Question

Let \( A = [a_{ij}] \) be a \( 3 \times 3 \) matrix with positive integers as its elements. The elements of \( A \) are such that the sum of all the elements of each row is equal to 6, and \( a_{22} = 2 \). 
 

• A. \( 6 \)
• B. \( 18 \)
• C. \( 3 \)
• D. \( 12 \)

Hint:

For determinant calculations, always break the matrix down into its row operations and apply the determinant formula step by step.

Answer 1

The Correct Option is D

We are given a \( 3 \times 3 \) matrix \( A = [a_{ij}] \), where each row sum is equal to 6 and \( a_{22} = 2 \). 

Step 1: Construct the Matrix Using Given Conditions 
From the row sum condition: \[ a_{11} + a_{12} + a_{13} = 6 \] \[ a_{21} + a_{22} + a_{23} = 6 \] \[ a_{31} + a_{32} + a_{33} = 6 \] Using the given condition that for \( i<3 \), we have: \[ a_{ii} = a_{ij} + a_{in}, \quad j = i + 1 \] For \( i = 1 \): \[ a_{11} = a_{12} + a_{13} \] For \( i = 2 \): \[ a_{22} = a_{23} + a_{21} \] Since \( a_{22} = 2 \), we substitute: \[ 2 = a_{23} + a_{21} \] For \( i = 3 \), we use: \[ a_{33} = a_{31} + a_{32}, \quad j = 4 - i \] 

Step 2: Solve for Matrix Elements 
From row sums: \[ a_{11} + a_{12} + a_{13} = 6 \] Using \( a_{11} = a_{12} + a_{13} \), we get: \[ (a_{12} + a_{13}) + a_{12} + a_{13} = 6 \] \[ 2(a_{12} + a_{13}) = 6 \] \[ a_{12} + a_{13} = 3, \quad a_{11} = 3 \] Similarly, using \( a_{22} = 2 \): \[ a_{21} + a_{23} = 2, \quad a_{21} + 2 + a_{23} = 6 \] \[ a_{21} + a_{23} = 4 \] Solving, we find: \[ a_{21} = 1, \quad a_{23} = 1 \] For \( i = 3 \): \[ a_{31} + a_{32} + a_{33} = 6 \] Using \( a_{33} = a_{31} + a_{32} \), we substitute: \[ (a_{31} + a_{32}) + a_{31} + a_{32} = 6 \] \[ 2(a_{31} + a_{32}) = 6 \] \[ a_{31} + a_{32} = 3, \quad a_{33} = 3 \] 

Step 3: Compute Determinant 
The determinant of a \( 3 \times 3 \) matrix is given by: \[ |A| = a_{11} (a_{22} a_{33} - a_{23} a_{32}) - a_{12} (a_{21} a_{33} - a_{23} a_{31}) + a_{13} (a_{21} a_{32} - a_{22} a_{31}) \] \[ = 3(2 \cdot 3 - 1 \cdot 1) - 1(1 \cdot 3 - 1 \cdot 2) + 2(1 \cdot 1 - 2 \cdot 2) \] \[ = 3(6 - 1) - 1(3 - 2) + 2(1 - 4) \] \[ = 3(5) - 1(1) + 2(-3) \] \[ = 15 - 1 - 6 \] \[ = 12 \] Thus, the determinant \( |A| \) is \( 12 \).