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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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$\text{Let }|A|=\begin{vmatrix}2 &0 &0 &0 \\ 8& 1& 7& 2\\ 2& 0&2 &0 \\ 9&0 & 6 & 1 \end{vmatrix}$

$\implies |A|=2\times\begin{vmatrix} 1& 7& 2\\ 0&2 &0 \\ 0 & 6 & 1 \end{vmatrix}$

$\implies |A|=2\times 1\times \begin{vmatrix} 2 &0 \\ 6 & 1 \end{vmatrix}$

$\implies |A|=2\times 1\times (2-0)$

$\implies |A|=2\times 1\times 2=4$

Answer$: A$

edited by

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 3 $\rightarrow$ 2
From row 4 column 4 $\rightarrow$ 1
$(\alpha,\beta,\gamma,\delta)=(1,2,3,4)$
As no swaps are required to get back to $(1,2,3,4)$, 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

Soumya29  the determinant you calculated as product of diagonal entries is after applying elimination method, right?

@meghna..  No..I applied it directly and this is a general procedure. You can apply it to find the determinant of any matrix that has lots of 0's in it. Upper triangular one is a special case of this method only. You can check the reference link in my above for it.

I reduced it to an upper triangular matrix.

### 1 comment

after reducing R4=R4-9/2R1 ,

in row 4th  column  3rd  there should be 6.

you wrote 0???

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

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

### 1 comment

if we do nornally by cofactor method we get 20 as ans why that's not true?
in this type of  question ask what is determinant of 4x4 matrix then first check which row or column has maximum number of zero’s.then find  co-factor of this row and it will answer.

for example  first i will take first row and first element

co-factor=2*(2*1-0)=4.

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