If `A^(-1)=[(1,-1,2),(0,3,1),(0,0,-1//3)]`, then

Options

A. |A|=-1
B. adjA=`[(-1,1,-2),(0,-3,-1),(0,0,1//3)]`
C. `A=[(1,1//3,7),(0,1//3,1),(0,0,-3)]`
D. `A=[(1,-1//3,-7),(0,-3,0),(0,0,1)]`

Answer

To solve the problem, we need to find the matrix \( A \), the determinant of \( A \), and the adjoint of \( A \) given that \( A^{-1} = \begin{pmatrix} 1 & -1 & 2 \\ 0 & 3 & 1 \\ 0 & 0 & -\frac{1}{3} \end{pmatrix} \).

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

We know that:

\[ A = (A^{-1})^{-1} \]

To find \( A \), we can multiply \( A^{-1} \) by the identity matrix \( I \):

\[ A = I \cdot A^{-1} \]

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 multiply \( I \) with \( A^{-1} \):

\[ A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & -1 & 2 \\ 0 & 3 & 1 \\ 0 & 0 & -\frac{1}{3} \end{pmatrix} \]

Using matrix multiplication:

\[ A = \begin{pmatrix} 1 \cdot 1 + 0 \cdot 0 + 0 \cdot 0 & 1 \cdot -1 + 0 \cdot 3 + 0 \cdot 0 & 1 \cdot 2 + 0 \cdot 1 + 0 \cdot -\frac{1}{3} \\ 0 \cdot 1 + 1 \cdot 0 + 0 \cdot 0 & 0 \cdot -1 + 1 \cdot 3 + 0 \cdot 0 & 0 \cdot 2 + 1 \cdot 1 + 0 \cdot -\frac{1}{3} \\ 0 \cdot 1 + 0 \cdot 0 + 1 \cdot 0 & 0 \cdot -1 + 0 \cdot 3 + 1 \cdot 0 & 0 \cdot 2 + 0 \cdot 1 + 1 \cdot -\frac{1}{3} \end{pmatrix} \]

Calculating each entry:

\[ A = \begin{pmatrix} 1 & -1 & 2 \\ 0 & 3 & 1 \\ 0 & 0 & -\frac{1}{3} \end{pmatrix} \]

### Step 2: Find 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 \( A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \).

For our matrix:

\[ A = \begin{pmatrix} 1 & -1 & 2 \\ 0 & 3 & 1 \\ 0 & 0 & -\frac{1}{3} \end{pmatrix} \]

Calculating the determinant:

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

This simplifies to:

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

### Step 3: Find the adjoint of \( A \)

The adjoint of a matrix \( A \) can be found using the formula:

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

We already have \( \text{det}(A) = -1 \) and \( A^{-1} \):

\[ A^{-1} = \begin{pmatrix} 1 & -1 & 2 \\ 0 & 3 & 1 \\ 0 & 0 & -\frac{1}{3} \end{pmatrix} \]

Now we multiply \( -1 \) with each element of \( A^{-1} \):

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

### Final Answers

1. The matrix \( A \) is:
\[ A = \begin{pmatrix} 1 & -1 & 2 \\ 0 & 3 & 1 \\ 0 & 0 & -\frac{1}{3} \end{pmatrix} \]

2. The determinant \( \text{det}(A) \) is:
\[ \text{det}(A) = -1 \]

3. The adjoint \( \text{adj}(A) \) is:
\[ \text{adj}(A) = \begin{pmatrix} -1 & 1 & -2 \\ 0 & -3 & -1 \\ 0 & 0 & \frac{1}{3} \end{pmatrix} \]