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, we will follow these steps:

### Step 1: Solve the equation for matrix A
We start with the equation given in the problem:
\[ A^2 - 4A + 4I = 0 \]

We can rearrange this equation:
\[ A^2 - 4A + 4I = 0 \]
This can be factored as:
\[ (A - 2I)^2 = 0 \]

### Step 2: Find the value of A
From the factored form, we have:
\[ A - 2I = 0 \]
Thus,
\[ A = 2I \]

### Step 3: Substitute A into matrix B
Now, we substitute \( A = 2I \) into the expression for matrix B:
\[ B = A^5 + 4A^4 + 6A^3 + 4A^2 + A - 162I \]

Substituting \( A \):
\[ B = (2I)^5 + 4(2I)^4 + 6(2I)^3 + 4(2I)^2 + (2I) - 162I \]

### Step 4: Calculate powers of A
Calculating each term:
- \( (2I)^5 = 32I \)
- \( 4(2I)^4 = 64I \)
- \( 6(2I)^3 = 48I \)
- \( 4(2I)^2 = 16I \)
- \( (2I) = 2I \)

### Step 5: Combine all terms
Now, we combine all these terms:
\[ B = 32I + 64I + 48I + 16I + 2I - 162I \]

### Step 6: Simplify B
Adding these together:
\[ B = (32 + 64 + 48 + 16 + 2 - 162)I \]
\[ B = (162 - 162)I \]
\[ B = 0I \]
\[ B = 0 \]

### Step 7: Find the determinant of B
The determinant of the zero matrix is:
\[ \text{det}(B) = 0 \]

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

---