Answer :
To solve the matrix equation
[tex]\[ \left[\begin{array}{cc} 2 & 7 \\ 2 & 6 \end{array}\right]\left[\begin{array}{c} x \\ y \end{array}\right]=\left[\begin{array}{c} 4 \\ 6 \end{array}\right], \][/tex]
we will use the inverse of the coefficient matrix.
### Step 1: Define the matrices
Let [tex]\( A \)[/tex] be the coefficient matrix, [tex]\( \mathbf{x} \)[/tex] be the variable vector, and [tex]\( \mathbf{b} \)[/tex] be the constant vector.
[tex]\[ A = \left[\begin{array}{cc} 2 & 7 \\ 2 & 6 \end{array}\right], \quad \mathbf{x} = \left[\begin{array}{c} x \\ y \end{array}\right], \quad \mathbf{b} = \left[\begin{array}{c} 4 \\ 6 \end{array}\right] \][/tex]
### Step 2: Calculate the inverse of matrix [tex]\( A \)[/tex]
We need to find [tex]\( A^{-1} \)[/tex], the inverse of matrix [tex]\( A \)[/tex]. Recall that if [tex]\( A \)[/tex] is a 2x2 matrix
[tex]\[ A = \left[\begin{array}{cc} a & b \\ c & d \end{array}\right], \][/tex]
then the inverse [tex]\( A^{-1} \)[/tex] is given by
[tex]\[ A^{-1} = \frac{1}{ad - bc} \left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right] \][/tex]
For our matrix [tex]\( A \)[/tex]:
[tex]\[ a = 2, \quad b = 7, \quad c = 2, \quad d = 6 \][/tex]
First, calculate the determinant [tex]\( \text{det}(A) \)[/tex]:
[tex]\[ \text{det}(A) = ad - bc = (2)(6) - (2)(7) = 12 - 14 = -2 \][/tex]
Now, calculate [tex]\( A^{-1} \)[/tex]:
[tex]\[ A^{-1} = \frac{1}{-2} \left[\begin{array}{cc} 6 & -7 \\ -2 & 2 \end{array}\right] = \left[\begin{array}{cc} -3 & \frac{7}{2} \\ 1 & -1 \end{array}\right] = \left[\begin{array}{cc} -3 & 3.5 \\ 1 & -1 \end{array}\right] \][/tex]
### Step 3: Multiply [tex]\( A^{-1} \)[/tex] by [tex]\( \mathbf{b} \)[/tex]
To find the vector [tex]\( \mathbf{x} \)[/tex], multiply [tex]\( A^{-1} \)[/tex] by [tex]\( \mathbf{b} \)[/tex]:
[tex]\[ \mathbf{x} = A^{-1} \mathbf{b} \][/tex]
Perform the matrix multiplication:
[tex]\[ \mathbf{x} = \left[\begin{array}{cc} -3 & 3.5 \\ 1 & -1 \end{array}\right] \left[\begin{array}{c} 4 \\ 6 \end{array}\right] \][/tex]
Calculate each component of [tex]\( \mathbf{x} \)[/tex]:
1. For [tex]\( x \)[/tex]:
[tex]\[ x = (-3)(4) + (3.5)(6) = -12 + 21 = 9 \][/tex]
2. For [tex]\( y \)[/tex]:
[tex]\[ y = (1)(4) + (-1)(6) = 4 - 6 = -2 \][/tex]
### Step 4: Verify the solution [tex]\( x = 9 \)[/tex] and [tex]\( y = -2 \)[/tex]
Thus, the solution to the system of equations is:
[tex]\[ x = 9, \quad y = -2 \][/tex]
So, the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy the given matrix equation are [tex]\( x = 9 \)[/tex] and [tex]\( y = -2 \)[/tex]. The inverse matrix [tex]\( A^{-1} \)[/tex] is
[tex]\[ \left[\begin{array}{cc} -3 & 3.5 \\ 1 & -1 \end{array}\right] \][/tex]
and the result calculated using the inverse is:
[tex]\[ \begin{pmatrix} 9 \\ -2 \end{pmatrix}. \][/tex]
[tex]\[ \left[\begin{array}{cc} 2 & 7 \\ 2 & 6 \end{array}\right]\left[\begin{array}{c} x \\ y \end{array}\right]=\left[\begin{array}{c} 4 \\ 6 \end{array}\right], \][/tex]
we will use the inverse of the coefficient matrix.
### Step 1: Define the matrices
Let [tex]\( A \)[/tex] be the coefficient matrix, [tex]\( \mathbf{x} \)[/tex] be the variable vector, and [tex]\( \mathbf{b} \)[/tex] be the constant vector.
[tex]\[ A = \left[\begin{array}{cc} 2 & 7 \\ 2 & 6 \end{array}\right], \quad \mathbf{x} = \left[\begin{array}{c} x \\ y \end{array}\right], \quad \mathbf{b} = \left[\begin{array}{c} 4 \\ 6 \end{array}\right] \][/tex]
### Step 2: Calculate the inverse of matrix [tex]\( A \)[/tex]
We need to find [tex]\( A^{-1} \)[/tex], the inverse of matrix [tex]\( A \)[/tex]. Recall that if [tex]\( A \)[/tex] is a 2x2 matrix
[tex]\[ A = \left[\begin{array}{cc} a & b \\ c & d \end{array}\right], \][/tex]
then the inverse [tex]\( A^{-1} \)[/tex] is given by
[tex]\[ A^{-1} = \frac{1}{ad - bc} \left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right] \][/tex]
For our matrix [tex]\( A \)[/tex]:
[tex]\[ a = 2, \quad b = 7, \quad c = 2, \quad d = 6 \][/tex]
First, calculate the determinant [tex]\( \text{det}(A) \)[/tex]:
[tex]\[ \text{det}(A) = ad - bc = (2)(6) - (2)(7) = 12 - 14 = -2 \][/tex]
Now, calculate [tex]\( A^{-1} \)[/tex]:
[tex]\[ A^{-1} = \frac{1}{-2} \left[\begin{array}{cc} 6 & -7 \\ -2 & 2 \end{array}\right] = \left[\begin{array}{cc} -3 & \frac{7}{2} \\ 1 & -1 \end{array}\right] = \left[\begin{array}{cc} -3 & 3.5 \\ 1 & -1 \end{array}\right] \][/tex]
### Step 3: Multiply [tex]\( A^{-1} \)[/tex] by [tex]\( \mathbf{b} \)[/tex]
To find the vector [tex]\( \mathbf{x} \)[/tex], multiply [tex]\( A^{-1} \)[/tex] by [tex]\( \mathbf{b} \)[/tex]:
[tex]\[ \mathbf{x} = A^{-1} \mathbf{b} \][/tex]
Perform the matrix multiplication:
[tex]\[ \mathbf{x} = \left[\begin{array}{cc} -3 & 3.5 \\ 1 & -1 \end{array}\right] \left[\begin{array}{c} 4 \\ 6 \end{array}\right] \][/tex]
Calculate each component of [tex]\( \mathbf{x} \)[/tex]:
1. For [tex]\( x \)[/tex]:
[tex]\[ x = (-3)(4) + (3.5)(6) = -12 + 21 = 9 \][/tex]
2. For [tex]\( y \)[/tex]:
[tex]\[ y = (1)(4) + (-1)(6) = 4 - 6 = -2 \][/tex]
### Step 4: Verify the solution [tex]\( x = 9 \)[/tex] and [tex]\( y = -2 \)[/tex]
Thus, the solution to the system of equations is:
[tex]\[ x = 9, \quad y = -2 \][/tex]
So, the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy the given matrix equation are [tex]\( x = 9 \)[/tex] and [tex]\( y = -2 \)[/tex]. The inverse matrix [tex]\( A^{-1} \)[/tex] is
[tex]\[ \left[\begin{array}{cc} -3 & 3.5 \\ 1 & -1 \end{array}\right] \][/tex]
and the result calculated using the inverse is:
[tex]\[ \begin{pmatrix} 9 \\ -2 \end{pmatrix}. \][/tex]
