The matrix equation represents a system of equations. A matrix with 2 rows and 2 columns, where row 1 is 3 and 10 and row 2 is 1 and 3, 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 6 and row 2 is 1. Solve for x and y using matrices. Show or explain all necessary steps.

To solve for \(x\) and \(y\) in the matrix equation using matrices, we need to follow these steps:

### Step 1: Write the Matrix Equation

Given the matrix equation:

\[
\begin{pmatrix} 3 & 10 \\ 1 & 3 \end{pmatrix} \times \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 6 \\ 1 \end{pmatrix}
\]

This represents the system of linear equations:
1. \(3x + 10y = 6\)
2. \(x + 3y = 1\)

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

Let:
- \(A = \begin{pmatrix} 3 & 10 \\ 1 & 3 \end{pmatrix}\)
- \(\mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix}\)
- \(\mathbf{b} = \begin{pmatrix} 6 \\ 1 \end{pmatrix}\)

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

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

To solve for \(\mathbf{v}\), we need to find the inverse of matrix \(A\), denoted as \(A^{-1}\), and multiply both sides of the equation by \(A^{-1}\):

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

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

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

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

where \(\text{det}(A)\) is the determinant of matrix \(A\), calculated as:

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

For our matrix \(A = \begin{pmatrix} 3 & 10 \\ 1 & 3 \end{pmatrix}\):

- \(a = 3\), \(b = 10\), \(c = 1\), \(d = 3\)
- \(\text{det}(A) = (3 \times 3) - (10 \times 1) = 9 - 10 = -1\)

Now, the inverse of \(A\) is:

\[
A^{-1} = \frac{1}{-1} \begin{pmatrix} 3 & -10 \\ -1 & 3 \end{pmatrix} = \begin{pmatrix} -3 & 10 \\ 1 & -3 \end{pmatrix}
\]

#### Step 3.2: Multiply \(A^{-1}\) by \(\mathbf{b}\)

Now, we calculate:

\[
\mathbf{v} = A^{-1} \mathbf{b} = \begin{pmatrix} -3 & 10 \\ 1 & -3 \end{pmatrix} \times \begin{pmatrix} 6 \\ 1 \end{pmatrix}
\]

Perform the matrix multiplication:

- First row: \((-3 \times 6) + (10 \times 1) = -18 + 10 = -8\)
- Second row: \((1 \times 6) + (-3 \times 1) = 6 - 3 = 3\)

Thus, the result is:

\[
\mathbf{v} = \begin{pmatrix} -8 \\ 3 \end{pmatrix}
\]

### Step 4: Interpret the Result

The solution is:

\[
x = -8 \quad \text{and} \quad y = 3
\]

Thus, the values of \(x\) and \(y\) that satisfy the system of equations are \(x = -8\) and \(y = 3\).

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

1. What is the geometric interpretation of solving a system of equations using matrices?
2. How can matrix multiplication be used to model real-world problems?
3. What are some methods for finding the inverse of a matrix other than the formula for a 2x2 matrix?
4. What happens if the determinant of a matrix is zero?
5. How can we solve larger systems of equations using matrices and Gaussian elimination?

**Tip:** The determinant of a matrix plays a crucial role in determining whether a matrix is invertible. If the determinant is zero, the matrix does not have an inverse, and the system of equations may have no solution or infinitely many solutions.

Learn how to solve a system of equations represented by a matrix equation. Detailed steps include forming the matrix equation, finding the inverse of matrix A, and solving for variables x and y. Suitable for high school students studying matrices and systems of linear equations.

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