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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. $20$
edited | 1.1k views

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

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

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

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

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

Answer$: A$
edited
0
+1

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

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

0
@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.

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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
0
if we do nornally by cofactor method we get 20 as ans why that's not true?

1