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

Options

A. `(1)/(27)[[-1, 26], [0, 27]]`
B. `(1)/(27)[[1, -26], [0, 27]]`
C. `(1)/(27)[[1, -26], [0, -27]]`
D. `(1)/(27)[[-1, -26], [0, -27]]`

Answer

To solve the problem, we need to find the cube of the inverse of the matrix \( A \), where \( A = \begin{bmatrix} 3 & 2 \\ 0 & 1 \end{bmatrix} \).

### Step 1: Calculate the Determinant of \( A \)
The determinant of a \( 2 \times 2 \) matrix \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is given by:
\[ \text{det}(A) = ad - bc \]
For our matrix \( A \):
\[ \text{det}(A) = (3)(1) - (2)(0) = 3 - 0 = 3 \]

### Step 2: Check if the Inverse Exists
Since the determinant \( \text{det}(A) = 3 \) is not equal to 0, the inverse of \( A \) exists.

### Step 3: Calculate the Adjoint of \( A \)
The adjoint of a \( 2 \times 2 \) matrix \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is given by:
\[ \text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \]
For our matrix \( A \):
\[ \text{adj}(A) = \begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix} \]

### Step 4: Calculate the Inverse of \( A \)
The inverse of the matrix \( A \) is given by the formula:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]
Substituting the values we found:
\[ A^{-1} = \frac{1}{3} \cdot \begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} \frac{1}{3} & -\frac{2}{3} \\ 0 & 1 \end{bmatrix} \]

### Step 5: Calculate \( (A^{-1})^3 \)
To find \( (A^{-1})^3 \), we first need to calculate \( (A^{-1})^2 \):
\[ A^{-1} = \begin{bmatrix} \frac{1}{3} & -\frac{2}{3} \\ 0 & 1 \end{bmatrix} \]
Now, we multiply \( A^{-1} \) by itself:
\[ (A^{-1})^2 = A^{-1} \cdot A^{-1} = \begin{bmatrix} \frac{1}{3} & -\frac{2}{3} \\ 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} \frac{1}{3} & -\frac{2}{3} \\ 0 & 1 \end{bmatrix} \]
Calculating the product:
- First row, first column: \( \frac{1}{3} \cdot \frac{1}{3} + (-\frac{2}{3}) \cdot 0 = \frac{1}{9} \)
- First row, second column: \( \frac{1}{3} \cdot (-\frac{2}{3}) + (-\frac{2}{3}) \cdot 1 = -\frac{2}{9} - \frac{2}{3} = -\frac{2}{9} - \frac{6}{9} = -\frac{8}{9} \)
- Second row, first column: \( 0 \cdot \frac{1}{3} + 1 \cdot 0 = 0 \)
- Second row, second column: \( 0 \cdot (-\frac{2}{3}) + 1 \cdot 1 = 1 \)

Thus, we have:
\[ (A^{-1})^2 = \begin{bmatrix} \frac{1}{9} & -\frac{8}{9} \\ 0 & 1 \end{bmatrix} \]

Now we calculate \( (A^{-1})^3 \):
\[ (A^{-1})^3 = (A^{-1})^2 \cdot A^{-1} = \begin{bmatrix} \frac{1}{9} & -\frac{8}{9} \\ 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} \frac{1}{3} & -\frac{2}{3} \\ 0 & 1 \end{bmatrix} \]
Calculating this product:
- First row, first column: \( \frac{1}{9} \cdot \frac{1}{3} + (-\frac{8}{9}) \cdot 0 = \frac{1}{27} \)
- First row, second column: \( \frac{1}{9} \cdot (-\frac{2}{3}) + (-\frac{8}{9}) \cdot 1 = -\frac{2}{27} - \frac{8}{9} = -\frac{2}{27} - \frac{24}{27} = -\frac{26}{27} \)
- Second row, first column: \( 0 \cdot \frac{1}{3} + 1 \cdot 0 = 0 \)
- Second row, second column: \( 0 \cdot (-\frac{2}{3}) + 1 \cdot 1 = 1 \)

Thus, we have:
\[ (A^{-1})^3 = \begin{bmatrix} \frac{1}{27} & -\frac{26}{27} \\ 0 & 1 \end{bmatrix} \]

### Final Result
The final result is:
\[ (A^{-1})^3 = \begin{bmatrix} \frac{1}{27} & -\frac{26}{27} \\ 0 & 1 \end{bmatrix} \]