Q.The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹ 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹ 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is ₹ 70. Find cost of each item per kg by matrix method.

The key idea is to represent the given word problem as a system of linear equations and solve it using the matrix method (inverting the coefficient matrix).

Let the cost per kg of onion, wheat, and rice be $x$, $y$, and $z$ respectively. The equations are:

In matrix form, $AX = B$, where:

The solution is $X = A^{-1}B$. First, find $\det(A)$:

Since $\det(A) \neq 0$, the inverse exists. Compute the adjoint (matrix of cofactors transposed) and then $A^{-1}$:

Now multiply $A^{-1}B$: …

The problem gives us three different combinations of purchases and their total costs. Each combination is a linear equation where the unknowns are the per‑kg prices of onion ($x$), wheat ($y$), and rice ($z$).

The matrix method is perfect here because it turns a messy system of equations into a clean multiplication problem: $A \mathbf{x} = \mathbf{b}$, where $A$ holds the quantities, $\mathbf{x}$ holds the unknown prices, and $\mathbf{b}$ holds the total costs. If $A$ is invertible, we can find $\mathbf{x}$ directly as $\mathbf{x} = A^{-1} \mathbf{b}$.

Let’s set it up and solve.

1. Write the equations from the problem statement

   Let the cost per kg of onion be $x$, wheat be $y$, and rice be $z$ (all in ₹).

   From the first purchase:

   $4x + 3y + 2z = 60$

   From the second purchase:

   $2x + 4y + 6z = 90$

   From the third purchase:

   $6x + 2y + 3z = 70$

2. Write the system in matrix form

So $A = \begin{bmatrix} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{bmatrix}$, $\mathbf{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, $\mathbf{b} = \begin{bmatrix} 60 \\ 90 \\ 70 \end{bmatrix}$.

3. Find the determinant of $A$ — we need to check if $A$ is invertible.

Compute each minor:

• $\begin{vmatrix} 4 & 6 \\ 2 & 3 \end{vmatrix} = (4)(3) - (6)(2) = 12 - 12 = 0$
• $\begin{vmatrix} 2 & 6 \\ 6 & 3 \end{vmatrix} = (2)(3) - (6)(6) = 6 - 36 = -30$
• $\begin{vmatrix} 2 & 4 \\ 6 & 2 \end{vmatrix} = (2)(2) - (4)(6) = 4 - 24 = -20$

So

Since $\det(A) = 50 \neq 0$, $A$ is invertible and the system has a unique solution.

4. Find the adjoint of $A$ (the transpose of the cofactor matrix).

   First, compute all cofactors $C_{ij} = (-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor.

   • $C_{11} = + \begin{vmatrix} 4 & 6 \\ 2 & 3 \end{vmatrix} = 0$

   • $C_{12} = - \begin{vmatrix} 2 & 6 \\ 6 & 3 \end{vmatrix} = -(-30) = 30$

   • $C_{13} = + \begin{vmatrix} 2 & 4 \\ 6 & 2 \end{vmatrix} = -20$

   • $C_{21} = - \begin{vmatrix} 3 & 2 \\ 2 & 3 \end{vmatrix} = -[(3)(3) - (2)(2)] = -(9 - 4) = -5$

   • $C_{22} = + \begin{vmatrix} 4 & 2 \\ 6 & 3 \end{vmatrix} = (4)(3) - (2)(6) = 12 - 12 = 0$

   • $C_{23} = - \begin{vmatrix} 4 & 3 \\ 6 & 2 \end{vmatrix} = -[(4)(2) - (3)(6)] = -(8 - 18) = 10$

   • $C_{31} = + \begin{vmatrix} 3 & 2 \\ 4 & 6 \end{vmatrix} = (3)(6) - (2)(4) = 18 - 8 = 10$

   • $C_{32} = - \begin{vmatrix} 4 & 2 \\ 2 & 6 \end{vmatrix} = -[(4)(6) - (2)(2)] = -(24 - 4) = -20$

   • $C_{33} = + \begin{vmatrix} 4 & 3 \\ 2 & 4 \end{vmatrix} = (4)(4) - (3)(2) = 16 - 6 = 10$

   The cofactor matrix is …

Method: Translating a Word Problem into $AX=B$ and Solving

Word problems that ask for the "cost of each item" or similar are just linear systems in disguise — the real skill is converting the sentences into equations correctly before any matrix work begins.

Steps

Step 1: Name the unknowns clearly

Assign a variable to each quantity you're asked to find (e.g. cost per kg of each item), and write down what each variable represents in words — this prevents mixing up which number belongs to which unknown later.

Step 2: Convert each stated scenario into one linear equation

Read each sentence carefully and match the quantities given (e.g. "$4$ kg of X, $3$ kg of Y") to coefficients of your variables, and the stated total to the constant on the right-hand side. One realistic purchase/scenario in the problem = one equation.

Step 3: Stack the equations into $AX=B$ …

Common Mistakes

Mistake 1: Misreading which quantity goes with which unknown

Why it's wrong: swapping, say, the coefficient meant for wheat with the one meant for rice in even one equation produces a completely different (and wrong) system, even though the arithmetic that follows is done correctly. Correct approach: label each column of $A$ with the variable it belongs to before filling in numbers, and re-read each sentence once after setting up its equation.

Mistake 2: Treating an implicit assumption as if it were a fourth data point …