Question

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

Answer 1

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

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=\(\frac{1}{\mid A \mid}\)(adj A)=\(\frac{1}{4}\)\(\begin{bmatrix}7&1&3\\-19&-1&11\\-11&1&7\end{bmatrix}\)

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

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

=\(\frac{1}{4}\)\(\begin{bmatrix}8\\4\\12\end{bmatrix}\)=\(\begin{bmatrix}2\\1\\3\end{bmatrix}\)

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