If `A^(-1)=[{:(,1,-1,0),(,0,-2,1),(,0,0,-1):}]` then

Options

A. `|A|=2`
B. A is non-singular
C. Adj. A=`=[{:(,1//2,-1//2,0),(,0,-1,1//2),(,0,0,-1//2):}]`
D. A is skew symmetric matrix

Answer

To solve the problem, we need to find the matrix \( A \) given its inverse \( A^{-1} \) and then determine the determinant of \( A \).

Given:
\[ A^{-1} = \begin{pmatrix} 1 & -1 & 0 \\ 0 & -2 & 1 \\ 0 & 0 & -1 \end{pmatrix} \]

### Step 1: Find the matrix \( A \)

We know that:
\[ A \cdot A^{-1} = I \]
where \( I \) is the identity matrix. Therefore, we can express \( A \) as:
\[ A = I \cdot A^{-1} \]
The identity matrix \( I \) for a 3x3 matrix is:
\[ I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

Now, we will compute \( A \) by multiplying \( I \) with \( A^{-1} \):
\[ A = I \cdot A^{-1} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & -1 & 0 \\ 0 & -2 & 1 \\ 0 & 0 & -1 \end{pmatrix} \]

### Step 2: Perform the multiplication

Calculating the product:
\[ A = \begin{pmatrix} 1 \cdot 1 + 0 \cdot 0 + 0 \cdot 0 & 1 \cdot (-1) + 0 \cdot (-2) + 0 \cdot 0 & 1 \cdot 0 + 0 \cdot 1 + 0 \cdot (-1) \\ 0 \cdot 1 + 1 \cdot 0 + 0 \cdot 0 & 0 \cdot (-1) + 1 \cdot (-2) + 0 \cdot 0 & 0 \cdot 0 + 1 \cdot 1 + 0 \cdot (-1) \\ 0 \cdot 1 + 0 \cdot 0 + 1 \cdot 0 & 0 \cdot (-1) + 0 \cdot (-2) + 1 \cdot 0 & 0 \cdot 0 + 0 \cdot 1 + 1 \cdot (-1) \end{pmatrix} \]

This simplifies to:
\[ A = \begin{pmatrix} 1 & -1 & 0 \\ 0 & -2 & 1 \\ 0 & 0 & -1 \end{pmatrix} \]

### Step 3: Calculate the determinant of \( A \)

To find the determinant of \( A \), we can use the formula for the determinant of a 3x3 matrix:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is:
\[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]

For our matrix \( A \):
\[ A = \begin{pmatrix} 1 & -1 & 0 \\ 0 & -2 & 1 \\ 0 & 0 & -1 \end{pmatrix} \]
we have:
- \( a = 1, b = -1, c = 0 \)
- \( d = 0, e = -2, f = 1 \)
- \( g = 0, h = 0, i = -1 \)

Calculating the determinant:
\[ \text{det}(A) = 1((-2)(-1) - (1)(0)) - (-1)(0(-1) - (1)(0)) + 0(0(0) - (-2)(0)) \]
\[ = 1(2 - 0) - (-1)(0 - 0) + 0 \]
\[ = 2 - 0 + 0 = 2 \]

### Conclusion

Thus, the determinant of \( A \) is:
\[ \text{det}(A) = 2 \]