Question
Determine the number of solutions of this system of linear equations: −x+y=3 −2x+2y=3 This question requires you to show your work. (1 point) Responses No Solutions No Solutions One Solution One Solution Infinitely Many Solutions
Answers
There are no human answers yet.
Answered by
GPT 3.5
To determine the number of solutions of this system of linear equations, we can rewrite the equations in the form Ax = B, where A is the matrix of coefficients and B is the matrix of constants.
The system of equations can be represented as:
A = \begin{bmatrix} -1 & 1 \\ -2 & 2 \end{bmatrix}
B = \begin{bmatrix} 3 \\ 3 \end{bmatrix}
To find the determinant of matrix A, we calculate:
det(A) = (-1)(2) - (-2)(1) = -2 + 2 = 0
Since the determinant of matrix A is 0, the system of equations has either no solution, or infinitely many solutions. To determine which is the case, we can reduce the matrix A to its reduced row-echelon form:
\begin{bmatrix} -1 & 1 & 3 \\ -2 & 2 & 3 \end{bmatrix}
R2 = R2 + 2R1
\begin{bmatrix} -1 & 1 & 3 \\ 0 & 4 & 9 \end{bmatrix}
R2 = (1/4)R2
\begin{bmatrix} -1 & 1 & 3 \\ 0 & 1 & 9/4 \end{bmatrix}
R1 = R1 + R2
\begin{bmatrix} -1 & 0 & 3 + 1 \\ 0 & 1 & 9/4 \end{bmatrix}
\begin{bmatrix} -1 & 0 & 4 \\ 0 & 1 & 9/4 \end{bmatrix}
The system of equations has been reduced to:
-x = 4
y = 9/4
This implies that the system of equations is inconsistent, as the equations cannot be simultaneously true. Therefore, the system of equations has no solution.
The system of equations can be represented as:
A = \begin{bmatrix} -1 & 1 \\ -2 & 2 \end{bmatrix}
B = \begin{bmatrix} 3 \\ 3 \end{bmatrix}
To find the determinant of matrix A, we calculate:
det(A) = (-1)(2) - (-2)(1) = -2 + 2 = 0
Since the determinant of matrix A is 0, the system of equations has either no solution, or infinitely many solutions. To determine which is the case, we can reduce the matrix A to its reduced row-echelon form:
\begin{bmatrix} -1 & 1 & 3 \\ -2 & 2 & 3 \end{bmatrix}
R2 = R2 + 2R1
\begin{bmatrix} -1 & 1 & 3 \\ 0 & 4 & 9 \end{bmatrix}
R2 = (1/4)R2
\begin{bmatrix} -1 & 1 & 3 \\ 0 & 1 & 9/4 \end{bmatrix}
R1 = R1 + R2
\begin{bmatrix} -1 & 0 & 3 + 1 \\ 0 & 1 & 9/4 \end{bmatrix}
\begin{bmatrix} -1 & 0 & 4 \\ 0 & 1 & 9/4 \end{bmatrix}
The system of equations has been reduced to:
-x = 4
y = 9/4
This implies that the system of equations is inconsistent, as the equations cannot be simultaneously true. Therefore, the system of equations has no solution.
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.