Question

Find inverse of matrices ​

Answer 1

INVERSE BY ADJOINT METHOD :

check the determinant, whether it's equal to zero or not

$|A|= \begin{vmatrix}3&4&2\\5&5&2\\5&7&3\end{vmatrix}$

$= 3(15 - 14)-4(15-10)+2(35-25)$

$=3-20+20 = 3 \neq 0$

$\implies \sf \:therefore\:A^{-1}\:Exist$

Now, find the co-factors ,

$A11=M11\:(-1)^{1+1}=\begin{vmatrix}5&2\\7&3\end{vmatrix}=(15-14)=1$

$A12=M12\:(-1)^{1+2}=\begin{vmatrix}5&2\\5&3\end{vmatrix}(-1)^{1+2}=-(15-10)=-5$

$A13=M13\:(-1)^{1+3}=\begin{vmatrix}5&5\\5&7\end{vmatrix}(-1)^{1+3}=(35-25)=10$

$A21=M21\:(-1)^{2+1}=\begin{vmatrix}4&2\\7&3\end{vmatrix}(-1)^3=-(12-14)=2$

$A22=M22\:(-1)^{2+2}=\begin{vmatrix}3&2\\5&3\end{vmatrix}(-1)^{4}=(9-10)=-1$

$A23=M23\:(-1)^{2+3}=\begin{vmatrix}3&4\\5&7\end{vmatrix}(-1)^5=-(21-20)=-1$

$A31=M31\:(-1)^{3+1}=\begin{vmatrix}4&2\\5&2\end{vmatrix}(-1)^4=(8-10)=-2$

$A32=M32\:(-1)^{3+2}=\begin{vmatrix}3&2\\5&2\end{vmatrix}(-1)^5=-(6-10)=4$

$A33=M33\:(-1)^{3+3}=\begin{vmatrix}3&4\\5&5\end{vmatrix}(-1)^{3+3}=(15-20)=-5$

Therefore co-factors of matrix is

$\sf \displaystyle{\left[A_{ij}\right]_{3\times3}}=\left[\begin{array}{ccc}A11&A12&A13\\A21&A22&A23\\A31&A32&A33\end{array}\right]=\left[\begin{array}{ccc}1&-5&10\\2&-1&-1\\-2&4&5\end{array}\right]$

$\sf Adj(A)=\sf \displaystyle{\left[A_{ij}\right]^{T}_{3\times3}}=\left[\begin{array}{ccc}1&2&-2\\-5&-1&4\\10&-1&-5\end{array}\right]$

$\sf A^{-1}=\frac{1}{|A|}(adjA)=\frac{1}{|3|}\left[\begin{array}{ccc}1&2&-2\\-5&-1&4\\10&-1&-5\end{array}\right]$