question_answer

Two schools A and B decided to award prizes to their students for three values honesty (x), punctuality (y) and obedience (z). School A decided to award a total of Rs. 11000 for the three values to 5, 4 and 3 students, respectively, while school B decided to award Rs. 10700 for the three values to 4, 3 and 5 students, respectively. If all the three prizes together amount to Rs. 2700, then (i) Represent the above situation by a matrix equation and form linear equations using matrix multiplication. (ii) Is it possible to solve the system of equations, so obtained using matrices? (iii) Which value you prefer to be rewarded most and why?

Answer:

(i) The given information can be written in the matrix form as                         \[\left[ \begin{matrix}    5 & 4 & 3  \\    4 & 3 & 5  \\    1 & 1 & 1  \\ \end{matrix} \right]\,\,\left[ \begin{matrix}    x  \\    y  \\    z  \\ \end{matrix} \right]=\left[ \begin{matrix}    11000  \\    10700  \\    2700  \\ \end{matrix} \right]\] Now, the linear equations are             \[5x+4y+3z=11000\]             \[4x+3y+5z=10700\] and       \[x+y+z=2700\] (ii) Let   \[A=\left[ \begin{matrix}    5 & 4 & 3  \\    4 & 3 & 5  \\    1 & 1 & 1  \\ \end{matrix} \right]\] Then, \[|A|=\left| \begin{matrix}    5 & 4 & 3  \\    4 & 3 & 5  \\    1 & 1 & 1  \\ \end{matrix} \right|\]             \[=5\,(3-5)-4\,(4-5)+3\,(4-3)\]             \[=-10+4+3=-\,3\ne 0\] \[\Rightarrow \] \[{{A}^{-1}}\] exists, so the equation have a unique solution. Hence, it is possible to solve the system of equations by using matrices.             (iii) In my view, obedience should be rewarded most, because at school level, first of all students should be obedient to his/her teachers, then only he/she can liked by everyone.