What is the triangle rule for determinants?

2 Views|Posted 5 months ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

5 months ago

This is a mnemonic method that computes the determinant of 3*3 times. It involves drawing diagonal lines for remembring how to multiply and sum elements of the matrix. Say there is a matrix:

| a b c |
| d e f |
| g h i |

We will first multiply the elements that are connected by three diagonals that run

Upvote

Similar Questions for you

$A = [\begin{matrix} 1 & 0 & 0 \\ 0 & α & β \\ 0 & β & α \end{matrix}]$ then α is (if α, β Î I)

Alok Kumar Singh

|2A| = 2⁷

8|A| = 2⁷

Now |A| = α²–β² = 2⁴

α² = 16 + β²

α²– β² = 16

(α–β) (α+β) = 16

->α + β = 8 and

α – β = 2

->α = 5 and β = 3

Alok Kumar Singh

|A| = 3

|B| = 1

->|C| = |ABA^T| = |A|B|A⁷| = |A|²|B|

= 9

->|X| = |A|C|²|A^T|

= 3 * 9² * 3 = 9 * 9² = 729

Alok Kumar Singh

|A| = 2

$\underset{2024 t i m e s}{\underset{?}{a d j (a d j (a d j . . . . (a)))}} = {| A |}^{{(n - 1)}^{2024}}$

$= {| A |}^{{(n - 1)}^{2024}}$

$= 2^{2^{2024}}$ &nb

Vishal Baghel

$\int_{0}^{2} \sqrt[]{2 x d x} - \int_{0}^{2} \sqrt[]{2 x - x^{2} d x} = \int_{0}^{1} d y - \int_{0}^{1} \sqrt[]{1 - y^{2}} d y - \int_{0}^{2} \frac{y^{2}}{2} d y + \int_{1}^{2} 2 d y + I$

$\Rightarrow \frac{8}{3} - \int_{0}^{1} \sqrt[]{1 - t^{2}} d t = 1 - \frac{8}{6} + 2 + I$

$I = 1 - \int_{0}^{1} \sqrt[]{1 - t^{2}} d t$

If $\int_{0}^{2} (\sqrt[]{2 x} - \sqrt[]{2 x - x^{2}}) d x = \int_{0}^{1} (1 - \sqrt[]{1 - y^{2}} - \frac{y^{2}}{2}) d y + \int_{1}^{2} (2 - \frac{y^{2}}{2}) d y + I t h e n I e q u a l s$

Alok Kumar Singh

$\int_{0}^{2} \sqrt[]{2 x d x} - \int_{0}^{2} \sqrt[]{2 x - x^{2} d x} = \int_{0}^{1} d y - \int_{0}^{1} \sqrt[]{1 - y^{2}} d y - \int_{0}^{2} \frac{y^{2}}{2} d y + \int_{1}^{2} 2 d y + I$