Question

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

Answer 1

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

A=\(\begin{bmatrix}1&-1&1\\2&1&-3\\1&1&1\end{bmatrix}\),X= \(\begin{bmatrix}x\\y\\z\end{bmatrix}\)and B=\(\begin{bmatrix}4\\0\\2\end{bmatrix}\).

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

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

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

=\(\frac{1}{10}\)\(\begin{bmatrix}20\\-10\\10\end{bmatrix}\)=\(\begin{bmatrix}2\\-1\\1\end{bmatrix}\)

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