Question

A school wants to award its students for the values of Honesty, Regularity and Hard work with a total cash award of Rs \[6,000\]. Three times the award money for Hard work added to that given for honesty amounts to Rs \[11,000\]. The award money given for Honesty and Hard work together is double the one given for Regularity. Represent the above situation algebraically and find the award money for each value, using the matrix method. Apart from these values, namely, Honesty, Regularity and Hard work, suggest one more value which the school must include for awards.

Answer 1

Hint: First we have to write the given statement as a system of linear equations by considering the unknowns. To solve the system by using the matrix method we have expressed the system in matrix form. Consider the augmented matrix using the row operations converted into a row reduced echelon form. Then check if the rank of the augmented matrix is equal to the rank of the coefficient matrix. If it is equal then a solution exists. Obtain the solution by writing in a system of equations. Finally suggesting one more award to the school must include.

Complete step by step solution:
Let \[h\] be the award money for Honesty. Similarly, let \[r\] be the award money for Regularity and let \[w\]be the award money for Hard work.
The given system of statements can be written in the algebraic linear system of equations as follows
\[ h + r + w = 6000 \\
  3w + h = 11000 \\\]
\[ h + w = 2r \] ----(1)
The system (2) can be rewritten as
\[ h + r + w = 6000 \\
  h + 3w = 11000 \\ \]
\[ h - 2r + w = 0 \]-----(2)
The above system of equations (2) can be written in the matrix form as
\[\left[ {\begin{array}{*{20}{c}}
  \begin{gathered}
  1 \\
  1 \\
  1 \\
\end{gathered} &\begin{gathered}
  1 \\
  0 \\
   - 2 \\
\end{gathered} &\begin{gathered}
  1 \\
  3 \\
  1 \\
\end{gathered}
\end{array}} \right]\left[ \begin{gathered}
  h \\
  r \\
  w \\
\end{gathered} \right] = \left[ \begin{gathered}
  6000 \\
  11000 \\
  0 \\
\end{gathered} \right]\]---(3)
The equation (3) can be written as
\[AX = B\]---(4)
Where \[A = \left[ {\begin{array}{*{20}{c}}
  \begin{gathered}
  1 \\
  1 \\
  1 \\
\end{gathered} &\begin{gathered}
  1 \\
  0 \\
   - 2 \\
\end{gathered} &\begin{gathered}
  1 \\
  3 \\
  1 \\
\end{gathered}
\end{array}} \right]\],\[X = \left[ \begin{gathered}
  h \\
  r \\
  w \\
\end{gathered} \right]\],\[B = \left[ \begin{gathered}
  5000 \\
  11000 \\
  0 \\
\end{gathered} \right]\]
We use row reduced echelon form to find the values of \[h\],\[r\]and \[w\] by constructing the augmented matrix \[\left[ {A:B} \right]\].
Consider \[\left[ {A:B} \right] = \]\[\left[ {\begin{array}{*{20}{c}}
  \begin{gathered}
  1 \\
  1 \\
  1 \\
\end{gathered} &\begin{gathered}
  1 \\
  0 \\
   - 2 \\
\end{gathered} &\begin{gathered}
  1 \\
  3 \\
  1 \\
\end{gathered} &\begin{gathered}
  :\quad 6000 \\
  :\quad 11000 \\
  :\quad 0 \\
\end{gathered}
\end{array}} \right]\]
Using row operations (by row reduced echelon form)
\[\begin{gathered}
  {R_2} \to {R_2} - {R_1} \\
  {R_3} \to {R_3} - {R_1} \\
\end{gathered} \]
\[\left[ {A:B} \right] \approx \]\[\left[ {\begin{array}{*{20}{c}}
  \begin{gathered}
  1 \\
  0 \\
  0 \\
\end{gathered} &\begin{gathered}
  1 \\
   - 1 \\
   - 3 \\
\end{gathered} &\begin{gathered}
  1 \\
  2 \\
  0 \\
\end{gathered} &\begin{gathered}
  :\quad 6000 \\
  :\quad 5000 \\
  :\quad - 6000 \\
\end{gathered}
\end{array}} \right]\]
\[{R_3} \to {R_3} - 3{R_2}\]
\[\left[ {A:B} \right] \approx \]\[\left[ {\begin{array}{*{20}{c}}
  \begin{gathered}
  1 \\
  0 \\
  0 \\
\end{gathered} &\begin{gathered}
  1 \\
   - 1 \\
  0 \\
\end{gathered} &\begin{gathered}
  1 \\
  2 \\
   - 6 \\
\end{gathered} &\begin{gathered}
  :\quad 6000 \\
  :\quad 5000 \\
  :\quad - 21000 \\
\end{gathered}
\end{array}} \right]\]--(4)
\[ \Rightarrow \]\[r\left( {\left[ {A:B} \right]} \right) = 3 = r(A)\]. Hence the solution exists.
The equation (4) can be written as a system of equations as
\[h + r + w = 6000\]--(5)
\[ - r + 2w = 5000\]--(6)
\[ - 6w = - 21000\]--(7)
From the equation (7) we get
\[w = 3500\]--(8)
Using the equation (8) in the equation (6), we get
\[ - r + 7000 = 5000\]
Hence \[r = 2000\]--(9)
Using the equations (8) and (9) in the equation (5), we get
\[h + 2000 + 3500 = 6000\]
Hence \[h = 500\].
Finally, the award money for Honesty, Regularity and Hard work are Rs.\[500\], Rs.\[2000\] and Rs.\[3500\] respectively.
I suggest one more value which the school must include for awards was obedience.

Note:
Note that the rank of the matrix is the number of non-zero rows in the reduced echelon form. Also note that the given system of equations consistent if the number of unknowns can not exceed the number of equations means the number of unknowns is always less than or equal to the number of equations of the given system.