If `A=[(-2,3),(-1,1)]` then `A^(3) ` is equal to

Options

A. 2A
B. A
C. 2I.
D. I

Answer

To find \( A^3 \) for the matrix \( A = \begin{pmatrix} -2 & 3 \\ -1 & 1 \end{pmatrix} \), we will follow these steps:

### Step 1: Calculate \( A^2 \)
To find \( A^2 \), we need to multiply matrix \( A \) by itself:

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

Calculating the elements of \( A^2 \):

- First row, first column:
\[ (-2) \cdot (-2) + (3) \cdot (-1) = 4 - 3 = 1 \]

- First row, second column:
\[ (-2) \cdot (3) + (3) \cdot (1) = -6 + 3 = -3 \]

- Second row, first column:
\[ (-1) \cdot (-2) + (1) \cdot (-1) = 2 - 1 = 1 \]

- Second row, second column:
\[ (-1) \cdot (3) + (1) \cdot (1) = -3 + 1 = -2 \]

Thus, we have:
\[ A^2 = \begin{pmatrix} 1 & -3 \\ 1 & -2 \end{pmatrix} \]

### Step 2: Calculate \( A^3 \)
Now, we will calculate \( A^3 \) by multiplying \( A^2 \) by \( A \):

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

Calculating the elements of \( A^3 \):

- First row, first column:
\[ (1) \cdot (-2) + (-3) \cdot (-1) = -2 + 3 = 1 \]

- First row, second column:
\[ (1) \cdot (3) + (-3) \cdot (1) = 3 - 3 = 0 \]

- Second row, first column:
\[ (1) \cdot (-2) + (-2) \cdot (-1) = -2 + 2 = 0 \]

- Second row, second column:
\[ (1) \cdot (3) + (-2) \cdot (1) = 3 - 2 = 1 \]

Thus, we have:
\[ A^3 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

### Conclusion
The matrix \( A^3 \) is equal to the identity matrix \( I \):
\[ A^3 = I \]