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 when the matrix \( \begin{bmatrix} t^2 & t \\ t + 5 & t - 3 \end{bmatrix} \) is invertible, we need to calculate its determinant and set it not equal to zero. The determinant of a 2x2 matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is given by \( ad - bc \). For our matrix, we have: - \( a = t^2 \) - \( b = t \) - \( c = t + 5 \) - \( d = t - 3 \) The determinant is \[ \det = t^2(t - 3) - t(t + 5) = t^3 - 3t^2 - t^2 - 5t = t^3 - 4t^2 - 5t. \] Now, we need to set this determinant not equal to zero: \[ t^3 - 4t^2 - 5t \neq 0. \] Factoring out \( t \): \[ t(t^2 - 4t - 5) \neq 0. \] Next, we can factor the quadratic \( t^2 - 4t - 5 \) further: \[ t^2 - 4t - 5 = (t - 5)(t + 1). \] Thus, we have: \[ t(t - 5)(t + 1) \neq 0. \] This will be true as long as: 1. \( t \neq 0 \) 2. \( t \neq 5 \) 3. \( t \neq -1 \) So, the matrix is invertible for all values of \( t \) except \( t = 0, 5, -1 \).