Two factories decided to award their employees for three values of (a) adaptable tonew techniques, (b) careful and alert in difficult situations and (c) keeping clam in tense situations, at the rate of ₹ x, ₹ y and ₹ z per person respectively. The first factory decided to honour respectively 2, 4 and 3 employees with a total prize money of ₹ 29000. The second factory decided to honour respectively 5, 2 and 3 employees with the prize money of ₹ 30500. If the three prizes per person together cost ₹ 9500, then
i) represent the above situation by matrix equation and form linear equation using matrix multiplication.
ii) Solve these equation by matrix method.
iii) Which values are reflected in the questions?
According to question,
$2 x+3 y+4 z=29000 \quad \ldots . .(1)$
$5 x+2 y+3 z=30500 \quad \ldots . .(2)$
$x+y+z=9500 \quad \ldots . .(3)$
From (1), (2) and (3) we get the matrix equation,
$\left[\begin{array}{lll}2 & 3 & 4 \\ 5 & 2 & 3 \\ 1 & 1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}29000 \\ 30500 \\ 9500\end{array}\right]$
$\Rightarrow\left[\begin{array}{ccc}-2 & -1 & 0 \\ 2 & -1 & 0 \\ 1 & 1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}-9000 \\ 2000 \\ 9500\end{array}\right] \quad\left(R_{1} \rightarrow R_{1}-4 R_{3}\right.$ and $\left.R_{2} \rightarrow R_{2}-3 R_{3}\right)$
$\Rightarrow\left[\begin{array}{ccc}-2 & -1 & 0 \\ -2 & 1 & 0 \\ 1 & 1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}-9000 \\ -2000 \\ 9500\end{array}\right] \quad\left(R_{2} \rightarrow-R_{2}\right)$
$\Rightarrow\left[\begin{array}{ccc}-4 & 0 & 0 \\ -2 & 1 & 0 \\ 3 & 0 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}-11000 \\ -2000 \\ 11500\end{array}\right] \quad\left(R_{1} \rightarrow R_{1}+R_{2}\right.$ and $\left.R_{3} \rightarrow R_{3}-R_{2}\right)$
$\Rightarrow\left[\begin{array}{ccc}1 & 0 & 0 \\ -2 & 1 & 0 \\ 3 & 0 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}2750 \\ -2000 \\ 11500\end{array}\right] \quad\left(R_{1} \rightarrow \frac{-1}{4} R_{1}\right)$
$\Rightarrow\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}2750 \\ 3500 \\ 3250\end{array}\right] \quad\left(R_{2} \rightarrow R_{2}+2 R_{1}\right.$ and $\left.R_{3} \rightarrow R_{3}-3 R_{1}\right)$
$\therefore x=2750, y=3500$ and $z=3250$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.