Answer :
To solve the system of equations using the matrix method, we need to first rearrange the equations into a system of linear equations. The given equations are:
[tex]\[ \frac{3}{y} = \frac{2}{x} + 2 \][/tex]
[tex]\[ \frac{6}{x} - \frac{2}{y} = 1 \][/tex]
First, let's rewrite each equation by finding a common variable term. Let's take [tex]\( \frac{1}{x} = a \)[/tex] and [tex]\( \frac{1}{y} = b \)[/tex]. Therefore, the equations become:
[tex]\[ \frac{3}{y} = 2a + 2 \][/tex]
[tex]\[ \frac{6}{x} - \frac{2}{y} = 1 \][/tex]
Substituting [tex]\( b \)[/tex] for [tex]\( \frac{1}{y} \)[/tex] and [tex]\( a \)[/tex] for [tex]\( \frac{1}{x} \)[/tex]:
[tex]\[ 3b = 2a + 2 \][/tex]
[tex]\[ 6a - 2b = 1 \][/tex]
We need to convert these into a system of linear equations that can be expressed in matrix form.
Simplifying both equations:
[tex]\[ 3b - 2a = 2 \][/tex]
[tex]\[ 6a - 2b = 1 \][/tex]
Now, represent the above system in the matrix form [tex]\( AX = B \)[/tex] where:
[tex]\[ A = \begin{pmatrix} -2 & 3 \\ 6 & -2 \end{pmatrix}, \quad X = \begin{pmatrix} a \\ b \end{pmatrix}, \quad B = \begin{pmatrix} 2 \\ 1 \end{pmatrix} \][/tex]
The system can be expressed as:
[tex]\[ \begin{pmatrix} -2 & 3 \\ 6 & -2 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix} \][/tex]
To solve for [tex]\( X \)[/tex], we use the formula [tex]\( X = A^{-1}B \)[/tex], where [tex]\( A^{-1} \)[/tex] is the inverse of matrix [tex]\( A \)[/tex].
First, find the determinant of [tex]\( A \)[/tex]:
[tex]\[ \text{det}(A) = (-2)(-2) - (3)(6) = 4 - 18 = -14 \][/tex]
The inverse [tex]\( A^{-1} \)[/tex] is given by:
[tex]\[ A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} -2 & -3 \\ -6 & -2 \end{pmatrix}^T \][/tex]
Here, [tex]\( \begin{pmatrix} -2 & -3 \\ -6 & -2 \end{pmatrix} \)[/tex] is the matrix of cofactors transposed (adjugate matrix).
[tex]\[ A^{-1} = \frac{1}{-14} \begin{pmatrix} -2 & -6 \\ 3 & -2 \end{pmatrix} \][/tex]
Simplify the inverse matrix:
[tex]\[ A^{-1} = \begin{pmatrix} \frac{1}{7} & \frac{3}{7} \\ -\frac{3}{14} & \frac{1}{7} \end{pmatrix} \][/tex]
Now, multiply [tex]\( A^{-1} \)[/tex] with [tex]\( B \)[/tex]:
[tex]\[ X = A^{-1}B = \begin{pmatrix} \frac{1}{7} & \frac{3}{7} \\ -\frac{3}{14} & \frac{1}{7} \end{pmatrix} \begin{pmatrix} 2 \\ 1 \end{pmatrix} \][/tex]
[tex]\[ X = \begin{pmatrix} \frac{1}{7}(2) + \frac{3}{7}(1) \\ -\frac{3}{14}(2) + \frac{1}{7}(1) \end{pmatrix} = \begin{pmatrix} \frac{2}{7} + \frac{3}{7} \\ -\frac{6}{14} + \frac{1}{7} \end{pmatrix} \][/tex]
[tex]\[ X = \begin{pmatrix} \frac{5}{7} \\ -\frac{6}{14} + \frac{2}{14} \end{pmatrix} = \begin{pmatrix} \frac{5}{7} \\ -\frac{4}{14} \end{pmatrix} = \begin{pmatrix} \frac{5}{7} \\ -\frac{2}{7} \end{pmatrix} \][/tex]
Thus,
[tex]\[ a = \frac{5}{7}, \quad b = -\frac{2}{7} \][/tex]
Remember that [tex]\( a = \frac{1}{x} \)[/tex] and [tex]\( b = \frac{1}{y} \)[/tex]:
[tex]\[ \frac{1}{x} = \frac{5}{7} \implies x = \frac{7}{5} \][/tex]
[tex]\[ \frac{1}{y} = -\frac{2}{7} \implies y = -\frac{7}{2} \][/tex]
Therefore, the solution to the system of equations is:
[tex]\[ x = \frac{7}{5}, \quad y = -\frac{7}{2} \][/tex]
[tex]\[ \frac{3}{y} = \frac{2}{x} + 2 \][/tex]
[tex]\[ \frac{6}{x} - \frac{2}{y} = 1 \][/tex]
First, let's rewrite each equation by finding a common variable term. Let's take [tex]\( \frac{1}{x} = a \)[/tex] and [tex]\( \frac{1}{y} = b \)[/tex]. Therefore, the equations become:
[tex]\[ \frac{3}{y} = 2a + 2 \][/tex]
[tex]\[ \frac{6}{x} - \frac{2}{y} = 1 \][/tex]
Substituting [tex]\( b \)[/tex] for [tex]\( \frac{1}{y} \)[/tex] and [tex]\( a \)[/tex] for [tex]\( \frac{1}{x} \)[/tex]:
[tex]\[ 3b = 2a + 2 \][/tex]
[tex]\[ 6a - 2b = 1 \][/tex]
We need to convert these into a system of linear equations that can be expressed in matrix form.
Simplifying both equations:
[tex]\[ 3b - 2a = 2 \][/tex]
[tex]\[ 6a - 2b = 1 \][/tex]
Now, represent the above system in the matrix form [tex]\( AX = B \)[/tex] where:
[tex]\[ A = \begin{pmatrix} -2 & 3 \\ 6 & -2 \end{pmatrix}, \quad X = \begin{pmatrix} a \\ b \end{pmatrix}, \quad B = \begin{pmatrix} 2 \\ 1 \end{pmatrix} \][/tex]
The system can be expressed as:
[tex]\[ \begin{pmatrix} -2 & 3 \\ 6 & -2 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix} \][/tex]
To solve for [tex]\( X \)[/tex], we use the formula [tex]\( X = A^{-1}B \)[/tex], where [tex]\( A^{-1} \)[/tex] is the inverse of matrix [tex]\( A \)[/tex].
First, find the determinant of [tex]\( A \)[/tex]:
[tex]\[ \text{det}(A) = (-2)(-2) - (3)(6) = 4 - 18 = -14 \][/tex]
The inverse [tex]\( A^{-1} \)[/tex] is given by:
[tex]\[ A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} -2 & -3 \\ -6 & -2 \end{pmatrix}^T \][/tex]
Here, [tex]\( \begin{pmatrix} -2 & -3 \\ -6 & -2 \end{pmatrix} \)[/tex] is the matrix of cofactors transposed (adjugate matrix).
[tex]\[ A^{-1} = \frac{1}{-14} \begin{pmatrix} -2 & -6 \\ 3 & -2 \end{pmatrix} \][/tex]
Simplify the inverse matrix:
[tex]\[ A^{-1} = \begin{pmatrix} \frac{1}{7} & \frac{3}{7} \\ -\frac{3}{14} & \frac{1}{7} \end{pmatrix} \][/tex]
Now, multiply [tex]\( A^{-1} \)[/tex] with [tex]\( B \)[/tex]:
[tex]\[ X = A^{-1}B = \begin{pmatrix} \frac{1}{7} & \frac{3}{7} \\ -\frac{3}{14} & \frac{1}{7} \end{pmatrix} \begin{pmatrix} 2 \\ 1 \end{pmatrix} \][/tex]
[tex]\[ X = \begin{pmatrix} \frac{1}{7}(2) + \frac{3}{7}(1) \\ -\frac{3}{14}(2) + \frac{1}{7}(1) \end{pmatrix} = \begin{pmatrix} \frac{2}{7} + \frac{3}{7} \\ -\frac{6}{14} + \frac{1}{7} \end{pmatrix} \][/tex]
[tex]\[ X = \begin{pmatrix} \frac{5}{7} \\ -\frac{6}{14} + \frac{2}{14} \end{pmatrix} = \begin{pmatrix} \frac{5}{7} \\ -\frac{4}{14} \end{pmatrix} = \begin{pmatrix} \frac{5}{7} \\ -\frac{2}{7} \end{pmatrix} \][/tex]
Thus,
[tex]\[ a = \frac{5}{7}, \quad b = -\frac{2}{7} \][/tex]
Remember that [tex]\( a = \frac{1}{x} \)[/tex] and [tex]\( b = \frac{1}{y} \)[/tex]:
[tex]\[ \frac{1}{x} = \frac{5}{7} \implies x = \frac{7}{5} \][/tex]
[tex]\[ \frac{1}{y} = -\frac{2}{7} \implies y = -\frac{7}{2} \][/tex]
Therefore, the solution to the system of equations is:
[tex]\[ x = \frac{7}{5}, \quad y = -\frac{7}{2} \][/tex]
