Browse Questions

# write $A^{-1}$ for $A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}$

Toolbox:
• The determinant value of $2 \times 2$ matrix is
• $|A|=a_{11} \times a_{22}- a_{12} \times a_{21}$
• The inverse of a $2 \times 2$ matrix is
• $\frac{1}{ | A | } \begin{bmatrix} a_{22} & -a_{12} \\ -a_{21} & a_{11}\end{bmatrix}$
Given $A=\begin{bmatrix} 2 & 5\\ 1 & 3 \end{bmatrix}$
Let us find the value of the determinant of the matrix
$|A| = 2\times 3 - 5 \times 1$
$=6-5=1$
$|A| \neq 0$
Hence it is a non singuar matrix and $A^{-1} exists.$
Adjoint of $2 \times 2$ matrix can be found by intyerchanging the element $a_11$ and $a_12$ and change the symbols of the elements of $a_12$ and $a_21$
Hence adjoint of $A=\begin{bmatrix} 3 & -5\\ -1 & 2 \end{bmatrix}$
Therefore $A^{-1}=\frac{1}{1} \begin{bmatrix} 3 & -5\\ -1 & 2 \end{bmatrix}=\begin{bmatrix} 3 & -5\\ -1 & 2 \end{bmatrix}$
Solution:$A^{-1}=\begin{bmatrix} 3 & -5\\ -1 & 2 \end{bmatrix}$