what is inverse of this 3x3 matrix (im listing first a11 then a12...) :1,1,2,1,0,1,-1,1,-4

To find the inverse of the given \(3 \times 3\) matrix:

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

We can use the formula for the inverse of a matrix, which is given by:

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

### Step 1: Calculate the determinant of \(A\)

The determinant of a \(3 \times 3\) matrix is calculated as follows:

\[
\text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg)
\]

For our matrix \(A\):

- \(a = 1, b = 1, c = 2\)
- \(d = 1, e = 0, f = 1\)
- \(g = -1, h = 1, i = -4\)

Calculating the determinant:

\[
\text{det}(A) = 1(0 \cdot -4 - 1 \cdot 1) - 1(1 \cdot -4 - 1 \cdot -1) + 2(1 \cdot 1 - 0 \cdot -1)
\]
\[
= 1(0 - 1) - 1(-4 + 1) + 2(1)
\]
\[
= -1 + 3 + 2 = 4
\]

### Step 2: Calculate the adjugate of \(A\)

The adjugate is the transpose of the cofactor matrix. The cofactor matrix is obtained by calculating the determinant of the \(2 \times 2\) matrices formed by eliminating one row and one column.

1. **Cofactor of \(a_{11}\)**: 
   \[
   \begin{vmatrix}
   0 & 1 \\
   1 & -4
   \end{vmatrix} = (0)(-4) - (1)(1) = -1
   \]

2. **Cofactor of \(a_{12}\)**:
   \[
   \begin{vmatrix}
   1 & 1 \\
   -1 & -4
   \end{vmatrix} = (1)(-4) - (1)(-1) = -4 + 1 = -3
   \]

3. **Cofactor of \(a_{13}\)**:
   \[
   \begin{vmatrix}
   1 & 0 \\
   -1 & 1
   \end{vmatrix} = (1)(1) - (0)(-1) = 1
   \]

4. **Cofactor of \(a_{21}\)**:
   \[
   \begin{vmatrix}
   1 & 2 \\
   -1 & -4
   \end{vmatrix} = (1)(-4) - (2)(-1) = -4 + 2 = -2
   \]

5. **Cofactor of \(a_{22}\)**:
   \[
   \begin{vmatrix}
   1 & 2 \\
   -1 & -4
   \end{vmatrix} = (1)(-4) - (2)(-1) = -4 + 2 = -2
   \]

6. **Cofactor of \(a_{23}\)**:
   \[
   \begin{vmatrix}
   1 & 1 \\
   -1 & 1
   \end{vmatrix} = (1)(1) - (1)(-1) = 1 + 1 = 2
   \]

7. **Cofactor of \(a_{31}\)**:
   \[
   \begin{vmatrix}
   1 & 2 \\
   1 & 1
   \end{vmatrix} = (1)(1) - (2)(1) = 1 - 2 = -1
   \]

8. **Cofactor of \(a_{32}\)**:
   \[
   \begin{vmatrix}
   1 & 2 \\
   1 & 1
   \end{vmatrix} = (1)(1) - (2)(1) = 1 - 2 = -1
   \]

9. **Cofactor of \(a_{33}\)**:
   \[
   \begin{vmatrix}
   1 & 1 \\
   1 & 0
   \end{vmatrix} = (1)(0) - (1)(1) = -1
   \]

The cofactor matrix \(C\) is:

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

### Step 3: Transpose the cofactor matrix to get the adjugate

\[
\text{adj}(A) = C^T = \begin{pmatrix}
-1 & -2 & -1 \\
3 & -2 & -1 \\
1 & 2 & -1
\end{pmatrix}
\]

### Step 4: Calculate the inverse

Now we can find the inverse:

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

### Final Result

Thus, the inverse of the matrix is:

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

If you have any questions or need further details, let me know! Here are some related questions you might find interesting:

1. What is the geometric interpretation of a matrix's inverse?
2. How does the determinant relate to the invertibility of a matrix?
3. Can you explain the process of calculating the adjugate matrix in more detail?
4. What are the applications of inverse matrices in real-world scenarios?
5. How do you check if your calculated inverse is correct? 

**Tip:** When dealing with larger matrices, consider using numerical software for efficiency.

Discover how to find the inverse of a 3x3 matrix through detailed steps, including determinant calculation and adjugate matrix formation. This problem uses a matrix defined by the elements 1, 1, 2, 1, 0, 1, -1, 1, -4, suitable for Grades 10-12 students.

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