Sum of the values of t `in` C for which the matrix `((1+t,3,2),(2,5,t),(4,7-t,-6))` has no inverse is _______ .

Answer

To find the sum of the values of \( t \) in \( \mathbb{C} \) for which the matrix

\[ A = \begin{pmatrix} 1+t & 3 & 2 \\ 2 & 5 & t \\ 4 & 7-t & -6 \end{pmatrix} \]

has no inverse, we need to calculate the determinant of the matrix and set it equal to zero.

### Step 1: Calculate the determinant of the matrix

The determinant of a 3x3 matrix

\[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]

is given by the formula:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix \( A \):

- \( a = 1+t \)
- \( b = 3 \)
- \( c = 2 \)
- \( d = 2 \)
- \( e = 5 \)
- \( f = t \)
- \( g = 4 \)
- \( h = 7-t \)
- \( i = -6 \)

Substituting these values into the determinant formula:

\[ \text{det}(A) = (1+t) \left( 5 \cdot (-6) - t \cdot (7-t) \right) - 3 \left( 2 \cdot (-6) - t \cdot 4 \right) + 2 \left( 2(7-t) - 5 \cdot 4 \right) \]

### Step 2: Simplify the determinant expression

Calculating each part:

1. **First term**:
\[ 5 \cdot (-6) = -30 \]
\[ t \cdot (7-t) = 7t - t^2 \]
So,
\[ 5 \cdot (-6) - t \cdot (7-t) = -30 - (7t - t^2) = -30 - 7t + t^2 \]

2. **Second term**:
\[ 2 \cdot (-6) = -12 \]
\[ -12 - t \cdot 4 = -12 - 4t \]

3. **Third term**:
\[ 2(7-t) = 14 - 2t \]
\[ 5 \cdot 4 = 20 \]
So,
\[ 2(7-t) - 5 \cdot 4 = 14 - 2t - 20 = -6 - 2t \]

Now substituting these back into the determinant:

\[ \text{det}(A) = (1+t)(t^2 - 7t - 30) - 3(-12 - 4t) + 2(-6 - 2t) \]

Expanding this:

\[ = (1+t)(t^2 - 7t - 30) + 36 + 12t - 12 - 4t \]
\[ = (1+t)(t^2 - 7t - 30) + 24 + 8t \]

### Step 3: Set the determinant to zero

We need to set the determinant equal to zero:

\[ (1+t)(t^2 - 7t - 30) + 24 + 8t = 0 \]

### Step 4: Solve for \( t \)

Expanding \( (1+t)(t^2 - 7t - 30) \):

\[ = t^2 - 7t - 30 + t^3 - 7t^2 - 30t = t^3 - 6t^2 - 60t - 30 + 24 + 8t = t^3 - 6t^2 - 52t - 6 \]

Setting this equal to zero:

\[ t^3 - 6t^2 - 52t - 6 = 0 \]

### Step 5: Find the sum of the roots

Using Vieta's formulas, the sum of the roots \( t_1 + t_2 + t_3 \) of the polynomial \( t^3 + bt^2 + ct + d = 0 \) is given by \( -b \). Here \( b = -6 \), so:

\[ t_1 + t_2 + t_3 = 6 \]

### Final Answer

The sum of the values of \( t \) in \( \mathbb{C} \) for which the matrix has no inverse is \( \boxed{6} \).