Let `A=[(-1, 2, -3),(-2,0,3),(3,-3, 1)]` be a matrix, then `|a| adj(A^(-1))` is equal to

Options

A. `O_(3xx3)`
B. `[(-1,2,-3),(-2,0,3),(3,-3,1)]`
C. `I_(3)`
D. `[(-3, -3, 1),(3,0,-2),(-1, 2, -3)]`

Answer

To solve the problem, we need to find the value of \(|A| \cdot \text{adj}(A^{-1})\) where \(A = \begin{pmatrix} -1 & 2 & -3 \\ -2 & 0 & 3 \\ 3 & -3 & 1 \end{pmatrix}\).

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

The determinant of a \(3 \times 3\) matrix \(A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}\) is given by:

\[ |A| = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix \(A\):

\[ |A| = -1 \cdot (0 \cdot 1 - 3 \cdot (-3)) - 2 \cdot (-2 \cdot 1 - 3 \cdot 3) + (-3) \cdot (-2 \cdot (-3) - 0 \cdot 3) \]

Calculating each term:

1. First term: \(-1 \cdot (0 + 9) = -9\)
2. Second term: \(-2 \cdot (-2 - 9) = -2 \cdot (-11) = 22\)
3. Third term: \(-3 \cdot (6 - 0) = -18\)

Now, summing these values:

\[ |A| = -9 + 22 - 18 = -5 \]

### Step 2: Find the Adjoint of \(A^{-1}\)

We know that:

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

Thus, we can express \(\text{adj}(A^{-1})\) as:

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

Substituting the determinant we found:

\[ \text{adj}(A^{-1}) = \frac{1}{-5} A \]

### Step 3: Calculate \(|A| \cdot \text{adj}(A^{-1})\)

Now we can compute:

\[ |A| \cdot \text{adj}(A^{-1}) = |A| \cdot \left(\frac{1}{|A|} A\right) = A \]

Since \(|A| = -5\):

\[ |A| \cdot \text{adj}(A^{-1}) = -5 \cdot \left(\frac{1}{-5} A\right) = A \]

### Final Result

Thus, we find that:

\[ |A| \cdot \text{adj}(A^{-1}) = A = \begin{pmatrix} -1 & 2 & -3 \\ -2 & 0 & 3 \\ 3 & -3 & 1 \end{pmatrix} \]

### Conclusion

The answer is the matrix \(A\).

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