Question

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

Answer 1

The given system of equations can be written in the form of AX=B, where
A=$\begin{bmatrix}2&3&3\\\1&-2&1\\\3&-1&-2\end{bmatrix}$,X=$\begin{bmatrix}x\\\y\\\z\end{bmatrix}$and B=$\begin{bmatrix}5\\\\-4\\\3\end{bmatrix}$.
Now, |A|=2(4+1)-3(-2-3)+3(-1+6)=10+15+15=40≠0
Thus, A is non-singular.
Therefore, its inverse exists.
Now, A11=5, A12=5, A13=5
A21=3, A22=-13, A23=11
A31=9, A32=1, A33=-7
Now, A-1=$\frac{1}{\mid A\mid}$(adj A)=\frac{1}{40}$$\begin{bmatrix}5&3&9\\\5&-13&1\\\5&11&-7\end{bmatrix}

∴X=A-1 B=\frac{1}{40}$$\begin{bmatrix}5&3&9\\\5&-13&1\\\5&11&-7\end{bmatrix}$$\begin{bmatrix}5\\\\-4\\\3\end{bmatrix}

⇒ $\begin{bmatrix}x\\\y\\\z\end{bmatrix}$=$\frac{1}{40}$[$\begin{bmatrix}25-12+27\\\25+52+3\\\25-44-21\end{bmatrix}$=$\begin{bmatrix}1\\\2\\\\-1\end{bmatrix}$

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