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To solve the system of equations using inverse matrices, we should follow these steps:
Step 1: Represent the system of equations in matrix form
Given system:
[tex]\[ \begin{cases} 2x + 5y = -16 \\ -3x - y = 11 \end{cases} \][/tex]
We represent the system in the matrix form [tex]\(AX = B\)[/tex], where [tex]\(A\)[/tex] is the coefficient matrix, [tex]\(X\)[/tex] is the variable matrix, and [tex]\(B\)[/tex] is the constant matrix.
The coefficient matrix [tex]\(A\)[/tex], variable matrix [tex]\(X\)[/tex], and constant matrix [tex]\(B\)[/tex] are:
[tex]\[ A = \begin{pmatrix} 2 & 5 \\ -3 & -1 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \end{pmatrix}, \quad B = \begin{pmatrix} -16 \\ 11 \end{pmatrix} \][/tex]
Step 2: Calculate the determinant of the coefficient matrix [tex]\(A\)[/tex]
To find the inverse of [tex]\(A\)[/tex], we need to ensure it is invertible by checking if its determinant ([tex]\(\det(A)\)[/tex]) is non-zero.
[tex]\[ \det(A) = (2)(-1) - (5)(-3) = -2 + 15 = 13 \][/tex]
The determinant is 13, which is not zero, so [tex]\(A\)[/tex] is invertible.
Step 3: Find the inverse of the coefficient matrix [tex]\(A\)[/tex]
The inverse of a 2x2 matrix [tex]\( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex] is given by
[tex]\[ A^{-1} = \frac{1}{\det(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \][/tex]
Applying this formula, we get:
[tex]\[ A^{-1} = \frac{1}{13} \begin{pmatrix} -1 & -5 \\ 3 & 2 \end{pmatrix} = \begin{pmatrix} -\frac{1}{13} & -\frac{5}{13} \\ \frac{3}{13} & \frac{2}{13} \end{pmatrix} \][/tex]
Step 4: Solve for [tex]\(X\)[/tex] (i.e., [tex]\(x\)[/tex] and [tex]\(y\)[/tex])
We use the inverse of the matrix [tex]\(A\)[/tex] to solve for [tex]\(X\)[/tex] as follows:
[tex]\[ X = A^{-1}B \][/tex]
[tex]\[ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -\frac{1}{13} & -\frac{5}{13} \\ \frac{3}{13} & \frac{2}{13} \end{pmatrix} \begin{pmatrix} -16 \\ 11 \end{pmatrix} \][/tex]
Step 5: Perform the matrix multiplication
Multiplying the matrices, we get:
For [tex]\(x\)[/tex]:
[tex]\[ x = \left( -\frac{1}{13} \times (-16) \right) + \left( -\frac{5}{13} \times 11 \right) = \frac{16}{13} - \frac{55}{13} = -\frac{39}{13} = -3 \][/tex]
For [tex]\(y\)[/tex]:
[tex]\[ y = \left( \frac{3}{13} \times (-16) \right) + \left( \frac{2}{13} \times 11 \right) = -\frac{48}{13} + \frac{22}{13} = -\frac{26}{13} = -2 \][/tex]
Thus, the solution is:
[tex]\[ x = -3, \quad y = -2 \][/tex]
That's the solution to the system of equations.
Step 1: Represent the system of equations in matrix form
Given system:
[tex]\[ \begin{cases} 2x + 5y = -16 \\ -3x - y = 11 \end{cases} \][/tex]
We represent the system in the matrix form [tex]\(AX = B\)[/tex], where [tex]\(A\)[/tex] is the coefficient matrix, [tex]\(X\)[/tex] is the variable matrix, and [tex]\(B\)[/tex] is the constant matrix.
The coefficient matrix [tex]\(A\)[/tex], variable matrix [tex]\(X\)[/tex], and constant matrix [tex]\(B\)[/tex] are:
[tex]\[ A = \begin{pmatrix} 2 & 5 \\ -3 & -1 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \end{pmatrix}, \quad B = \begin{pmatrix} -16 \\ 11 \end{pmatrix} \][/tex]
Step 2: Calculate the determinant of the coefficient matrix [tex]\(A\)[/tex]
To find the inverse of [tex]\(A\)[/tex], we need to ensure it is invertible by checking if its determinant ([tex]\(\det(A)\)[/tex]) is non-zero.
[tex]\[ \det(A) = (2)(-1) - (5)(-3) = -2 + 15 = 13 \][/tex]
The determinant is 13, which is not zero, so [tex]\(A\)[/tex] is invertible.
Step 3: Find the inverse of the coefficient matrix [tex]\(A\)[/tex]
The inverse of a 2x2 matrix [tex]\( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex] is given by
[tex]\[ A^{-1} = \frac{1}{\det(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \][/tex]
Applying this formula, we get:
[tex]\[ A^{-1} = \frac{1}{13} \begin{pmatrix} -1 & -5 \\ 3 & 2 \end{pmatrix} = \begin{pmatrix} -\frac{1}{13} & -\frac{5}{13} \\ \frac{3}{13} & \frac{2}{13} \end{pmatrix} \][/tex]
Step 4: Solve for [tex]\(X\)[/tex] (i.e., [tex]\(x\)[/tex] and [tex]\(y\)[/tex])
We use the inverse of the matrix [tex]\(A\)[/tex] to solve for [tex]\(X\)[/tex] as follows:
[tex]\[ X = A^{-1}B \][/tex]
[tex]\[ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -\frac{1}{13} & -\frac{5}{13} \\ \frac{3}{13} & \frac{2}{13} \end{pmatrix} \begin{pmatrix} -16 \\ 11 \end{pmatrix} \][/tex]
Step 5: Perform the matrix multiplication
Multiplying the matrices, we get:
For [tex]\(x\)[/tex]:
[tex]\[ x = \left( -\frac{1}{13} \times (-16) \right) + \left( -\frac{5}{13} \times 11 \right) = \frac{16}{13} - \frac{55}{13} = -\frac{39}{13} = -3 \][/tex]
For [tex]\(y\)[/tex]:
[tex]\[ y = \left( \frac{3}{13} \times (-16) \right) + \left( \frac{2}{13} \times 11 \right) = -\frac{48}{13} + \frac{22}{13} = -\frac{26}{13} = -2 \][/tex]
Thus, the solution is:
[tex]\[ x = -3, \quad y = -2 \][/tex]
That's the solution to the system of equations.
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