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?

Multiplying the 1st row by $3$ and then adding it to the 4th row, and then multiplying the 3rd row of the resulting matrix by $-\frac 1 4$ and adding it to the 5th row, we obtain
$$\det \begin{bmatrix} 1 & 2 & 3 &  4 & 1\\ 0 & -1 & 2 & 4 & 2\\ 0 & 0 & 4 & 0 & 0\\ -3 & -6 & -9 & -12 & 4\\ 0 & 0 & 1 & 1 & 1\end{bmatrix} = \det \begin{bmatrix} 1 & 2 & 3 &  4 & 1\\ 0 & -1 & 2 & 4 & 2\\ 0 & 0 & 4 & 0 & 0\\ 0 & 0 & 0 & 0 & 7\\ 0 & 0 & 1 & 1 & 1\end{bmatrix} = \det \begin{bmatrix} 1 & 2 & 3 &  4 & 1\\ 0 & -1 & 2 & 4 & 2\\ 0 & 0 & 4 & 0 & 0\\ 0 & 0 & 0 & 0 & 7\\ 0 & 0 & 0 & 1 & 1\end{bmatrix}$$
We have obtained a block upper triangular matrix. Hence,
$$\det \begin{bmatrix} 1 & 2 & 3 &  4 & 1\\ 0 & -1 & 2 & 4 & 2\\ 0 & 0 & 4 & 0 & 0\\ 0 & 0 & 0 & 0 & 7\\ 0 & 0 & 0 & 1 & 1\end{bmatrix} = \det \begin{bmatrix} 1 & 2 & 3\\ 0 & -1 & 2\\ 0 & 0 & 4\end{bmatrix} \cdot \det \begin{bmatrix} 0 & 7\\ 1 & 1\end{bmatrix} = (-4) \cdot (-7) = 28$$
