Consider the following system of equations.

*x*1 + 2*x*2 + *x*3

=

*a*1

*x*1 + 2*x*2 − *x*3

=

*a*2

*x*1 − 2*x*2 + *x*3

=

*a*3

Find the inverse of the coefficient matrix *A*. (Do not perform any row operations when creating *A*.)

*A*−1

=

Use the inverse matrix to solve each of the following systems of linear equations.

(a)

*x*1 + 2*x*2 + *x*3

=

2

*x*1 + 2*x*2 − *x*3

=

0

*x*1 − 2*x*2 + *x*3

=

−2

(*x*1, *x*2, *x*3) =

(b)

*x*1 + 2*x*2 + *x*3

=

1

*x*1 + 2*x*2 − *x*3

=

3

*x*1 − 2*x*2 + *x*3

=

−3

(*x*1, *x*2, *x*3) =

We begin by writing the given system of equations in matrix form. The system of equations is:

\[
\begin{aligned}
x_1 + 2x_2 + x_3 &= a_1 \\
x_1 + 2x_2 - x_3 &= a_2 \\
x_1 - 2x_2 + x_3 &= a_3
\end{aligned}
\]

This can be written as:

\[
\mathbf{A} \mathbf{x} = \mathbf{a}
\]

where \(\mathbf{A}\) is the coefficient matrix, \(\mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}\), and \(\mathbf{a} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}\).

The coefficient matrix \(\mathbf{A}\) is:

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

### Step 1: Find the inverse of the matrix \(\mathbf{A}\)

We need to find \(\mathbf{A}^{-1}\), the inverse of matrix \(\mathbf{A}\). The inverse of a 3x3 matrix is calculated using the formula:

\[
\mathbf{A}^{-1} = \frac{1}{\det(\mathbf{A})} \text{adj}(\mathbf{A})
\]

where \(\det(\mathbf{A})\) is the determinant of \(\mathbf{A}\) and \(\text{adj}(\mathbf{A})\) is the adjugate (transpose of the cofactor matrix) of \(\mathbf{A}\).

Let's calculate \(\det(\mathbf{A})\) and \(\mathbf{A}^{-1}\).The determinant of matrix \(\mathbf{A}\) is \(\det(\mathbf{A}) = -8\). Since the determinant is non-zero, the matrix is invertible.

The inverse matrix \(\mathbf{A}^{-1}\) is:

\[
\mathbf{A}^{-1} = \begin{pmatrix} 
0 & 0.5 & 0.5 \\
0.25 & 0 & -0.25 \\
0.5 & -0.5 & 0 
\end{pmatrix}
\]

### Step 2: Solve the systems of linear equations

Now, we will use \(\mathbf{A}^{-1}\) to solve the systems of equations \(\mathbf{A} \mathbf{x} = \mathbf{b}\), where \(\mathbf{b}\) is different for each part of the problem.

#### (a) Solve for:
\[
\mathbf{b}_a = \begin{pmatrix} 2 \\ 0 \\ -2 \end{pmatrix}
\]
We will compute \(\mathbf{x}_a = \mathbf{A}^{-1} \mathbf{b}_a\).

#### (b) Solve for:
\[
\mathbf{b}_b = \begin{pmatrix} 1 \\ 3 \\ -3 \end{pmatrix}
\]
We will compute \(\mathbf{x}_b = \mathbf{A}^{-1} \mathbf{b}_b\).

Let’s compute the solutions.The solutions for the systems of linear equations are:

### (a)
\[
(x_1, x_2, x_3) = (-1, 1, 1)
\]

### (b)
\[
(x_1, x_2, x_3) = (0, 1, -1)
\]

Would you like further details on the process, or have any questions?

Here are 5 related questions:
1. How do you calculate the determinant of a 3x3 matrix?
2. What are the conditions for a matrix to be invertible?
3. How can the adjugate matrix be used in finding an inverse?
4. How does matrix multiplication solve systems of equations?
5. What are common methods for solving systems of equations if the matrix isn't invertible?

**Tip:** When calculating the inverse of a matrix manually, check if its determinant is zero first, as it will indicate if the matrix is non-invertible.

Consider the following system of equations.

x1 + 2x2 + x3 = a1
x1 + 2x2 − x3 = a2
x1 − 2x2 + x3 = a3

Find the inverse of the coefficient matrix A. (Do not perform any row operations when creating A.)

A^(-1) = 

Use the inverse matrix to solve each of the following systems of linear equations.

(a)

x1 + 2x2 + x3 = 2
x1 + 2x2 − x3 = 0
x1 − 2x2 + x3 = −2

(x1, x2, x3) = 

(b)

x1 + 2x2 + x3 = 1
x1 + 2x2 − x3 = 3
x1 − 2x2 + x3 = −3

(x1, x2, x3) =

Learn how to solve a system of linear equations using matrix inversion. This detailed example walks through finding the inverse of a 3x3 matrix, calculating its determinant, and using it to solve systems of equations. Ideal for students in Grades 11-12 or college level.

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