Question

If A=[4136]A=[4316]  then A−1=
matrices​

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{A=\left(\begin{array}{cc}4&1\\3&6\end{array}\right)}$

$\underline{\textbf{To find:}}$

$\mathsf{A^{-1}}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{|A|=\left|\begin{array}{cc}4&1\\3&6\end{array}\right|}$

$\mathsf{|A|=24-3}$

$\mathsf{|A|=21\;\neq\;0}$

$\therefore\mathsf{A^{-1}\;exists}$

$\mathsf{adj\,A=\left(\begin{array}{cc}6&-1\\-3&4\end{array}\right)}$

$\mathsf{Now,}$

$\mathsf{A^{-1}=\dfrac{1}{|A|}\;adj\,A}$

$\implies\boxed{\mathsf{A^{-1}=\dfrac{1}{21}\left(\begin{array}{cc}6&-1\\-3&4\end{array}\right)}}$

$\underline{\textbf{Find more:}}$

X + y + z = 6, 3x - y + 3z = 10, 5x + 5y - 4z = 3.

[Find A-¹ using adjoint method.]​

https://brainly.in/question/30325696  

Solve be matrix inversion method

x-3y-8z+10=0

3x+y=4

2x+5y+6z=13​

https://brainly.in/question/21075691