Question

The system of equations \( 4x + 6y = 5 \) and \( 8x + 12y = 10 \) has:

• A. Infinitely many solutions.
• B. A unique solution.
• C. Only two solutions.
• D. No solution.

Hint:

When the determinant of the coefficient matrix of a system of linear equations is 0, check the adjoint matrix product to determine if there are infinitely many solutions or no solution.

Answer 1

The Correct Option is A

The system of equations is: \[ 4x + 6y = 5 \] \[ 8x + 12y = 10 \] The coefficient matrix is: \[ A = \begin{bmatrix} 4 & 6
8 & 12 \end{bmatrix} \] Now, compute the determinant of matrix \( A \): \[ |A| = (4)(12) - (6)(8) = 48 - 48 = 0 \] Since the determinant is 0, the system has either infinitely many solutions or no solution. To determine which, calculate \( (adjA)B \): \[ (adjA)B = \begin{bmatrix} 12 & -6
-8 & 4 \end{bmatrix} \cdot \begin{bmatrix} 5
10 \end{bmatrix} = \begin{bmatrix} 60 - 60
-40 + 40 \end{bmatrix} = \begin{bmatrix} 0
0 \end{bmatrix} \] Since the result is the zero vector, the system has infinitely many solutions.