The matrix equation represents a system of equations. A matrix with 2 rows and 2 columns, where row 1 is 2 and 7 and row 2 is 2 and 6, is multiplied by matrix with 2 rows and 1 column, where row 1 is x and row 2 is y, equals a matrix with 2 rows and 1 column, where row 1 is 4 and row 2 is 6. Solve for y using matrices. Show or explain all necessary steps.

To solve for \( y \) using matrices, we will follow the same matrix-based approach as before. This involves writing the system of equations, representing it as a matrix equation, and solving it using matrix algebra. Here's a step-by-step breakdown:

### Step 1: Write the Matrix Equation

Given the matrix equation:

\[
\begin{pmatrix} 2 & 7 \\ 2 & 6 \end{pmatrix} \times \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 4 \\ 6 \end{pmatrix}
\]

This represents the system of linear equations:
1. \( 2x + 7y = 4 \)
2. \( 2x + 6y = 6 \)

### Step 2: Write the Equation in Matrix Form

Let:
- \( A = \begin{pmatrix} 2 & 7 \\ 2 & 6 \end{pmatrix} \)
- \( \mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix} \)
- \( \mathbf{b} = \begin{pmatrix} 4 \\ 6 \end{pmatrix} \)

The matrix equation is \( A \mathbf{v} = \mathbf{b} \).

### Step 3: Solve for \( \mathbf{v} \) Using the Inverse of Matrix \( A \)

To solve for \( y \), we need to find the inverse of matrix \( A \) (if it exists) and multiply both sides of the equation by \( A^{-1} \):

\[
\mathbf{v} = A^{-1} \mathbf{b}
\]

#### Step 3.1: Find the Determinant of Matrix \( A \)

For a 2x2 matrix \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \), the determinant is given by:

\[
\text{det}(A) = ad - bc
\]

For our matrix \( A = \begin{pmatrix} 2 & 7 \\ 2 & 6 \end{pmatrix} \):

- \( a = 2 \), \( b = 7 \), \( c = 2 \), \( d = 6 \)
- \(\text{det}(A) = (2 \times 6) - (7 \times 2) = 12 - 14 = -2 \)

Since the determinant is non-zero (\(\text{det}(A) = -2\)), matrix \( A \) is invertible, so we can proceed to find the inverse.

#### Step 3.2: Find the Inverse of Matrix \( A \)

The inverse of a 2x2 matrix \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by:

\[
A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}
\]

Substituting the values:

\[
A^{-1} = \frac{1}{-2} \begin{pmatrix} 6 & -7 \\ -2 & 2 \end{pmatrix} = \begin{pmatrix} -3 & \frac{7}{2} \\ 1 & -1 \end{pmatrix}
\]

#### Step 3.3: Multiply \( A^{-1} \) by \( \mathbf{b} \)

Now, we compute the product:

\[
\mathbf{v} = A^{-1} \mathbf{b} = \begin{pmatrix} -3 & \frac{7}{2} \\ 1 & -1 \end{pmatrix} \times \begin{pmatrix} 4 \\ 6 \end{pmatrix}
\]

Perform the matrix multiplication:

- First row: \( (-3 \times 4) + \left(\frac{7}{2} \times 6\right) = -12 + 21 = 9 \)
- Second row: \( (1 \times 4) + (-1 \times 6) = 4 - 6 = -2 \)

Thus, the result is:

\[
\mathbf{v} = \begin{pmatrix} 9 \\ -2 \end{pmatrix}
\]

### Step 4: Interpret the Result

The solution is:

\[
x = 9 \quad \text{and} \quad y = -2
\]

Thus, the value of \( y \) that satisfies the system of equations is \( y = -2 \).

---

Would you like further details or explanations on any of the steps? Here are some related questions to expand on this topic:

1. What would happen if the determinant of the matrix \( A \) was zero?
2. How does matrix inversion change if the matrix is larger than 2x2?
3. Can matrix multiplication be used to solve nonlinear systems of equations?
4. How can the properties of matrices help in simplifying systems of linear equations?
5. How can Gaussian elimination be used as an alternative to finding the inverse of a matrix?

**Tip:** When working with matrices, always check if the determinant is zero before attempting to find the inverse. A zero determinant means that the matrix is singular and does not have an inverse, so the system of equations may have no unique solution.

Learn how to solve a system of linear equations using matrices. Step-by-step solution to find the value of y using matrix algebra and inverse calculations. Suitable for high school students and anyone interested in matrix operations and linear algebra.

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