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Mathematics Question on Determinants

Solve system of linear equations, using matrix method.
x-y+z=4
2x+y-3z=0
x+y+z=2

Answer

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

A=[1−11\21−3\111]\begin{bmatrix}1&-1&1\\\2&1&-3\\\1&1&1\end{bmatrix},X= [x\y\z]\begin{bmatrix}x\\\y\\\z\end{bmatrix}and B=[4\0\2]\begin{bmatrix}4\\\0\\\2\end{bmatrix}.

Now, |A|=1(1+3)+1(2+3)+1(2-1)=4+5+1=10≠0
Thus, A is non-singular.
Therefore, its inverse exists.
Now,A11=4, A12=-5, A13=1, A21=2, A22=0, A23=-2, A31=2, A32=5, A33=3

Now, A-1=1∣A∣\frac{1}{\mid A\mid}(adj A)=\frac{1}{10}$$\begin{bmatrix}4&2&2\\\\-5&0&5\\\1&-2&3\end{bmatrix}

∴X=A-1B=\frac{1}{10}$$\begin{bmatrix}4&2&2\\\\-5&0&5\\\1&-2&3\end{bmatrix}$$\begin{bmatrix}4\\\0\\\2\end{bmatrix}

⇒[x\y\z]\begin{bmatrix}x\\\y\\\z\end{bmatrix}=\frac{1}{10}$$\begin{bmatrix}16+0+4\\\\-20+0+10\\\4+0+6\end{bmatrix}

=\frac{1}{10}$$\begin{bmatrix}20\\\\-10\\\10\end{bmatrix}=[2−1\1]\begin{bmatrix}2\\\\-1\\\1\end{bmatrix}

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