`(i) if A=[{:(4,-1,-4),(3,0,-4),(3,-1,-3):}],` then show that `A^(2)=I

Explanation

To solve the problem, we need to show that \( A^2 = I \), where \( A = \begin{pmatrix} 4 & -1 & -4 \\ 3 & 0 & -4 \\ 3 & -1 & -3 \end{pmatrix} \) and \( I \) is the identity matrix.

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

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

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

### Step 2: Perform the matrix multiplication

We will calculate each element of the resulting matrix \( A^2 \):

- **Element (1,1)**:
\[ 4 \cdot 4 + (-1) \cdot 3 + (-4) \cdot 3 = 16 - 3 - 12 = 1 \]

- **Element (1,2)**:
\[ 4 \cdot (-1) + (-1) \cdot 0 + (-4) \cdot (-1) = -4 + 0 + 4 = 0 \]

- **Element (1,3)**:
\[ 4 \cdot (-4) + (-1) \cdot (-4) + (-4) \cdot (-3) = -16 + 4 + 12 = 0 \]

- **Element (2,1)**:
\[ 3 \cdot 4 + 0 \cdot 3 + (-4) \cdot 3 = 12 + 0 - 12 = 0 \]

- **Element (2,2)**:
\[ 3 \cdot (-1) + 0 \cdot 0 + (-4) \cdot (-1) = -3 + 0 + 4 = 1 \]

- **Element (2,3)**:
\[ 3 \cdot (-4) + 0 \cdot (-4) + (-4) \cdot (-3) = -12 + 0 + 12 = 0 \]

- **Element (3,1)**:
\[ 3 \cdot 4 + (-1) \cdot 3 + (-3) \cdot 3 = 12 - 3 - 9 = 0 \]

- **Element (3,2)**:
\[ 3 \cdot (-1) + (-1) \cdot 0 + (-3) \cdot (-1) = -3 + 0 + 3 = 0 \]

- **Element (3,3)**:
\[ 3 \cdot (-4) + (-1) \cdot (-4) + (-3) \cdot (-3) = -12 + 4 + 9 = 1 \]

### Step 3: Combine the results

Putting all the calculated elements together, we get:

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

### Conclusion

Thus, we have shown that \( A^2 = I \).

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