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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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<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

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  5. Division
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  1. Substitute in points [Distance Formula]:                                                       \displaystyle d = \sqrt{(5-11)^2+(8-4)^2}
  2. [√Radical] (Parenthesis) Subtract:                                                                 \displaystyle d = \sqrt{(-6)^2+(4)^2}
  3. [√Radical] Evaluate exponents:                                                                    \displaystyle d = \sqrt{36+16}
  4. [√Radical] Add:                                                                                               \displaystyle d = \sqrt{52}
  5. [√Radical] Simplify:                                                                                         \displaystyle d = 2\sqrt{13}
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