Certainly! Let's solve the given system of equations using matrices step by step.
The system of equations given is:
[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 solve this using the matrix inversion method. Here's the process:
### Step 1: Define the coefficient matrix [tex]\( A \)[/tex] and the constants matrix [tex]\( B \)[/tex].
The coefficient matrix [tex]\( A \)[/tex] and the constants matrix [tex]\( B \)[/tex] are:
[tex]\[
A = \left[\begin{array}{cc}
2 & 7 \\
2 & 6
\end{array}\right], \quad B = \left[\begin{array}{c}
4 \\
6
\end{array}\right]
\][/tex]
### Step 2: Calculate the inverse of matrix [tex]\( A \)[/tex], denoted as [tex]\( A^{-1} \)[/tex].
The inverse of a 2x2 matrix [tex]\( \left[\begin{array}{cc} a & b \\ c & d \end{array}\right] \)[/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], we have:
[tex]\[
a = 2, \quad b = 7, \quad c = 2, \quad d = 6
\][/tex]
First, compute the determinant [tex]\( \text{det}(A) \)[/tex]:
[tex]\[
\text{det}(A) = (2)(6) - (7)(2) = 12 - 14 = -2
\][/tex]
Since the determinant is non-zero, the inverse exists. Now compute the inverse:
[tex]\[
A^{-1} = \frac{1}{-2} \left[\begin{array}{cc}
6 & -7 \\
-2 & 2
\end{array}\right]
= \left[\begin{array}{cc}
-\frac{6}{2} & \frac{7}{2} \\
\frac{2}{2} & -1
\end{array}\right]
= \left[\begin{array}{cc}
-3 & 3.5 \\
1 & -1
\end{array}\right]
\][/tex]
### Step 3: Use the inverse matrix to solve for [tex]\( \left[\begin{array}{c} x \\ y \end{array}\right] \)[/tex].
The solution is given by:
[tex]\[
\left[\begin{array}{c}
x \\
y
\end{array}\right] = A^{-1} \cdot B
\][/tex]
Compute the multiplication:
[tex]\[
\left[\begin{array}{cc}
-3 & 3.5 \\
1 & -1
\end{array}\right] \left[\begin{array}{c}
4 \\
6
\end{array}\right]
\][/tex]
### Step 4: Perform the matrix multiplication.
[tex]\[
\left[\begin{array}{cc}
-3 & 3.5 \\
1 & -1
\end{array}\right] \left[\begin{array}{c}
4 \\
6
\end{array}\right]
=
\left[\begin{array}{c}
(-3)(4) + (3.5)(6) \\
(1)(4) + (-1)(6)
\end{array}\right]
=
\left[\begin{array}{c}
-12 + 21 \\
4 - 6
\end{array}\right]
=
\left[\begin{array}{c}
9 \\
-2
\end{array}\right]
\][/tex]
Thus, we find:
[tex]\[
x = 9, \quad y = -2
\][/tex]
So the solution to the system of equations is:
[tex]\[
y = -2
\][/tex]
