Linear Algebra -finding determinant of a matrix

[Unusualskill]

the first row 1 0 0 2
the 2nd row 0 1 2 0
the 3rd row 0 2 1 0
the 4th row 2 0 0 1

I would like to ask which is the most efficient way of solving this ques.Though i can solve but is long method, I know there must have some quick 1, appreciate if u can share it. thank you

 

[SteamKing]

Expansion using minors is your best approach.

 

[Unusualskill]

Expansion using minors is your best approach.

minors?

 

[Mark44]

minors?

If a_ij is an element in matrix A, the minor of this element, A_ij, is the submatrix that contains all of the elements of A that are not in row i or in column j.
For example, a₁₁ in your matrix is 1. The minor of this element is A₁₁, which is
\begin{bmatrix} 1 & 2 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}

For more details of how this works to calculate the determinant, do a web search for "determinant minor".

 

[HallsofIvy]

Expanding by minors on the first row:
$$\left|\begin{array}{cccc}1 & 0 & 0 & 2 \\ 0 & 1 & 2 & 0 \\ 0 & 2 & 1 & 0 \\ 2 & 0 & 0 & 1 \end{array}\right|= 1\left|\begin{array}{ccc} 1 & 2 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|- 2 \left|\begin{array}{ccc}0 & 0 & 2 \\ 1 & 2 & 0 \\ 2 & 1 & 0 \end{array}\right|$$.
You could then expand the first of those two determinants on the last row and the second on the first row.

You can also use "row reduction" to reduce the underlying matrix to a triangular matrix. The determinant of a triangular matrix is just the product of the numbers on the diagonal and "row reduction" changes the determinant is a regular way. The basic "row operation" are
1) multiply every number on a row by the same number. This multiplies the determinant by that number.
2) swap two rows. This multiplies the determinant by -1.
3) add a multiple of a row to another row. This does not change the determinant- and it is always possible to reduce a matrix to a triangular matrix using only this.

Here, if you add -2 times the first row to the last row you get $$\begin{bmatrix}1& 0 & 0 & 2 \\ 0 & 1 & 2 & 0 \\ 0 & 2 & 1 & 0 \\ 0 & 0 & 0 & -3\end{bmatrix}$$.

Now, add -2 times the second row to the third row: $$\begin{bmatrix}1 & 0 & 0 & 2 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & -3 & 0 \\ 0 & 0 & 0 & -3\end{bmatrix}$$

Since we used only "add a multiple of one row to another" that matrix has the same determinant as the original.