Write the given system of equations as a matrix equation and solve using inverses. Complete parts a through c.
x 1
plus
2 x 2
equals
k 1
x 2
plus
x 3
equals
k 2
5 x 1
minus
x 2
minus
12 x 3
equals
k 3
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Part 1
a. What are x 1​, x 2​, and x 3 when k 1equals3​, k 2equals0​, and k 3equals15​?

To solve the given system of equations using matrix methods, we need to follow these steps:

### Step 1: Write the System of Equations in Matrix Form

The given system of equations is:

\[
\begin{aligned}
x_1 + 2x_2 & = k_1 \quad \text{(Equation 1)} \\
x_2 + x_3 & = k_2 \quad \text{(Equation 2)} \\
5x_1 - x_2 - 12x_3 & = k_3 \quad \text{(Equation 3)}
\end{aligned}
\]

This system can be written in matrix form as:

\[
\begin{bmatrix}
1 & 2 & 0 \\
0 & 1 & 1 \\
5 & -1 & -12
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}
=
\begin{bmatrix}
k_1 \\
k_2 \\
k_3
\end{bmatrix}
\]

### Step 2: Find the Inverse of the Coefficient Matrix

The coefficient matrix \( A \) is:

\[
A = \begin{bmatrix}
1 & 2 & 0 \\
0 & 1 & 1 \\
5 & -1 & -12
\end{bmatrix}
\]

We need to find the inverse of \( A \), denoted as \( A^{-1} \).

### Step 3: Multiply the Inverse by the Constant Matrix

The constant matrix \( \mathbf{k} \) is:

\[
\mathbf{k} = \begin{bmatrix}
k_1 \\
k_2 \\
k_3
\end{bmatrix}
\]

The solution to the system is given by:

\[
\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}
= A^{-1} \begin{bmatrix}
k_1 \\
k_2 \\
k_3
\end{bmatrix}
\]

### Step 4: Solve for \( x_1 \), \( x_2 \), and \( x_3 \) When \( k_1 = 3 \), \( k_2 = 0 \), and \( k_3 = 15 \)

Substitute the values \( k_1 = 3 \), \( k_2 = 0 \), and \( k_3 = 15 \) into the constant matrix:

\[
\mathbf{k} = \begin{bmatrix}
3 \\
0 \\
15
\end{bmatrix}
\]

Now, I'll calculate the inverse of the matrix \( A \), then multiply it by the constant matrix \( \mathbf{k} \) to find \( x_1 \), \( x_2 \), and \( x_3 \).

Let's calculate this now.The solution to the system of equations when \( k_1 = 3 \), \( k_2 = 0 \), and \( k_3 = 15 \) is:

\[
\begin{aligned}
x_1 & = 3 \\
x_2 & \approx 0 \quad \text{(close to zero due to numerical approximation)} \\
x_3 & \approx 0 \quad \text{(close to zero due to numerical approximation)}
\end{aligned}
\]

This means that \( x_1 = 3 \), \( x_2 = 0 \), and \( x_3 = 0 \) is the solution for the given values of \( k_1 \), \( k_2 \), and \( k_3 \).

Would you like more details or have any questions? Here are some related questions you might consider:

1. How do you determine the inverse of a matrix?
2. What is the significance of the determinant in finding the inverse?
3. How would the solution change if \( k_1 \), \( k_2 \), and \( k_3 \) were different?
4. What if the determinant of the coefficient matrix \( A \) was zero?
5. How can you verify the solution obtained from the matrix equation?

**Tip:** When solving a system of equations using matrix inverses, always check the determinant first. If it’s zero, the matrix is non-invertible, meaning the system might have no solutions or infinitely many solutions.

Learn how to solve a system of equations using matrix inverses with detailed step-by-step instructions. This problem involves matrix algebra and finding inverse matrices to determine unknown variables x1, x2, and x3. Suitable for advanced high school students (Grades 11-12).

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