`if A=[{:(1,-1,1),(2,1,-1),(-1,-2,2):}]` then show that `A^(-1)` Does not exist.

Explanation

To show that the inverse of the matrix \( A \) does not exist, we need to calculate the determinant of the matrix \( A \) and check if it equals zero. If the determinant is zero, then the inverse does not exist.

Given the matrix:
\[ A = \begin{pmatrix} 1 & -1 & 1 \\ 2 & 1 & -1 \\ -1 & -2 & 2 \end{pmatrix} \]

### Step 1: Calculate the Determinant of Matrix A

The determinant of a \( 3 \times 3 \) matrix can be calculated using the formula:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is represented as:
\[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]

For our matrix \( A \):
- \( a = 1, b = -1, c = 1 \)
- \( d = 2, e = 1, f = -1 \)
- \( g = -1, h = -2, i = 2 \)

Substituting these values into the determinant formula:
\[ \text{det}(A) = 1 \cdot (1 \cdot 2 - (-1) \cdot (-2)) - (-1) \cdot (2 \cdot 2 - (-1) \cdot (-1)) + 1 \cdot (2 \cdot (-2) - 1 \cdot (-1)) \]

### Step 2: Simplify the Determinant Calculation

Calculating each term:
1. \( ei - fh = 1 \cdot 2 - (-1)(-2) = 2 - 2 = 0 \)
2. \( di - fg = 2 \cdot 2 - (-1)(-1) = 4 - 1 = 3 \)
3. \( dh - eg = 2 \cdot (-2) - 1 \cdot (-1) = -4 + 1 = -3 \)

Now substituting these results back into the determinant calculation:
\[ \text{det}(A) = 1 \cdot 0 - (-1) \cdot 3 + 1 \cdot (-3) \]
\[ = 0 + 3 - 3 = 0 \]

### Step 3: Conclusion

Since the determinant of \( A \) is \( 0 \):
\[ \text{det}(A) = 0 \]
This implies that the inverse of matrix \( A \) does not exist.

### Final Statement

Thus, we conclude that \( A^{-1} \) does not exist because the determinant of \( A \) is zero.

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