If `A=[(1,-2,3),(0,-1,4),(-2,2,1)]`, find `(A')^(-1)`

Answer

To find \((A')^{-1}\) where \(A' = A^T\) (the transpose of matrix \(A\)), we will follow these steps:

1. **Find the transpose of matrix \(A\)**.
2. **Calculate the cofactor matrix of \(A'\)**.
3. **Find the adjoint of \(A'\)**.
4. **Calculate the determinant of \(A'\)**.
5. **Use the formula for the inverse of a matrix**.

Let's go through these steps in detail.

### Step 1: Find the Transpose of Matrix \(A\)

Given:
\[ A = \begin{bmatrix} 1 & -2 & 3 \\ 0 & -1 & 4 \\ -2 & 2 & 1 \end{bmatrix} \]

The transpose \(A'\) is obtained by swapping rows and columns:
\[ A' = \begin{bmatrix} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{bmatrix} \]

### Step 2: Calculate the Cofactor Matrix of \(A'\)

The cofactor \(C_{ij}\) is calculated using the formula:
\[ C_{ij} = (-1)^{i+j} \cdot M_{ij} \]
where \(M_{ij}\) is the minor of the element in the \(i\)-th row and \(j\)-th column.

Let's calculate the cofactors:

- \(C_{11} = (-1)^{1+1} \cdot \text{det}\begin{bmatrix}-1 & 2 \\ 4 & 1\end{bmatrix} = 1 \cdot (-1 \cdot 1 - 4 \cdot 2) = -9\)
- \(C_{12} = (-1)^{1+2} \cdot \text{det}\begin{bmatrix}0 & 2 \\ 3 & 1\end{bmatrix} = -1 \cdot (0 \cdot 1 - 3 \cdot 2) = 6\)
- \(C_{13} = (-1)^{1+3} \cdot \text{det}\begin{bmatrix}0 & -1 \\ -2 & 2\end{bmatrix} = 1 \cdot (0 \cdot 2 - (-1) \cdot (-2)) = -2\)

Continuing this process for all elements, we find the cofactor matrix:
\[ C = \begin{bmatrix} -9 & 6 & -2 \\ -8 & 7 & -4 \\ -2 & 2 & -1 \end{bmatrix} \]

### Step 3: Find the Adjoint of \(A'\)

The adjoint of a matrix is the transpose of the cofactor matrix:
\[ \text{adj}(A') = C^T = \begin{bmatrix} -9 & -8 & -2 \\ 6 & 7 & 2 \\ -2 & -4 & -1 \end{bmatrix} \]

### Step 4: Calculate the Determinant of \(A'\)

The determinant of a \(3 \times 3\) matrix can be calculated using the formula:
\[ \text{det}(A') = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix \(A'\):
\[ \text{det}(A') = 1 \cdot (-1 \cdot 1 - 2 \cdot 4) - 0 + (-2) \cdot (0 - (-1) \cdot -2) = 1 \cdot (-1 - 8) - 0 - 2 \cdot 0 = -9 \]

### Step 5: Use the Formula for the Inverse of a Matrix

The inverse of a matrix is given by:
\[ (A')^{-1} = \frac{1}{\text{det}(A')} \cdot \text{adj}(A') \]
Substituting the values we found:
\[ (A')^{-1} = \frac{1}{-9} \cdot \begin{bmatrix} -9 & -8 & -2 \\ 6 & 7 & 2 \\ -2 & -4 & -1 \end{bmatrix} \]
This simplifies to:
\[ (A')^{-1} = \begin{bmatrix} 1 & \frac{8}{9} & \frac{2}{9} \\ -\frac{2}{3} & -\frac{7}{9} & -\frac{2}{9} \\ \frac{2}{9} & \frac{4}{9} & \frac{1}{9} \end{bmatrix} \]

### Final Answer
\[ (A')^{-1} = \begin{bmatrix} 1 & \frac{8}{9} & \frac{2}{9} \\ -\frac{2}{3} & -\frac{7}{9} & -\frac{2}{9} \\ \frac{2}{9} & \frac{4}{9} & \frac{1}{9} \end{bmatrix} \]