Let A be a square matrix of order 2 such that `A^(2)-4A+4I=0`, where I is an identity matrix of order 2.
If `B=A^(5)+4A^(4)+6A^(3)+4A^(2)+A-162I` , then det(B) is equal to _________

Explanation

To solve the problem step by step, we begin with the equation given for the matrix \( A \):

1. **Given Equation**:
\[ A^2 - 4A + 4I = 0 \]

2. **Rearranging the Equation**:
We can rewrite the equation as:
\[ A^2 - 4A + 4I = (A - 2I)^2 = 0 \]

3. **Finding Matrix A**:
From the equation \((A - 2I)^2 = 0\), we conclude that:
\[ A - 2I = 0 \implies A = 2I \]

4. **Substituting A into B**:
Now, we substitute \( A = 2I \) into the expression for \( B \):
\[ B = A^5 + 4A^4 + 6A^3 + 4A^2 + A - 162I \]

5. **Calculating Powers of A**:
Since \( A = 2I \), we can calculate the powers:
- \( A^2 = (2I)^2 = 4I \)
- \( A^3 = (2I)^3 = 8I \)
- \( A^4 = (2I)^4 = 16I \)
- \( A^5 = (2I)^5 = 32I \)

6. **Substituting Powers into B**:
Now substituting these values into \( B \):
\[ B = 32I + 4 \cdot 16I + 6 \cdot 8I + 4 \cdot 4I + 2I - 162I \]

7. **Calculating Each Term**:
- \( 4 \cdot 16I = 64I \)
- \( 6 \cdot 8I = 48I \)
- \( 4 \cdot 4I = 16I \)

8. **Combining Terms**:
Now combine all the terms:
\[ B = 32I + 64I + 48I + 16I + 2I - 162I \]
\[ B = (32 + 64 + 48 + 16 + 2 - 162)I \]
\[ B = (162 - 162)I = 0I \]
\[ B = 0 \]

9. **Finding the Determinant**:
Since \( B = 0 \), the determinant of \( B \) is:
\[ \det(B) = \det(0) = 0 \]

Thus, the final answer is:
\[ \text{det}(B) = 0 \]