Use an inverse matrix to solve each system of linear equations.

(a)    *x*1

+

*x*2

−

2*x*3

=

0

*x*1

−

2*x*2

+

*x*3

=

0

*x*1

−

*x*2

−

*x*3

=

−1

(*x*1, *x*2, *x*3) =

(b)    *x*1

+

*x*2

−

2*x*3

=

−3

*x*1

−

2*x*2

+

*x*3

=

6

*x*1

−

*x*2

−

*x*3

=

0

(*x*1, *x*2, *x*3) =

To solve these systems of linear equations using an inverse matrix, we follow these steps:

1. **Write the system of equations in matrix form**: \( A \mathbf{x} = \mathbf{b} \), where \( A \) is the coefficient matrix, \( \mathbf{x} \) is the column vector of variables, and \( \mathbf{b} \) is the column vector of constants.

2. **Find the inverse of matrix \( A \)**, denoted as \( A^{-1} \).

3. **Multiply the inverse matrix \( A^{-1} \) by the vector \( \mathbf{b} \)** to find the solution vector \( \mathbf{x} \): \( \mathbf{x} = A^{-1} \mathbf{b} \).

### Part (a)

Given system of equations:

\[
\begin{aligned}
x_1 + x_2 - 2x_3 & = 0, \\
x_1 - 2x_2 + x_3 & = 0, \\
x_1 - x_2 - x_3 & = -1.
\end{aligned}
\]

This system can be written in matrix form as:

\[
\begin{pmatrix}
1 & 1 & -2 \\
1 & -2 & 1 \\
1 & -1 & -1
\end{pmatrix}
\begin{pmatrix}
x_1 \\
x_2 \\
x_3
\end{pmatrix}
=
\begin{pmatrix}
0 \\
0 \\
-1
\end{pmatrix}
\]

Now, let's find the inverse of the coefficient matrix \( A \):

\[
A = \begin{pmatrix}
1 & 1 & -2 \\
1 & -2 & 1 \\
1 & -1 & -1
\end{pmatrix}
\]

To find \( A^{-1} \), we can use various methods (e.g., Gauss-Jordan elimination, or formula for the inverse of a 3x3 matrix). I'll calculate it.

### Part (b)

Given system of equations:

\[
\begin{aligned}
x_1 + x_2 - 2x_3 & = -3, \\
x_1 - 2x_2 + x_3 & = 6, \\
x_1 - x_2 - x_3 & = 0.
\end{aligned}
\]

This system can be written in matrix form as:

\[
\begin{pmatrix}
1 & 1 & -2 \\
1 & -2 & 1 \\
1 & -1 & -1
\end{pmatrix}
\begin{pmatrix}
x_1 \\
x_2 \\
x_3
\end{pmatrix}
=
\begin{pmatrix}
-3 \\
6 \\
0
\end{pmatrix}
\]

The coefficient matrix is the same as in part (a), so we will use the same inverse matrix \( A^{-1} \).

### Let's calculate the solutions for both parts (a) and (b).The solutions to the systems of linear equations are:

### Part (a)
The solution vector \((x_1, x_2, x_3)\) is:
\[
(x_1, x_2, x_3) = (1, 1, 1)
\]

### Part (b)
The solution vector \((x_1, x_2, x_3)\) is:
\[
(x_1, x_2, x_3) = (3, 0, 3)
\]
(Note: The value \( x_2 = 0 \) is represented as a very small number close to zero, \( 1.11 \times 10^{-16} \), which is due to numerical precision.)

Would you like further details or have any other questions?

Here are 8 follow-up questions related to this topic:

1. How do you calculate the inverse of a 3x3 matrix manually?
2. What are the conditions for a matrix to have an inverse?
3. How would you solve the system if the matrix was not invertible?
4. What is the geometric interpretation of a system of linear equations in three variables?
5. Can a system of linear equations have more than one solution? Under what conditions?
6. How would you solve these systems using Gaussian elimination instead of the inverse matrix?
7. What role does the determinant play in finding the inverse of a matrix?
8. How do the solutions change if the right-hand side vector \( \mathbf{b} \) changes?

**Tip:** When solving systems of equations using matrices, ensure the coefficient matrix is invertible; otherwise, the system may have no unique solution.

Learn how to solve systems of linear equations using inverse matrices. Detailed step-by-step solutions for two examples: (a) x1 + x2 - 2x3 = 0, x1 - 2x2 + x3 = 0, x1 - x2 - x3 = -1 and (b) x1 + x2 - 2x3 = -3, x1 - 2x2 + x3 = 6, x1 - x2 - x3 = 0. Suitable for college-level students studying linear algebra and matrix theory.

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