A square matrix A of order 3 satisfies `A^(2)=I-2A`, where I is an identify matrix of order 3. If `A^(n)=29A-12I`, then the value of n is equal to

Options

A. 3
B. 4
C. 5
D. 6

Answer

To solve the problem, we need to find the value of \( n \) such that \( A^n = 29A - 12I \) given that \( A^2 = I - 2A \).

### Step 1: Find \( A^3 \)
We start with the equation given for \( A^2 \):
\[ A^2 = I - 2A \]
To find \( A^3 \), we can use the relation:
\[ A^3 = A \cdot A^2 \]
Substituting the expression for \( A^2 \):
\[ A^3 = A \cdot (I - 2A) = A - 2A^2 \]
Now substitute \( A^2 \) again:
\[ A^3 = A - 2(I - 2A) = A - 2I + 4A = 5A - 2I \]

### Step 2: Find \( A^4 \)
Next, we calculate \( A^4 \):
\[ A^4 = A \cdot A^3 = A \cdot (5A - 2I) = 5A^2 - 2A \]
Substituting \( A^2 \) again:
\[ A^4 = 5(I - 2A) - 2A = 5I - 10A - 2A = 5I - 12A \]

### Step 3: Find \( A^5 \)
Now we find \( A^5 \):
\[ A^5 = A \cdot A^4 = A \cdot (5I - 12A) = 5A - 12A^2 \]
Substituting \( A^2 \) again:
\[ A^5 = 5A - 12(I - 2A) = 5A - 12I + 24A = 29A - 12I \]

### Step 4: Conclusion
We have found that:
\[ A^5 = 29A - 12I \]
Thus, the value of \( n \) is:
\[ \boxed{5} \]