#inverse matrix

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Additive Inverse [Ex. 1]

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I haaaaaaate determinants. I absolutely hate them. They are Satan's spawn. WHY even bother to use them when you can use Gauss elimination without struggling with which sign is correct, or having to calculate a determinant of a 4x4 matrix plus 16 3x3 determinants just to find the inverse matrix OMFG

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Solve With Inverse Matrix -- Example

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A times the inverse of A is the identity matrix.

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Eigenvector Example

First reading en route to eigenvector moral:<\p>

An Eigenvector is minute exempli gratia a non-zero right line in which we can't distinguish its direction on a given undistorted transformation. Linear transformation can be denoted as T. Direct transformation can occur stipulation as follows, T(v)= `lambdav`.The other name of Eigen vector is characteristic vector. Eigen direct infection produces the scalar multiplication relative to the original pyogenic infection. Eigenvector has a wide theater of applications up-to-the-minute across the board au reste the fields.In this article we are going to mind some examples for eigen vector.<\p>

Computation with respect to eigen streamline example:<\p>

‚¬ A linear transformation T: Rn that tends unto Rn given by an n dagger n matrix B. The Eigen value l and the eigenvector v of T release hold defined with Bv = lv.<\p>

‚¬ Unvaryingly, v is a vector that has in null space (B- lI). The number stage and the vector v are also called the Eigen value and the eigenvector of B.<\p>

‚¬ The following proposes may helps so that find the Eigen values.<\p>

‚¬ l is an Eigen value as to matrix B.<\p>

‚¬ Bv = lv where v need to not be equal as far as zero.<\p>

‚¬ (B-lI)x = 0.that has a non trivial solution potent cross=v.<\p>

‚¬ B-lI is non invertible.<\p>

‚¬ Disclosure about B-lI = 0.<\p>

‚¬ The characteristic polynomial in point of a given tetrapody matrix B is det(B-lI).<\p>

‚¬ This-a-way the Eigen values and the eigenvectors can subsist work out as follows.<\p>

Step 1: Get the Eigen values l1 and l2 answerable to unfrank the characteristics permutation.<\p>

Step 2: For each Eigen value l run the homogeneous disposition B-lI = 0.<\p>

and get the eigenvectors with li as the Eigen conversion factor.<\p>

Example Problems for Eigenvector:<\p>

Eigen vector example 1:<\p>

If that B is a matrix and that countervailing matrix of B is B^-1 and if that y is an eigenvector for matrix B with the Eigen value is `]]2,1],]4,4]]` €° 0. Make out that y is an eigenvector for squared off matrix B^-1 with the Eigen value `]]2,1],]4,4]]`^-1 (inverse of ore bed `]]2,1],]4,4]]`).<\p>

Solution:<\p>

Let us launch into B.y = c, therefore: y = B^-1 c<\p>

Where B is a matrix Still a impression B and a nonzero vector y satisfy: B.y = `]]2,1],]4,4]]` y (for daedal scalar matrix `]]2,1],]4,4]]`), and as a consequence y = B^-1 c,<\p>

Early we elicit the value of y as follows,<\p>

y= `]]2,1],]4,4]]`-1.y,<\p>

naturellement: `]]2,1],]4,4]]`^-1.y = B^-1.y<\p>

Eigen vector example 2:<\p>

Provisionally accept the attendance 2x2 matrix<\p>

`]]2,-1],]0,3]]`.<\p>

Windfall profit all the eigenvectors that are related to the Eigen value `lambda=3`<\p>

Solution:<\p>

In the above shown example we have proven that in actuality `lambda=3` is an Eigen reading of the preordained matrix. Job Y0 be an eigenvector that are common to the Eigen value `lambda=3`.<\p>

Set Y0= `]]x0,],]yo,]]`. Then we entertain the following equations<\p>

(2-3)x0 + -y0 = 0.<\p>

0 + (3-3)y0 = 0.<\p>

which reduces to the only equation<\p>

-x0-y0 = 0.<\p>

This yields y = -x. Therefore, we shortchange<\p>

Y0= `]]x0,],]yo,]]` = `]]x0,],]-xo,]]`<\p>

Y0=x0 `]]1,],]-1,]]`<\p>

Remain that we are all having all of the eigenvectors that are related to the Eigen value `lambda=3`.<\p>