Find all the possible values of \( t \) if this matrix \( \left[\begin{array}{cc}2 t & 3 \\ t-1 & t+1\end{array}\right] \) is singular.

To determine when the matrix \( \left[\begin{array}{cc}2 t & 3 \\ t-1 & t+1\end{array}\right] \) is singular, we need to find when its determinant is equal to zero. The determinant of a 2x2 matrix \( \left[\begin{array}{cc}a & b \\ c & d\end{array}\right] \) is calculated as \( ad - bc \). For our matrix, we have: - \( a = 2t \) - \( b = 3 \) - \( c = t - 1 \) - \( d = t + 1 \) Now, calculating the determinant: \[ \text{Determinant} = (2t)(t + 1) - (3)(t - 1) \] \[ = 2t^2 + 2t - 3t + 3 \] \[ = 2t^2 - t + 3 \] We set the determinant equal to zero to find the values of \( t \) that make the matrix singular: \[ 2t^2 - t + 3 = 0 \] Next, we use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): Here, \( a = 2 \), \( b = -1 \), and \( c = 3 \). Calculating the discriminant: \[ b^2 - 4ac = (-1)^2 - 4(2)(3) = 1 - 24 = -23 \] Since the discriminant is negative, there are no real solutions for \( t \). Therefore, the matrix \( \left[\begin{array}{cc}2 t & 3 \\ t-1 & t+1\end{array}\right] \) is never singular for real values of \( t \). Thus, the final conclusion is that there are no possible real values of \( t \) that make the matrix singular. So the answer is: No real values of \( t \) make the matrix singular. The determinant is always greater than zero for real \( t \).