Question

Two institutions decided to award their employees for the three values of resourcefulness, competence and determination in the form of prices at the rate of Rs. x, y and z respectively per person . The first institution decided to award respectively 4,3 and 2 employees with a total price money of Rs. 37000 and the second institution decided to award respectively 5,3 and 4 employees with a total price money of Rs. 47000. If all the three prices per person together amount to Rs. 12000 then using matrix method find the value of x,y and z.

Answer 1

$$\begin{gathered} \mathrm{A}.\mathrm{T}.\mathrm{Q} \\ 4\mathrm{x}+3\mathrm{y}+2\mathrm{z}=37000 \\ 5\mathrm{x}+3\mathrm{y}+4\mathrm{z}=47000 \\ \mathrm{x}+\mathrm{y}+\mathrm{z}=12000 \\ \mathrm{We}\mathrm{can}\mathrm{expressed}\mathrm{these}\mathrm{equations}\mathrm{as}\mathrm{AX}=\mathrm{B}. \\ \mathrm{Where}\mathrm{A}=\left[\begin{array}{ccc}4& 3& 2\\ 5& 3& 4\\ 1& 1& 1\end{array}\right],\mathrm{X}=\left[\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{c}37000\\ 47000\\ 12000\end{array}\right] \\ \left|\mathrm{A}\right|=4\left(3-4\right)-3\left(5-4\right)+2\left(5-3\right)=-4-3+4=-3\ne 0 \\ \mathrm{So},\mathrm{A}\mathrm{is}\mathrm{non}\mathrm{singular}\mathrm{therefore}\mathrm{inverse}\mathrm{exists}. \\ {\mathrm{A}}_{11}=-1{\mathrm{A}}_{12}=-1{\mathrm{A}}_{13}=2 \\ {\mathrm{A}}_{21}=-1{\mathrm{A}}_{22}=2{\mathrm{A}}_{23}=-1 \\ {\mathrm{A}}_{31}=6{\mathrm{A}}_{32}=-6{\mathrm{A}}_{33}=-3 \\ \mathrm{adj}\mathrm{A}=\left[\begin{array}{ccc}-1& -1& 6\\ -1& 2& -6\\ 2& -1& -3\end{array}\right] \\ {\mathrm{A}}^{-1}=\frac{1}{\left|\mathrm{A}\right|}\mathrm{adj}\mathrm{A}=-\frac{1}{3}\left[\begin{array}{ccc}-1& -1& 6\\ -1& 2& -6\\ 2& -1& -3\end{array}\right] \\ \mathrm{X}={\mathrm{A}}^{-1}\mathrm{B}=-\frac{1}{3}\left[\begin{array}{ccc}-1& -1& 6\\ -1& 2& -6\\ 2& -1& -3\end{array}\right]\left[\begin{array}{c}37000\\ 47000\\ 12000\end{array}\right] \\ =-\frac{1}{3}\left[\begin{array}{c}-37000-47000+72000\\ -37000+94000-72000\\ 74000-47000-36000\end{array}\right]=-\frac{1}{3}\left[\begin{array}{c}-12000\\ -15000\\ -9000\end{array}\right]=\left[\begin{array}{c}4000\\ 5000\\ 3000\end{array}\right] \\ \mathrm{So},\mathrm{x}=4000,\mathrm{y}=5000\mathrm{and}\mathrm{z}=3000. \end{gathered}$$