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 calculate its determinant and set it not equal to zero. The determinant of a \( 2 \times 2 \) matrix \( \left[\begin{array}{cc} a & b \\ c & d \end{array}\right] \) is given by: \[ \text{det} = ad - bc \] For our matrix, we have: - \( a = t^{2} \) - \( b = t \) - \( c = t + 5 \) - \( d = t - 3 \) Now, we compute the determinant: \[ \text{det} = t^{2}(t - 3) - t(t + 5) \] Expanding this expression: \[ = t^{3} - 3t^{2} - (t^{2} + 5t) \] \[ = t^{3} - 3t^{2} - t^{2} - 5t \] \[ = t^{3} - 4t^{2} - 5t \] For the matrix to be invertible, we need the determinant to be non-zero: \[ t^{3} - 4t^{2} - 5t \neq 0 \] Factoring out \( t \): \[ t(t^{2} - 4t - 5) \neq 0 \] Next, we factor the quadratic \( t^{2} - 4t - 5 \). To do this, we look for two numbers that multiply to \(-5\) and add to \(-4\). These numbers are \(-5\) and \(1\). Thus, we can factor the quadratic as follows: \[ t^{2} - 4t - 5 = (t - 5)(t + 1) \] Putting it all together, we have: \[ t(t - 5)(t + 1) \neq 0 \] This gives us critical points at \( t = 0 \), \( t = 5 \), and \( t = -1 \). The values of \( t \) where the matrix is not invertible are \( t = 0 \), \( t = 5 \), and \( t = -1 \). Therefore, the matrix is invertible for all values of \( t \) except: \[ t \neq 0, 5, -1 \]