Let `A_(t)=((1,3,2),(2,5,t),(4,7-t,-6))` then the values (s) of t for which inverse of `A_(t)` does not exist.

Options

A. `-2 ,1 `
B. 3,2
C. 2,-3
D. 3,-1

Answer

To find the values of \( t \) for which the inverse of the matrix \( A(t) = \begin{pmatrix} 1 & 3 & 2 \\ 2 & 5 & t \\ 4 & 7-t & -6 \end{pmatrix} \) does not exist, we need to determine when the determinant of \( A(t) \) is equal to zero.

### Step-by-Step Solution:

1. **Write down the matrix**:
\[ A(t) = \begin{pmatrix} 1 & 3 & 2 \\ 2 & 5 & t \\ 4 & 7-t & -6 \end{pmatrix} \]

2. **Calculate the determinant of \( A(t) \)**:
The determinant of a \( 3 \times 3 \) matrix can be calculated using the formula:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is represented as:
\[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]

For our matrix:
- \( a = 1, b = 3, c = 2 \)
- \( d = 2, e = 5, f = t \)
- \( g = 4, h = 7-t, i = -6 \)

Plugging in these values, we get:
\[ \text{det}(A(t)) = 1 \cdot (5 \cdot (-6) - t \cdot (7-t)) - 3 \cdot (2 \cdot (-6) - t \cdot 4) + 2 \cdot (2 \cdot (7-t) - 5 \cdot 4) \]

3. **Simplify the determinant**:
\[ = 1 \cdot (-30 - t(7-t)) - 3 \cdot (-12 - 4t) + 2 \cdot (14 - 2t - 20) \]
\[ = -30 - 7t + t^2 + 36 + 12t + 2(-6 - 2t) \]
\[ = -30 + 36 - 12 - 7t + 12t - 4t \]
\[ = -6 + t^2 + t \]

Therefore, the determinant simplifies to:
\[ \text{det}(A(t)) = t^2 + 5t - 6 \]

4. **Set the determinant to zero**:
To find when the inverse does not exist, set the determinant equal to zero:
\[ t^2 + 5t - 6 = 0 \]

5. **Factor the quadratic equation**:
\[ (t + 6)(t - 1) = 0 \]

6. **Solve for \( t \)**:
Setting each factor to zero gives:
\[ t + 6 = 0 \quad \Rightarrow \quad t = -6 \]
\[ t - 1 = 0 \quad \Rightarrow \quad t = 1 \]

### Conclusion:
The values of \( t \) for which the inverse of \( A(t) \) does not exist are:
\[ t = -6 \quad \text{and} \quad t = 1 \]