## 5 Answers

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

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

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

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

Answer$: A$

### 4 Comments

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

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