If `A = [(0,4,3),(1,-3,-3),(-1,4,4)]`, then find `A^(2)` and hence find `A^(-1)`

Answer

To find \( A^2 \) and then \( A^{-1} \) for the matrix

\[ A = \begin{pmatrix} 0 & 4 & 3 \\ 1 & -3 & -3 \\ -1 & 4 & 4 \end{pmatrix} \]

we will follow these steps:

### Step 1: Calculate \( A^2 \)

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

\[ A^2 = A \times A = \begin{pmatrix} 0 & 4 & 3 \\ 1 & -3 & -3 \\ -1 & 4 & 4 \end{pmatrix} \times \begin{pmatrix} 0 & 4 & 3 \\ 1 & -3 & -3 \\ -1 & 4 & 4 \end{pmatrix} \]

#### Step 1.1: Calculate the first row of \( A^2 \)

- First element: \( 0 \times 0 + 4 \times 1 + 3 \times (-1) = 0 + 4 - 3 = 1 \)
- Second element: \( 0 \times 4 + 4 \times (-3) + 3 \times 4 = 0 - 12 + 12 = 0 \)
- Third element: \( 0 \times 3 + 4 \times (-3) + 3 \times 4 = 0 - 12 + 12 = 0 \)

So, the first row of \( A^2 \) is \( (1, 0, 0) \).

#### Step 1.2: Calculate the second row of \( A^2 \)

- First element: \( 1 \times 0 + (-3) \times 1 + (-3) \times (-1) = 0 - 3 + 3 = 0 \)
- Second element: \( 1 \times 4 + (-3) \times (-3) + (-3) \times 4 = 4 + 9 - 12 = 1 \)
- Third element: \( 1 \times 3 + (-3) \times (-3) + (-3) \times 4 = 3 + 9 - 12 = 0 \)

So, the second row of \( A^2 \) is \( (0, 1, 0) \).

#### Step 1.3: Calculate the third row of \( A^2 \)

- First element: \( -1 \times 0 + 4 \times 1 + 4 \times (-1) = 0 + 4 - 4 = 0 \)
- Second element: \( -1 \times 4 + 4 \times (-3) + 4 \times 4 = -4 - 12 + 16 = 0 \)
- Third element: \( -1 \times 3 + 4 \times (-3) + 4 \times 4 = -3 - 12 + 16 = 1 \)

So, the third row of \( A^2 \) is \( (0, 0, 1) \).

### Conclusion for Step 1

Putting it all together, we have:

\[ A^2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = I \]

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

Since \( A^2 = I \), we can conclude that:

\[ A \times A = I \implies A^{-1} = A \]

Thus, we have:

\[ A^{-1} = \begin{pmatrix} 0 & 4 & 3 \\ 1 & -3 & -3 \\ -1 & 4 & 4 \end{pmatrix} \]

### Final Answer

\[ A^2 = I \quad \text{and} \quad A^{-1} = A = \begin{pmatrix} 0 & 4 & 3 \\ 1 & -3 & -3 \\ -1 & 4 & 4 \end{pmatrix} \]