 Solution

Solve $Ax=b$ for a general right-hand side $b=(b_1,b_2,b_3,b_4{)}^T$. Subtracting the fourth equation from the second gives $x_3=b_2-b_4$. The first and fourth then giveSubstitution into the third equation reduces the remaining condition toIf $t\ne 2$, every right-hand side therefore has a unique solution, and collecting its four coefficients yields the matrix inverseFor $t=2$, the nonzero vector $(-1,-1,0,1{)}^T$ lies in the kernel of a linear map represented by $A$, so $A$ is singular. Hence $A$ is invertible exactly when $t\ne 2$. The same calculation gives $\det A=t-2$; the formula above can also be checked by multiplying its numerator by $A$ to obtain $(t-2)I$.