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Mathematics Question on Determinants

Solve system of linear equations, using matrix method.
2x+y+z=1
x-2y-z=32\frac{3}{2}
3y-5z=9

Answer

The given system of equations can be written in the form of AX=B, where
A=[211\1−2−1\03−5]\begin{bmatrix}2&1&1\\\1&-2&-1\\\0&3&-5\end{bmatrix},X=[x\y\z]\begin{bmatrix}x\\\y\\\z\end{bmatrix}and B=[132\9]\begin{bmatrix}1\\\\\frac{3}{2}\\\9\end{bmatrix}.
Now, |A|=2(10+3)-1(-5-3)+0=2(13)-1(-8)=26+8=34≠0
Thus, A is non-singular.
Therefore, its inverse exists.
Now, A11=13, A12=5, A13=3
A21=8, A22=-10, A23=-6
A31=1, A32=3, A33=-5
Now, A-1=1∣A∣\frac{1}{\mid A \mid}(adj A)=\frac{1}{34}$$\begin{bmatrix}13&8&1\\\5&-10&3\\\3&-6&-5\end{bmatrix}

∴X=A-1 B=\frac{1}{34}$$\begin{bmatrix}13&8&1\\\5&-10&3\\\3&-6&-5\end{bmatrix}$$\begin{bmatrix}1\\\\\frac{3}{2}\\\9\end{bmatrix}

⇒[x\y\z]\begin{bmatrix}x\\\y\\\z\end{bmatrix}=\frac{1}{34}$$\begin{bmatrix}13+12+9\\\5-15+27\\\3-9-45\end{bmatrix}

=\frac{1}{34}$$\begin{bmatrix}34\\\17\\\\-51\end{bmatrix}=[112−32]\begin{bmatrix}1\\\\\frac{1}{2}\\\\-\frac{3}{2}\end{bmatrix}

Hence, x=1,y=12\frac{1}{2} and z=−32-\frac{3}{2}.