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Fix punctuation marks and articles (#20274)

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Sudhakar Kumar
2018-10-20 02:10:10 +05:30
committed by Jonathan Grah
parent f97fec8cf5
commit 3a0974f7e7
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guide/english/mathematics/2-by-2-determinants/index.md
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@@ -4,7 +4,7 @@ title: 2 by 2 Determinants
## 2 by 2 Determinants ## 2 by 2 Determinants
In linear algebra, the determinant of a two-by-two matrix is a useful quantity.Mostly it is used to calculate the area of the given quadilateral(convex polygons only) and is also a easy representation of a quadilateral(convex polygons only) to be stored in computers as arrays. Scientists, engineers, and mathematicians use determinants in many everyday applications including image and graphic processing. In linear algebra, the determinant of a two-by-two matrix is a useful quantity. Mostly it is used to calculate the area of the given quadilateral (convex polygons only) and is also an easy representation of a quadilateral(convex polygons only) to be stored in computers as arrays. Scientists, engineers, and mathematicians use determinants in many everyday applications including image and graphic processing.
Calculating the determinant of a square two-by-two matrix is simple, and is the basis of the [Laplace formula](https://en.wikipedia.org/wiki/Laplace_expansion) used for calculating determinants for larger square matrices. Calculating the determinant of a square two-by-two matrix is simple, and is the basis of the [Laplace formula](https://en.wikipedia.org/wiki/Laplace_expansion) used for calculating determinants for larger square matrices.
@@ -14,7 +14,7 @@ Given a matrix A, the determinant of A (written as |A|) is given by the followin
The rows and vectors of a 2 by 2 matrix can be associated with points on a cartesian plane, such that each row forms a 2D vector. These two vectors form a parallelogram, as shown in the image below. The rows and vectors of a 2 by 2 matrix can be associated with points on a cartesian plane, such that each row forms a 2D vector. These two vectors form a parallelogram, as shown in the image below.
PROOF: PROOF:
Let the vectors be M(a,b),N(c,d) originating from origin in a 2-D plane with an angle (*theta*>0) between them(head of one vector touching tail of another vector) . But in here it doesn't matter because sin(theta)=sin(2(pi)-theta).Then the other point is P(a+c,b+d).The area of the parallelogram is perpendicular distance from one point say N(c,d) to the base vector, M(a,b) multiplied by the length of the base vector, |M(a,b)|. The parallelogram consists of two triangles hence, the area is two times of a triangle. Let the vectors be M(a,b),N(c,d) originating from origin in a 2-D plane with an angle (*theta*>0) between them(head of one vector touching tail of another vector). But in here it doesn't matter because sin(theta)=sin(2(pi)-theta). Then the other point is P(a+c,b+d). The area of the parallelogram is perpendicular distance from one point say N(c,d) to the base vector, M(a,b) multiplied by the length of the base vector, |M(a,b)|. The parallelogram consists of two triangles hence, the area is two times of a triangle.
Let the perpendicular distance be h Let the perpendicular distance be h
h=|N(c,d)|* sin(*theta*(angle between two vectors)) h=|N(c,d)|* sin(*theta*(angle between two vectors))
b=|M(a,b)| b=|M(a,b)|