To solve the system of equations using a matrix, we first write the system in the form \(AX = B\), where \(A\) is the coefficient matrix, \(X\) is the variable matrix, and \(B\) is the constant matrix.
Given the system of equations:
- \(1.5x - 0.5y = -3.5\)
- \(1.5x + 1.5y = -7.5\)
The matrices can be defined as follows:
\[ A = \begin{pmatrix} 1.5 & -0.5 \ 1.5 & 1.5 \end{pmatrix}, \quad X = \begin{pmatrix} x \ y \end{pmatrix}, \quad B = \begin{pmatrix} -3.5 \ -7.5 \end{pmatrix} \]
Now, we can express the system in matrix form as:
\[ A \cdot X = B \]
Next, we can solve for \(X\) by computing \(X = A^{-1}B\) if \(A\) is invertible.
Step 1: Compute the determinant of A
\[ \text{det}(A) = (1.5)(1.5) - (1.5)(-0.5) = 2.25 + 0.75 = 3 \]
Since the determinant is not zero, \(A\) is invertible.
Step 2: Calculate the inverse of A
The inverse of a 2x2 matrix \(\begin{pmatrix} a & b \ c & d \end{pmatrix}\) is given by:
\[ \frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \]
Applying this to our matrix \(A\):
\[ A^{-1} = \frac{1}{3} \begin{pmatrix} 1.5 & 0.5 \ -1.5 & 1.5 \end{pmatrix} = \begin{pmatrix} 0.5 & \frac{1}{6} \ -0.5 & 0.5 \end{pmatrix} \]
Step 3: Multiply the inverse of A by B
Now we compute \(X = A^{-1}B\):
\[ X = \begin{pmatrix} 0.5 & \frac{1}{6} \ -0.5 & 0.5 \end{pmatrix} \begin{pmatrix} -3.5 \ -7.5 \end{pmatrix} \]
Calculating the product:
For \(x\):
\[ x = 0.5(-3.5) + \frac{1}{6}(-7.5) = -1.75 - 1.25 = -3 \]
For \(y\):
\[ y = -0.5(-3.5) + 0.5(-7.5) = 1.75 - 3.75 = -2 \]
So,
\[ X = \begin{pmatrix} -3 \ -2 \end{pmatrix} \]
Thus, the solution to the system of equations is \(x = -3\) and \(y = -2\).
Final Answer:
\[ \begin{pmatrix} -3 \ -2 \end{pmatrix} \]
In the notation of your choices, this means the solution is:
[−3 −2] (the 2 by 1 column matrix row 1 negative 3 row 2 negative 2)