Find the inverse of the matrix

1 answer:

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:

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

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

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

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