How to find the determinant of this $5 \times 5$ matrix? How can I find the determinant of this matrix?

I know in matrix $3 \times 3$ 
$$A= 1(5\cdot 9-8\cdot 6)-2 (4\cdot 9-7\cdot 6)+3(4\cdot 8-7\cdot 5) $$
but how to work with a $5\times 5$ matrix?

Once the matrix starts getting large, it can be easier to use row- or column-reduction to find the determinant, especially if there aren’t many sparse rows or columns to take advantage of in iterated Laplace expansions. This technique makes use of the facts that swapping two rows/columns changes the sign of the determinant, multiplying a row/column by a scalar multiplies the determinant by the same amount, and adding a scalar multiple of a row/column to another leaves the determinant unchanged.  
For your matrix, we can start by adding $3$ times the first row to the fourth: $$\begin{vmatrix}
1 & 2 & 3 & 4 & 1 \\ 
0 & -1 & 2 & 4 & 2 \\ 
0 & 0 & 4 & 0 & 0 \\ 
0 & 0 & 0 & 0 & 7 \\ 
0 & 0 & 1 & 1 & 1  \notag
\end{vmatrix}$$
Clear the first and second rows:$$\begin{vmatrix}
1 & 0 & 0 & 0 & 0 \\ 
0 & -1 & 0 & 0 & 0 \\ 
0 & 0 & 4 & 0 & 0 \\ 
0 & 0 & 0 & 0 & 7 \\ 
0 & 0 & 1 & 1 & 1  \notag
\end{vmatrix}$$
Clear the third and last columns:$$\begin{vmatrix}
1 & 0 & 0 & 0 & 0 \\ 
0 & -1 & 0 & 0 & 0 \\ 
0 & 0 & 4 & 0 & 0 \\ 
0 & 0 & 0 & 0 & 7 \\ 
0 & 0 & 0 & 1 & 0  \notag
\end{vmatrix}$$
Swap the fourth and fifth rows:$$-1\cdot\begin{vmatrix}
1 & 0 & 0 & 0 & 0 \\ 
0 & -1 & 0 & 0 & 0 \\ 
0 & 0 & 4 & 0 & 0 \\ 
0 & 0 & 0 & 1 & 0 \\ 
0 & 0 & 0 & 0 & 7  \notag
\end{vmatrix}$$
At this point, we can stop and multiply the diagonal elements together to find the determinant, which is 28.
Update: I should note that I did much more work than necessary above. You don’t need to perform a complete row-reduction—it’s enough to get the matrix into upper-triangular form since the determinant of such a matrix is also the product of its main diagonal elements. After the first step above, we can skip directly to adding $-\frac14$ times the third row to the last: $$\begin{vmatrix}
1 & 2 & 3 & 4 & 1 \\ 
0 & -1 & 2 & 4 & 2 \\ 
0 & 0 & 4 & 0 & 0 \\ 
0 & 0 & 0 & 0 & 7 \\ 
0 & 0 & 0 & 1 & 1 \notag
\end{vmatrix}$$ and then swap the last two rows$$-1\cdot\begin{vmatrix}
1 & 2 & 3 & 4 & 1 \\ 
0 & -1 & 2 & 4 & 2 \\ 
0 & 0 & 4 & 0 & 0 \\ 
0 & 0 & 0 & 1 & 1 \\ 
0 & 0 & 0 & 0 & 7 \notag
\end{vmatrix}.$$ This obviously has the same determinant as the result of the full row-reduction above.
