Consider the following two systems. a. {-6+3y=1
{x+3y=-1
b. {-6+3y=3
{x+3y=-4
(i) Find the inverse of the (common) coefficient matrix of the two systems. A⁻¹=[]
(ii)Find the solutions to the two systems by using the inverse, i.e. by evaluating A⁻¹B where B represents the right hand side (i.e.
Previous question
B=[1 -1]for system (a) and B=[3 -4] for system (b))
solution to system (a):x= ,y=
solution to system (b):x= ,y=

Respuesta :

Answer:

  (i)

  [tex]A^{-1}=\left[\begin{array}{cc}-\dfrac{1}{7}&\dfrac{1}{7}\\\\\dfrac{1}{21}&\dfrac{2}{7}\end{array}\right][/tex]

  (ii) (a) x = -2/7, y = -5/21; (b) x = -1, y = -1

Step-by-step explanation:

Given the following systems of equations, you want the inverse of the coefficient matrix, and the solution to each system found by multiplying that coefficient matrix by the constant vector.

  • -6x +3y = 1
  • x +3y = -1
  • -6x +3y = 3
  • x +3y = -4

Inverse matrix

The calculator display in the attachment shows the coefficient matrix and its inverse. The inverse of a matrix is the transpose of the cofactor matrix, divided by the determinant. For a 2×2 matrix, the transpose of the cofactor matrix is simply the matrix obtained by swapping the diagonal elements, and negating the off-diagonal elements.

Here the determinant is (-6)(3) -(1)(3) = -21. So, the upper left element of the inverse matrix, for example, is 3/(-21) = -1/7, as shown in the attachment.

  [tex]A^{-1}=\left[\begin{array}{cc}-\dfrac{1}{7}&\dfrac{1}{7}\\\\\dfrac{1}{21}&\dfrac{2}{7}\end{array}\right][/tex]

Solutions

Multiplying the inverse matrix (A⁻¹) by each constant column vector (B) gives a result that is a column vector. We can append the constant vectors to form a matrix of the two column vectors, saving a little work in computing the solutions to the two systems. The columns of the result are the solutions to the two systems.

  system (a):  x = -2/7, y = -5/21

  system (b):  x = -1, y = -1

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Additional comment

The second attachment shows the use of an augmented matrix to find both the inverse of the coefficient matrix and the solutions to the systems of equations. The input is the coefficient matrix augmented by a 2×2 identity matrix and the two constant vectors. The output is the identity matrix, the the inverse of the coefficient matrix, and the two solution vectors.

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