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 \( A = \left[\begin{array}{cc} 2t & 3 \\ t-1 & t+1 \end{array}\right] \) is singular, we need to compute its determinant and set it to zero. The determinant of a \( 2 \times 2 \) matrix \( \left[\begin{array}{cc} a & b \\ c & d \end{array}\right] \) is given by the formula: \[ \text{det}(A) = ad - bc \] For our matrix \( A \), we have: - \( a = 2t \) - \( b = 3 \) - \( c = t - 1 \) - \( d = t + 1 \) Now, substituting these values into the determinant formula, we get: \[ \text{det}(A) = (2t)(t + 1) - (3)(t - 1) \] \[ \text{det}(A) = 2t^2 + 2t - 3t + 3 \] \[ \text{det}(A) = 2t^2 - t + 3 \] To find the values of \( t \) for which the matrix is singular, we need to set the determinant equal to zero: \[ 2t^2 - t + 3 = 0 \] Now, we can use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 2 \), \( b = -1 \), and \( c = 3 \): First, we calculate the discriminant \( b^2 - 4ac \): \[ b^2 - 4ac = (-1)^2 - 4(2)(3) = 1 - 24 = -23 \] Since the discriminant is negative, there are no real solutions for the equation \( 2t^2 - t + 3 = 0 \). This implies that the matrix \( A \) is never singular for any real value of \( t \). Hence, there are no possible values of \( t \) that make the matrix singular.