Find the determinent of a $4 \times 4$ matrix with the letter $a$ in it Any idea how to compute the determinant of $4 \times 4$ matrix $A$ when
\begin{equation}
A = \begin{bmatrix} 1 & 4 & 8 & 1\\ 0 & 30 & 1 & 0 \\ 0 & 2 & 0 & 0 \\ 1 & 2 & 9 & a \end{bmatrix}.
\end{equation}
The $a$ in the $A_{44}$ location is really confusing me! 
 A: Just expand across the third row, as @Surb suggested:
\begin{equation}
|A| = -2 \left|
\begin{pmatrix}
1 & 8 & 1\\
0 & 1 & 0 \\
1 & 9 & a
\end{pmatrix} \right|
=-2 \left| 
\begin{pmatrix}
1 & 1 \\ 1 & a 
\end{pmatrix} \right| = -2(a - 1) = 2 - 2a.
\end{equation}
Another option is to expand down the fourth column:
\begin{align}
| A| &=
-1 \left|\begin{pmatrix} 0 & 30 & 1 \\ 0 & 2 & 0 \\ 1 & 2 & 9 \end{pmatrix} \right| + a \left |
\begin{pmatrix} 1 & 4 & 8 \\ 0 & 30 & 1 \\ 0 & 2 & 0 \end{pmatrix} \right |.
\end{align}
The two determinants on the right simplify greatly by expanding down the first column:
\begin{equation}
\left|\begin{pmatrix} 0 & 30 & 1 \\ 0 & 2 & 0 \\ 1 & 2 & 9 \end{pmatrix} \right|
= 1\left| \begin{pmatrix} 30 & 1 \\ 2 & 0 \end{pmatrix} \right| = -2
\end{equation}
and
\begin{equation}
\left |
\begin{pmatrix} 1 & 4 & 8 \\ 0 & 30 & 1 \\ 0 & 2 & 0 \end{pmatrix} \right |
= 1 \left| \begin{pmatrix} 30 & 1 \\ 2 & 0 \end{pmatrix} \right| = -2.
\end{equation}
So we see that $|A| = 2 - 2a$.
