Question

Find inverse of the matrix
2 1
4 3​

Answer 1

Given :

$A=\left[\begin{array}{cc}\sf2&1\\\sf4&3\end{array}\right]$

To Find :

• Find inverse of the Given matrix.

Formula Applied :

$\displaystyle {\sf A^{-1}=\frac{1}{|A|}adjA}$

Solution :

$\left[\begin{array}{cc}\sf2&1\\\sf4&3\end{array}\right]$

$\sf |A|=(6-4)\implies 2 \neq 0$

• Since A is a nonsingular matrix $\sf A^{-1}$ exist.
• Let $\sf C_{ij}$ be the co factor of $\sf a_{ij}$ in $\sf A = [Aij]^T$

$\sf C_{11}\implies (-1)^{1+1}(3)\implies 3$

$\sf C_{12}\implies (-1)^{1+2}(4)\implies -4$

$\sf C_{11}\implies (-1)^{2+1}(1)\implies -1$

$\sf C_{11}\implies (-1)^{2+2}(2)\implies 2$

$\sf \implies \left[\begin{array}{cc}\sf 3&-4\\\sf-1&2\end{array}\right]^T$

$\sf \therefore adjA = \left[\begin{array}{cc}\sf3&-1\\\sf-4&2\end{array}\right]$

$\displaystyle{\sf A^{-1}=\frac{1}{|A|}adjA}$

$\implies \sf \frac{1}{2}\left[\begin{array}{cc}\sf3&-1\\\sf-4&2\end{array}\right]$

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