Question

Find the inverse of the matrix using elementary row operations on the augmented matrix

Answer 1

The inverse of the matrix $\vec A = \left[\begin{array}{cc}1&2\\3&4\end{array}\right]$ is represented by the matrix $\vec A^{-1} = \left[\begin{array}{cc}-2&1\\\frac{3}{2} &-\frac{1}{2} \end{array}\right]$.

How to determine the inverse matrix

A matrix A has an inverse matrix if and only if its determinant is different than 0. Given that we have a matrix formed by 2 rows and 2 columns, we can obtain the following inverse matrix by using the following formula:

$\vec A ^{-1} = \frac{1}{\det (\vec A)} \cdot adj (\vec A)$   (1)

Where $adj (\vec A)$ is the adjoint of the matrix, which is the transposed of the cofactor matrix.

If we know that $\vec A = \left[\begin{array}{cc}1&2\\3&4\end{array}\right]$, then the inverse of the matrix is determined below:

$\det (\vec A) = -2$

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

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

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

The inverse of the matrix $\vec A = \left[\begin{array}{cc}1&2\\3&4\end{array}\right]$ is represented by the matrix $\vec A^{-1} = \left[\begin{array}{cc}-2&1\\\frac{3}{2} &-\frac{1}{2} \end{array}\right]$. $\blacksquare$

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