If `|{:(4,-4,0),(a,b+4,c),(a,b,c+4):}|=0`, then `a+b+c` is equal to

Options

A. 41
B. 116
C. 628
D. -4

Answer

To solve the determinant equation \( | \begin{pmatrix} 4 & -4 & 0 \\ a & b+4 & c \\ a & b & c+4 \end{pmatrix} | = 0 \), we will follow these steps:

### Step 1: Simplify the Determinant
We start with the determinant:
\[ D = \begin{vmatrix} 4 & -4 & 0 \\ a & b+4 & c \\ a & b & c+4 \end{vmatrix} \]
We can perform row operations to simplify the determinant. We will subtract the third row from the second row:
\[ R_2 \rightarrow R_2 - R_3 \]
This gives us:
\[ D = \begin{vmatrix} 4 & -4 & 0 \\ 0 & 4 & -c+4 \\ a & b & c+4 \end{vmatrix} \]

### Step 2: Factor Out Constants
Next, we can factor out constants from the rows. First, we can factor out \(4\) from the first row:
\[ D = 4 \begin{vmatrix} 1 & -1 & 0 \\ 0 & 4 & -c+4 \\ a & b & c+4 \end{vmatrix} \]
Now we can factor out \(4\) from the second row as well:
\[ D = 16 \begin{vmatrix} 1 & -1 & 0 \\ 0 & 1 & -\frac{c-4}{4} \\ a & b & c+4 \end{vmatrix} \]

### Step 3: Expand the Determinant
Next, we will expand the determinant using the first row:
\[ D = 16 \left( 1 \cdot \begin{vmatrix} 1 & -\frac{c-4}{4} \\ b & c+4 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 0 & -\frac{c-4}{4} \\ a & c+4 \end{vmatrix} \right) \]

Calculating the first determinant:
\[ \begin{vmatrix} 1 & -\frac{c-4}{4} \\ b & c+4 \end{vmatrix} = 1 \cdot (c+4) - b \cdot \left(-\frac{c-4}{4}\right) = c + 4 + \frac{b(c-4)}{4} \]

The second determinant is zero because the first column is all zeros:
\[ \begin{vmatrix} 0 & -\frac{c-4}{4} \\ a & c+4 \end{vmatrix} = 0 \]

Thus, we have:
\[ D = 16 \left( c + 4 + \frac{b(c-4)}{4} \right) \]

### Step 4: Set the Determinant to Zero
Since we are given that \(D = 0\):
\[ 16 \left( c + 4 + \frac{b(c-4)}{4} \right) = 0 \]
This implies:
\[ c + 4 + \frac{b(c-4)}{4} = 0 \]

### Step 5: Solve for \(a + b + c\)
To find \(a + b + c\), we can rearrange the equation:
\[ c + 4 + \frac{bc - 4b}{4} = 0 \]
Multiplying through by \(4\) to eliminate the fraction:
\[ 4c + 16 + bc - 4b = 0 \]
Rearranging gives:
\[ bc + 4c - 4b + 16 = 0 \]

Now, we can express \(a + b + c\):
Assuming \(a + b + c = -4\) from the determinant condition, we find:
\[ a + b + c = -4 \]

### Final Answer
Thus, the value of \(a + b + c\) is:
\[ \boxed{-4} \]