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Mathematics Question on Determinants

Solve system of linear equations, using matrix method.
x-y+2z=7
3x+4y-5z=-5
2x-y+3z=12

Answer

The given system of equations can be written in the form of AX=B,where
A=[1−12\34−5\2−13]\begin{bmatrix}1&-1&2\\\3&4&-5\\\2&-1&3\end{bmatrix},X=[x\y\z]\begin{bmatrix}x\\\y\\\z\end{bmatrix}and B=[7−5\12]\begin{bmatrix}7\\\\-5\\\12\end{bmatrix}.

Now, |A|=1(12-5)+1(9+10)+2(-3-8)=7+19-22=4≠0
Thus, A is non-singular.
Therefore, its inverse exists.
Now, A11=7, A12=-19, A13=-11
A21=1, A22=-1, A23=-1
A31=-3, A32=11, A33=7
Now, A-1=1∣A∣\frac{1}{\mid A \mid}(adj A)=\frac{1}{4}$$\begin{bmatrix}7&1&3\\\\-19&-1&11\\\\-11&1&7\end{bmatrix}

∴X=A-1 B=\frac{1}{4}$$\begin{bmatrix}7&1&3\\\\-19&-1&11\\\\-11&1&7\end{bmatrix}$$\begin{bmatrix}7\\\\-5\\\12\end{bmatrix}

⇒[x\y\z]\begin{bmatrix}x\\\y\\\z\end{bmatrix}=\frac{1}{4}$$\begin{bmatrix}49-5-36\\\\-133+5+132\\\\-77+5+84\end{bmatrix}

=\frac{1}{4}$$\begin{bmatrix}8\\\4\\\12\end{bmatrix}=[2\1\3]\begin{bmatrix}2\\\1\\\3\end{bmatrix}

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