Let `A = [[0, alpha],[0,0]] and(A+I)^(70) - 70 A = [[a-1,b-1],[c-1,d-1]],` the
value of ` a + b + c + d ` is

Explanation

To solve the problem, we need to evaluate the expression \((A + I)^{70} - 70A\) and compare it to the matrix \(\begin{bmatrix} a - 1 & b - 1 \\ c - 1 & d - 1 \end{bmatrix}\).

### Step 1: Define the matrices
Let \( A = \begin{bmatrix} 0 & \alpha \\ 0 & 0 \end{bmatrix} \) and \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \) (the identity matrix).

### Step 2: Calculate \( A + I \)
\[ A + I = \begin{bmatrix} 0 & \alpha \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & \alpha \\ 0 & 1 \end{bmatrix} \]

### Step 3: Find \( (A + I)^{70} \)
To find \((A + I)^{70}\), we can use the binomial theorem:
\[ (A + I)^{70} = I^{70} + \binom{70}{1} A I^{69} + \binom{70}{2} A^2 I^{68} + \ldots + A^{70} \]
Since \( A^2 = 0 \) (as shown in the video), all terms involving \( A^2 \) or higher powers will be zero. Thus, we only need to consider the first two terms:
\[ (A + I)^{70} = I + 70A \]

### Step 4: Substitute back into the equation
Now we substitute this back into the expression:
\[ (A + I)^{70} - 70A = I \]

### Step 5: Compare with the given matrix
We know from the problem statement that:
\[ (A + I)^{70} - 70A = \begin{bmatrix} a - 1 & b - 1 \\ c - 1 & d - 1 \end{bmatrix} \]
Since we found that \( (A + I)^{70} - 70A = I \), we have:
\[ I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]
Thus, we can equate:
\[ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} a - 1 & b - 1 \\ c - 1 & d - 1 \end{bmatrix} \]

### Step 6: Solve for \( a, b, c, d \)
From the above matrices, we can derive the following equations:
1. \( a - 1 = 1 \) → \( a = 2 \)
2. \( b - 1 = 0 \) → \( b = 1 \)
3. \( c - 1 = 0 \) → \( c = 1 \)
4. \( d - 1 = 1 \) → \( d = 2 \)

### Step 7: Calculate \( a + b + c + d \)
Now we can find \( a + b + c + d \):
\[ a + b + c + d = 2 + 1 + 1 + 2 = 6 \]

### Final Answer
Thus, the value of \( a + b + c + d \) is \( \boxed{6} \).

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