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The determinant of the matrix $$\begin{bmatrix}2 &0 &0 &0 \\ 8& 1& 7& 2\\ 2& 0&2 &0 \\ 9&0 & 6 & 1 \end{bmatrix}$$

1. $4$
2. $0$
3. $15$
4. $20$
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$2*\begin{bmatrix} 1 & 7 & 2\\ 0 & 2 & 0\\ 0 & 6 & 1 \end{bmatrix} = 4$

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A really cool way to find the determinant of an nxn matrix when many of its entries are 0.

From row 1 column 1 $\rightarrow$ 2
From row 2 column 2 $\rightarrow$ 1
From row 3 column 2 $\rightarrow$ 2
From row 4 column 2 $\rightarrow$ 1
As no swaps are required so the sign is positive and all the other terms are 0.
Determinant = 2*1*2*1=4

METHOD

The big formula for computing the determinant of any square matrix is:

$det A =$ $\sum_{n! terms}$ $±a_{1\alpha }a_{2\beta }a_{3\gamma }a_{n\omega }$
where (α, β, γ, ...ω) is some permutation of (1, 2, 3, ..., n)

There are n ways to choose an element from the first row (i.e. a value for α), after which there are only n − 1 ways to choose an element from the second row that avoids a zero determinant. Then there are n − 2 choices from the third row, n − 3 from the fourth, and so on. So there are total n! terms in the summation.

Sign of a term is determined by the number of swaps of permutation of 1,2,3,...n required to get back to 1,2,3.. n.
For even number of swaps - positive
For an odd number of swaps - negative
For example- 1,3,2
Only 1 swap is required to get back to 1,2,3

Reference

I reduced it to an upper triangular matrix.

2*1*2*1=4 break it into determinant of smaller pieces where you are doing it in such a way that you should have to do lesser number of operations
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if we do nornally by cofactor method we get 20 as ans why that's not true?

2*1*2*1 = 4 so option A is right