Questions: Solve the following system of equations by using the inverse of the coefficient matrix if it exists, and by the echelon method if the inverse doesn't exist. -x-3y = 12 4x+12y = -48 Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The solution of the system is □ . (Simplify your answer. Type an ordered pair.) B. The system has infinitely many solutions. The solution is □ 1. y), where y is any real number. (Use integers or fractions for any numbers in the expression.)

Solution Steps

To solve the given system of equations, we will first attempt to find the inverse of the coefficient matrix. If the inverse exists, we will use it to find the solution. If the inverse does not exist, we will use the echelon method to determine if the system has infinitely many solutions or no solution.

1. Define the coefficient matrix \( A \) and the constant matrix \( B \).
2. Check if the determinant of \( A \) is non-zero. If it is, compute the inverse of \( A \) and use it to find the solution.
3. If the determinant is zero, use the echelon method to determine the nature of the solutions.

We are given the system of equations: \[ \begin{aligned} -x - 3y &= 12 \\ 4x + 12y &= -48 \end{aligned} \]

The coefficient matrix \( A \) and the constant matrix \( B \) are: \[ A = \begin{bmatrix} -1 & -3 \\ 4 & 12 \end{bmatrix}, \quad B = \begin{bmatrix} 12 \\ -48 \end{bmatrix} \]

The determinant of \( A \) is calculated as follows: \[ \text{det}(A) = (-1)(12) - (-3)(4) = -12 + 12 = 0 \] Since \(\text{det}(A) = 0\), the matrix \( A \) is singular, and its inverse does not exist.

Since the determinant is zero, we use the echelon method to determine the nature of the solutions. We form the augmented matrix and reduce it to row echelon form: \[ \left[\begin{array}{cc|c} -1 & -3 & 12 \\ 4 & 12 & -48 \end{array}\right] \] Perform row operations to reduce it:

1. Multiply the first row by 4 and add to the second row: \[ \left[\begin{array}{cc|c} -1 & -3 & 12 \\ 0 & 0 & 0 \end{array}\right] \]

The row echelon form indicates that the second row is all zeros, which implies that the system has infinitely many solutions. We can express \( x \) in terms of \( y \): \[ -x - 3y = 12 \implies x = -3y - 12 \] Thus, the solution can be written as: \[ (x, y) = (-3y - 12, y) \] where \( y \) is any real number.

Final Answer