A = ₹ 150/kg, Fertilizer B = ₹ 50/kg, and Fertilizer C = ₹ 100/kg.