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Mathematics Question on Determinants

If A=[235\32−4\11−2]\begin{bmatrix}2&3&5\\\3&2&-4\\\1&1&-2\end{bmatrix},find A-1.UsingA-1 solve the system of equations
2x-3y+5z=11
3x+2y-4z=-5
x+y-2z=-3

Answer

A=[235\32−4\11−2]\begin{bmatrix}2&3&5\\\3&2&-4\\\1&1&-2\end{bmatrix}

∴ |A|=2(+4+4)+3(-6+4)+5(3-2)=0-6+5=-1≠0
Now, A11=0, A12=2, A13=1
A21=-1, A22=-9, A23=-5
A31=2. A32=23, A33=13

∴ A-1=1∣A∣\frac{1}{\mid A\mid}(adj A)=-[0−12\2−923\1−513]\begin{bmatrix}0&-1&2\\\2&-9&23\\\1&-5&13\end{bmatrix}=[01−2−29−23−15−13]\begin{bmatrix}0&1&-2\\\\-2&9&-23\\\\-1&5&-13\end{bmatrix} ...(1)

Now, the given system of equations can be written in the form of AX=B, where

A=[235\32−4\11−2]\begin{bmatrix}2&3&5\\\3&2&-4\\\1&1&-2\end{bmatrix},X=[x\y\z]\begin{bmatrix}x\\\y\\\z\end{bmatrix} and B=[11−5−3]\begin{bmatrix}11\\\\-5\\\\-3\end{bmatrix}

The solution of the system of equations is given by X=A-1 B

X=A-1 B
⇒[x\y\z]\begin{bmatrix}x\\\y\\\z\end{bmatrix}=\begin{bmatrix}0&1&-2\\\\-2&9&-23\\\\-1&5&-13\end{bmatrix}$$\begin{bmatrix}11\\\\-5\\\\-3\end{bmatrix} [using(1)]

=[0−5+6−22−45+69−11−25+39]\begin{bmatrix}0-5+6\\\\-22-45+69\\\\-11-25+39\end{bmatrix} =[1\2\3]\begin{bmatrix}1\\\2\\\3\end{bmatrix}

Hence x=1,y=2 and z=3.