Q45. - Algebra 2

Question

What is the inverse of $\left[\begin{array}{cc}4& 1\\ 10& 2\end{array}\right]$?

A $\left[\begin{array}{cc}-1& \frac{1}{2}\\ 5& -2\end{array}\right]$ B$\left[\begin{array}{cc}2& -1\\ -10& 4\end{array}\right]$ C$\left[\begin{array}{cc}1& 5\\ \frac{1}{2}& 2\end{array}\right]$  D $\left[\begin{array}{cc}1& 5\\ \frac{1}{2}& 2\end{array}\right]$

Step-by-Step Solution

Answer

The inverse of the given matrix is: $\mathbf{A}\left[\begin{array}{cc}\mathbf{-}\mathbf{1}& \frac{\mathbf{1}}{\mathbf{2}}\\ \mathbf{5}& \mathbf{-}\mathbf{2}\\ & \end{array}\right]$

Step 1 ­- Description of step.

A square matrix $A$ does not have its inverse if $\left|A\right|=0$.

A square matrix $A$ have its inverse if $\left|A\right|\ne 0$.

Step 2 ­- Find the determinant of the given matrix.

The determinant of the given matrix is:

$\begin{array}{c}\left|\begin{array}{cc}4& 1\\ 10& 2\end{array}\right|=\left(4\right)\left(2\right)-\left(10\right)\left(1\right)\\ =8-10\\ =-2\end{array}$

As the determinant of the given matrix is not equal to zero, therefore the inverse of the given matrix exists. 

Step 3 ­- Description of step.

The inverse of $2\times 2$matrix $A=\left[\begin{array}{cc}a& b\\ c& d\\ & \end{array}\right]$ is given by$A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\\ & \end{array}\right]$

and$ad-bc\ne 0$.

Step 4 ­- Description of step.

The inverse of the given matrix is given by:

\

$\begin{array}{c}\frac{1}{\left(4\right)\left(2\right)-\left(10\right)\left(1\right)}\left[\begin{array}{cc}2& -\left(1\right)\\ -\left(10\right)& 4\end{array}\right]=\frac{1}{8-10}\left[\begin{array}{cc}2& -1\\ -10& 4\end{array}\right]\\ =\frac{1}{-2}\left[\begin{array}{cc}2& -1\\ -10& 4\end{array}\right]\\ =\left[\begin{array}{cc}\frac{2}{-2}& \frac{-1}{-2}\\ \frac{-10}{-2}& \frac{4}{-2}\end{array}\right]\\ =\left[\begin{array}{cc}-1& \frac{1}{2}\\ 5& -2\end{array}\right]\end{array}$

Therefore, the inverse of the given matrix is $\left[\begin{array}{cc}-1& \frac{1}{2}\\ 5& -2\\ & \end{array}\right]$

Therefore, the option A is correct.