Find the inverse of the matrix   |1 1 1

-1 3 2

2 2 1|


Let's find the inverse of the matrix \( A \):

\[
A = \begin{pmatrix}
1 & 1 & 1 \\
-1 & 3 & 2 \\
2 & 2 & 1
\end{pmatrix}
\]

### Step 1: Find the Determinant of \( A \)

The determinant of \( A \) is calculated as follows:

\[
\text{det}(A) = 1 \cdot \left(3 \cdot 1 - 2 \cdot 2\right) - 1 \cdot \left(-1 \cdot 1 - 2 \cdot 2\right) + 1 \cdot \left(-1 \cdot 2 - 3 \cdot 2\right)
\]

\[
\text{det}(A) = 1 \cdot (3 - 4) - 1 \cdot (-1 - 4) + 1 \cdot (-2 - 6)
\]

\[
\text{det}(A) = 1 \cdot (-1) - 1 \cdot (-5) + 1 \cdot (-8)
\]

\[
\text{det}(A) = -1 + 5 - 8 = -4
\]

So, \(\text{det}(A) = -4\).

### Step 2: Find the Adjugate of \( A \)

Next, we calculate the cofactor matrix and then take its transpose to find the adjugate matrix.

**Cofactors:**

\[
\text{Cofactor}_{11} = (-1) \cdot \begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} = (-1) \cdot (3 - 4) = 1
\]
\[
\text{Cofactor}_{12} = (-1)^{1+2} \cdot \begin{vmatrix} -1 & 2 \\ 2 & 1 \end{vmatrix} = (-1) \cdot (-5) = 5
\]
\[
\text{Cofactor}_{13} = (-1) \cdot \begin{vmatrix} -1 & 3 \\ 2 & 2 \end{vmatrix} = -8
\]
\[
\text{Cofactor}_{21} = (-1) \cdot \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} = 1
\]
\[
\text{Cofactor}_{22} = \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} = -1
\]
\[
\text{Cofactor}_{23} = (-1)^{2+3} \cdot \begin{vmatrix} 1 & 1 \\ 2 & 2 \end{vmatrix} = 0
\]
\[
\text{Cofactor}_{31} = (-1) \cdot \begin{vmatrix} 1 & 1 \\ 3 & 2 \end{vmatrix} = 1
\]
\[
\text{Cofactor}_{32} = (-1)^{3+2} \cdot \begin{vmatrix} 1 & 1 \\ -1 & 2 \end{vmatrix} = -3
\]
\[
\text{Cofactor}_{33} = \begin{vmatrix} 1 & 1 \\ -1 & 3 \end{vmatrix} = 4
\]

The cofactor matrix is:

\[
\text{Cofactor}(A) = \begin{pmatrix}
1 & 5 & -8 \\
1 & -1 & 0 \\
1 & -3 & 4
\end{pmatrix}
\]

Taking the transpose of the cofactor matrix gives the adjugate matrix:

\[
\text{adj}(A) = \begin{pmatrix}
1 & 1 & 1 \\
5 & -1 & -3 \\
-8 & 0 & 4
\end{pmatrix}
\]

### Step 3: Find the Inverse of \( A \)

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

\[
A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A)
\]
\[
A^{-1} = \frac{1}{-4} \cdot \begin{pmatrix}
1 & 1 & 1 \\
5 & -1 & -3 \\
-8 & 0 & 4
\end{pmatrix}
\]

\[
A^{-1} = \begin{pmatrix}
-\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} \\
-\frac{5}{4} & \frac{1}{4} & \frac{3}{4} \\
2 & 0 & -1
\end{pmatrix}
\]

So, the inverse of the matrix \( A \) is:

\[
A^{-1} = \begin{pmatrix}
-\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} \\
-\frac{5}{4} & \frac{1}{4} & \frac{3}{4} \\
2 & 0 & -1
\end{pmatrix}
\]

Do you want details on any part of the calculation?

---

Here are 5 related questions to explore:

1. How can you determine if a matrix is invertible without calculating its determinant?
2. What is the geometric interpretation of a matrix inverse?
3. How would you compute the inverse of a \(2 \times 2\) matrix?
4. Can a non-square matrix have an inverse? Why or why not?
5. How does the inverse of a product of two matrices relate to the inverses of the individual matrices?

**Tip:** When finding the inverse of larger matrices, consider using computational tools to verify your results and save time.

Learn how to find the inverse of a 3x3 matrix with detailed step-by-step solution. This problem covers matrix algebra concepts including determinants, cofactor matrices, and matrix inverses. Suitable for advanced high school and college students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation