2. Find all the possible values of \( t \) so that the matrix \( \left[\begin{array}{cc}t^{2} & t \\ t+5 & t-3\end{array}\right] \) is invertible.

To determine the values of \( t \) that make the matrix \( \left[\begin{array}{cc}t^{2} & t \\ t+5 & t-3\end{array}\right] \) invertible, we need to find when the determinant of the matrix is non-zero. The determinant for a \( 2 \times 2 \) matrix \( \left[\begin{array}{cc} a & b \\ c & d\end{array}\right] \) is given by \( ad - bc \). For our matrix, we have: - \( a = t^2 \) - \( b = t \) - \( c = t + 5 \) - \( d = t - 3 \) Thus, the determinant is calculated as follows: \[ \text{det} = t^2(t-3) - t(t+5) \] Expanding this, we get: \[ \text{det} = t^3 - 3t^2 - (t^2 + 5t) = t^3 - 4t^2 - 5t \] To find when the determinant is non-zero, we set the determinant equal to zero: \[ t^3 - 4t^2 - 5t = 0 \] Factoring out a \( t \): \[ t(t^2 - 4t - 5) = 0 \] This gives us one solution \( t = 0 \). Now, we need to factor the quadratic \( t^2 - 4t - 5 \): \[ t^2 - 4t - 5 = (t - 5)(t + 1) \] Setting each factor equal to zero gives us the other solutions: \[ t - 5 = 0 \quad \Rightarrow \quad t = 5 \] \[ t + 1 = 0 \quad \Rightarrow \quad t = -1 \] Thus, the values at which the determinant is zero (and thus the matrix is non-invertible) are \( t = 0, 5, -1 \). The matrix is invertible for all other values of \( t \). Therefore, the possible values of \( t \) such that the matrix is invertible are: \[ t \in \mathbb{R} \setminus \{0, 5, -1\} \]