The sum of three numbers is 2.

Solution:

Let the three numbers be x, y and z.

According to the question,

$x+y+2$

$x+2 y+z=1$

 

$5 x+y+z=6$

The given system of equations can be written in matrix form as follows:

$\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 1 \\ 5 & 1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}2 \\ 1 \\ 6\end{array}\right]$

$A X=B$

Here,

$A=\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 1 \\ 5 & 1 & 1\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}2 \\ 1 \\ 6\end{array}\right]$

$|A|=1(2-1)-1(1-5)+1(1-10)$

$=1+4-9$

 

$=-4$

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}=\left[a_{i j}\right] .$ Then,

$C_{11}=(-1)^{1+1}\left|\begin{array}{ll}2 & 1 \\ 1 & 1\end{array}\right|=1, C_{12}=(-1)^{1+2}\left|\begin{array}{ll}1 & 1 \\ 5 & 1\end{array}\right|=4, C_{13}=(-1)^{1+3}\left|\begin{array}{ll}1 & 2 \\ 5 & 1\end{array}\right|=-9$

$C_{21}=(-1)^{2+1}\left|\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right|=0, C_{22}=(-1)^{2+2}\left|\begin{array}{ll}1 & 1 \\ 5 & 1\end{array}\right|=-4, C_{23}=(-1)^{2+3}\left|\begin{array}{ll}1 & 1 \\ 5 & 1\end{array}\right|=4$

 

$C_{31}=(-1)^{3+1}\left|\begin{array}{ll}1 & 1 \\ 2 & 1\end{array}\right|=-1, C_{32}=(-1)^{3+2}\left|\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right|=0, C_{33}=(-1)^{3+3}\left|\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right|=1$

$\operatorname{adj} A=\left[\begin{array}{ccc}1 & 4 & -9 \\ 0 & -4 & 4 \\ -1 & 0 & 1\end{array}\right]^{T}$

$=\left[\begin{array}{ccc}1 & 0 & -1 \\ 4 & -4 & 0 \\ -9 & 4 & 1\end{array}\right]$

$\Rightarrow A^{-1}=\frac{1}{|A|} \operatorname{adj} A$

$=\frac{1}{-4}\left[\begin{array}{ccc}1 & 0 & -1 \\ 4 & -4 & 0 \\ -9 & 4 & 1\end{array}\right]$

$X=A^{-1} B$

$\Rightarrow X=\frac{1}{-4}\left[\begin{array}{ccc}1 & 0 & -1 \\ 4 & -4 & 0 \\ -9 & 4 & 1\end{array}\right]\left[\begin{array}{l}2 \\ 1 \\ 6\end{array}\right]$

$\Rightarrow X=\frac{1}{-4}\left[\begin{array}{c}2+0-6 \\ 8-4+0 \\ -18+4+6\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{-4}\left[\begin{array}{c}-4 \\ 4 \\ -8\end{array}\right]$

$\therefore x=1, y=-1$ and $z=2$