Three students run on a racing track such that their speeds add up to $\displaystyle 6 \mathrm{~km} / \mathrm{h}$. However, double the speed of the third runner added to the speed of the first results in $\displaystyle 7 \mathrm{~km} / \mathrm{h}$. If thrice the speed of the first runner is added to the original speeds of the other two, the result is $\displaystyle 12 \mathrm{~km} / \mathrm{h}$. Using matrix method, find the original speed of each runner.

Let original speed of three runners be $\displaystyle \mathrm{x}, \mathrm{y}$ and z respectively. Then $\displaystyle \mathrm{x}+\mathrm{y}+\mathrm{z}=6 ; \mathrm{x}+2 \mathrm{z}=7 ; 3 \mathrm{x}+\mathrm{y}+\mathrm{z}=12$ Let $\displaystyle \mathrm{A}=\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1 \end{array}\right], X=\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ \mathrm{z}\end{array}\right], B=\left[\begin{array}{c}6 \\ 7 \\ 12\end{array}\right]$ $\displaystyle |\mathrm{A}|=4 \neq 0 \Rightarrow \mathrm{~A}^{-1}$ exists $\displaystyle \mathbf{A X}=\mathbf{B} \Rightarrow \mathbf{X}=\mathbf{A}^{-\mathbf{1}} \mathbf{B}$ $\displaystyle \operatorname{adj}(\mathrm{A})=\left[\begin{array}{ccc} -2 & 0 & 2 \\ 5 & -2 & -1 \\ 1 & 2 & -1 \end{array}\right]$ $\displaystyle \mathrm{A}^{-1}=\frac{1}{4}\left[\begin{array}{ccc} -2 & 0 & 2 \\ 5 & -2 & -1 \\ 1 & 2 & -1 \end{array}\right]$ $\displaystyle \therefore\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ \mathrm{z}\end{array}\right]=\frac{1}{4}\left[\begin{array}{ccc} -2 & 0 & 2 \\ 5 & -2 & -1 \\ 1 & 2 & -1 \end{array}\right]\left[\begin{array}{c}6 \\ 7 \\ 12\end{array}\right]=\left[\begin{array}{l}3 \\ 2\end{array}\right]$ Hence the original speed of three runners are $\displaystyle 3 \mathrm{~km} / \mathrm{h}, 1 \mathrm{~km} / \mathrm{h}$ and $\displaystyle 2 \mathrm{~km} / \mathrm{h}$ respectively.