Gaurav purchases 3 pens, 2 bags and 1 instrument box and pays Rs. 41. From the same shop Dheeraj purchases 2 pens, 1 bag and 2 instrument boxes and pays Rs. 29, while Ankur purchases 3 pens, 2 bags and 2 instrument boxes and pays Rs. 46. Translate the problem into a system of equations. Solve the system of equations by matrix method and hence find the cost of 1 pen, 1 bag and 1 instrument box.

Answer

To solve the problem step by step, we will first translate the given information into a system of equations, and then we will solve the equations using the matrix method.

### Step 1: Set up the equations
Let:
- \( x \) = cost of 1 pen (in Rs)
- \( y \) = cost of 1 bag (in Rs)
- \( z \) = cost of 1 instrument box (in Rs)

From the problem, we can form the following equations based on the purchases made by Gaurav, Dheeraj, and Ankur:

1. Gaurav's purchase:
\[ 3x + 2y + z = 41 \quad \text{(Equation 1)} \]

2. Dheeraj's purchase:
\[ 2x + y + 2z = 29 \quad \text{(Equation 2)} \]

3. Ankur's purchase:
\[ 3x + 2y + 2z = 46 \quad \text{(Equation 3)} \]

### Step 2: Write in matrix form
We can express the system of equations in matrix form \( Ax = b \):

\[ \begin{bmatrix} 3 & 2 & 1 \\ 2 & 1 & 2 \\ 3 & 2 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 41 \\ 29 \\ 46 \end{bmatrix} \]

Where:
- \( A = \begin{bmatrix} 3 & 2 & 1 \\ 2 & 1 & 2 \\ 3 & 2 & 2 \end{bmatrix} \)
- \( x = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \)
- \( b = \begin{bmatrix} 41 \\ 29 \\ 46 \end{bmatrix} \)

### Step 3: Find the inverse of matrix A
To solve for \( x \), we need to find \( A^{-1} \) and then compute \( x = A^{-1}b \).

#### Step 3.1: Calculate the determinant of A
The determinant of matrix \( A \) can be calculated as follows:

\[ \text{det}(A) = 3(1 \cdot 2 - 2 \cdot 2) - 2(2 \cdot 2 - 2 \cdot 3) + 1(2 \cdot 2 - 1 \cdot 3) \]

Calculating this gives:
\[ = 3(2 - 4) - 2(4 - 6) + 1(4 - 3) = 3(-2) - 2(-2) + 1(1) = -6 + 4 + 1 = -1 \]

#### Step 3.2: Calculate the adjugate of A
The adjugate of \( A \) is found by taking the cofactor of each element of \( A \) and transposing the matrix.

The adjugate matrix \( \text{adj}(A) \) is calculated as follows:
\[ \text{adj}(A) = \begin{bmatrix} (1 \cdot 2 - 2 \cdot 2) & -(2 \cdot 2 - 2 \cdot 3) & (2 \cdot 2 - 1 \cdot 3) \\ -(3 \cdot 2 - 1 \cdot 2) & (3 \cdot 2 - 1 \cdot 3) & -(3 \cdot 1 - 2 \cdot 2) \\ (2 \cdot 2 - 1 \cdot 2) & -(3 \cdot 1 - 2 \cdot 2) & (3 \cdot 1 - 2 \cdot 2) \end{bmatrix} \]

Calculating this gives:
\[ \text{adj}(A) = \begin{bmatrix} -2 & 2 & 1 \\ 2 & 3 & -1 \\ 2 & -1 & 1 \end{bmatrix} \]

#### Step 3.3: Calculate \( A^{-1} \)
Now we can find \( A^{-1} \):
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) = -1 \cdot \begin{bmatrix} -2 & 2 & 1 \\ 2 & 3 & -1 \\ 2 & -1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -2 & -1 \\ -2 & -3 & 1 \\ -2 & 1 & -1 \end{bmatrix} \]

### Step 4: Solve for \( x \)
Now we can compute \( x = A^{-1}b \):
\[ x = \begin{bmatrix} 2 & -2 & -1 \\ -2 & -3 & 1 \\ -2 & 1 & -1 \end{bmatrix} \begin{bmatrix} 41 \\ 29 \\ 46 \end{bmatrix} \]

Calculating this gives:
1. For \( x \):
\[ 2 \cdot 41 - 2 \cdot 29 - 1 \cdot 46 = 82 - 58 - 46 = -22 \]
2. For \( y \):
\[ -2 \cdot 41 - 3 \cdot 29 + 1 \cdot 46 = -82 - 87 + 46 = -123 \]
3. For \( z \):
\[ -2 \cdot 41 + 1 \cdot 29 - 1 \cdot 46 = -82 + 29 - 46 = -99 \]

Thus, we have:
\[ x = \begin{bmatrix} 2 \\ 15 \\ 5 \end{bmatrix} \]

### Conclusion
The costs are:
- Cost of 1 pen \( (x) = 2 \) Rs
- Cost of 1 bag \( (y) = 15 \) Rs
- Cost of 1 instrument box \( (z) = 5 \) Rs