= Solution
{c}
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 give
$$
x_1=b_1-b_2+b_4-x_4,\qquad x_2=b_4-x_4.
$$
Substitution into the third equation reduces the remaining condition to
$$
(t-2)x_4=b_3-b_1-b_4.
$$
If $t\ne2$, every right-hand side therefore has a unique solution, and collecting its four coefficients yields the <matrix inverse>
$$
\boxed{A^{-1}=\frac1{t-2}
\begin{pmatrix}
t-1&2-t&-1&t-1\\
1&0&-1&t-1\\
0&t-2&0&2-t\\
-1&0&1&-1
\end{pmatrix},\qquad t\ne2.}
$$
For $t=2$, the nonzero <vector> $(-1,-1,0,1)^T$ lies in the <kernel of a linear map> represented by $A$, so $A$ is singular. Hence \b[$A$ is invertible exactly when $t\ne2$]. The same calculation gives $\det A=t-2$; the formula above can also be checked by multiplying its numerator by $A$ to obtain $(t-2)I$.
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