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balu736 [363]
1 year ago
7

Find the inverse of the matrix

Mathematics
1 answer:
mamaluj [8]1 year ago
5 0

By applying inverse of a matrix, we find that the solution of the system of <em>linear</em> equations is (x, y) = (5/7, - 2/7).

<h3>How to solve a system of equation with inverse matrices</h3>

In linear algebra, systems of <em>linear</em> equations with a unique solution can be represented by the following expression:

\vec A \cdot \vec x = \vec B      (1)

Where:

  • \vec A - Matrix of dependent constants.
  • \vec x - Vector column of variables.
  • \vec B - Vector column of independent constants.

The solution of such systems is defined by:

\vec x = \vec A^{-1}\cdot \vec B

\vec x = \frac{1}{\det(\vec A)}\cdot adj\left(\vec A\right) \cdot \vec B, where \det \left(\vec A\right) \ne 0.

Where:

  • \det \left(\vec A\right) - Determinant of the matrix of dependent constants.
  • adj \left(\vec A\right) - Adjoint of the matrix of dependent constants.

For the case of \vec A \in \mathbb{R}_{2\times 2}, the inverse of \vec A is:

\vec A^{-1} = \frac{1}{\det \left(\vec A\right)} \cdot \left[\begin{array}{cc}a_{22}&-a_{12}\\-a_{21}&a_{11}\end{array}\right]     (2)

If we know that \vec A = \left[\begin{array}{cc}3&4\\5&2\end{array}\right] and \vec B = \left[\begin{array}{cc}1\\3\end{array}\right], then the solution of the system of linear equations is:

\vec A^{-1}= \frac{1}{(3)\cdot (2) - (5) \cdot (4)}\cdot \left[\begin{array}{cc}2&-4\\-5&3\end{array}\right]

\vec A^{-1} = -\frac{1}{14}\cdot \left[\begin{array}{cc}2&-4\\-5&3\end{array}\right]

\vec A^{-1} = \left[\begin{array}{cc}-\frac{1}{7} &\frac{2}{7} \\\frac{5}{14} &-\frac{3}{14} \end{array}\right]

\vec x = \left[\begin{array}{cc}-\frac{1}{7} &\frac{2}{7} \\\frac{5}{14} &-\frac{3}{14} \end{array}\right] \cdot \left[\begin{array}{cc}1\\3\end{array}\right]

\vec x = \left[\begin{array}{cc}\frac{5}{7} \\-\frac{2}{7} \end{array}\right]

By applying inverse of a matrix, we find that the solution of the system of <em>linear</em> equations is (x, y) = (5/7, - 2/7).

To learn more on inverse of matrices: brainly.com/question/4017205

#SPJ1

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Yes they are

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             => JL = \sqrt{(5-2)^{2} + (2-5)^{2}  } = \sqrt{3^{2}+(-3)^{2}  } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}

  • K(1;1); L(5;2)

             =>  KL = \sqrt{(5-1)^{2} + (2-1)^{2}  } = \sqrt{4^{2}+1^{2}  } = \sqrt{1+16}=\sqrt{17}

In the triangle QNP, the sides can be calculate as following:

  • Q(-4;4); N(-3;0)

             => QN = \sqrt{[-3-(-4)]^{2} + (0-4)^{2}  } = \sqrt{1^{2}+(-4)^{2}  } = \sqrt{1+16}=\sqrt{17}

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   => QP = \sqrt{[-7-(-4)]^{2} + (1-4)^{2}  } = \sqrt{(-3)^{2}+(-3)^{2}  } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}

  • N(-3;0); P(-7;1)

             =>  NP = \sqrt{[-7-(-3)]^{2} + (1-0)^{2}  } = \sqrt{(-4)^{2}+1^{2}  } = \sqrt{16+1}=\sqrt{17}

It can be seen that QPN and JKL have: JK = QN; JL = QP; KL = NP

=> They are congruent triangles

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