Question

Two schools P and Q want to award their selected students on the values of Discipline, Politeness, and Punctuality. The school P wants to award Rs $x$ each, Rs $y$ each and Rs $z$ each for the three respective values to its 3,2 and 1 students with a total award money of Rs 1000. School Q wants to spend Rs 1500 to award its 4,1 and 3 students on the respective values (by giving the same award money for the three values as before). If the total amount of awards for one prize on each value is Rs 600, using matrices, find the award money for each value. Apart from the above three values, suggest one more value for awards.

Answer 1

In this question it is given that $x,y$ and $z$ is the prize amount per student for the values of Discipline, Politeness, and Punctuality respectively. Now using these values and the statements given in the question form equations. The first statement is school P is awarding a total amount of Rs 1000 to its
3,2 and 1 students for three respective values and hence the equation corresponding to this statement is $3x+2y+z=1000$. Similarly the second statement for school Q and its prize money will give us the equation $4x+y+3z=1500$. The final and last statement is the total amount of awards for one prize on each value is Rs 600 gives the third equation which is $x+y+z=600$. Represent these three equation in matrix form and using the formula of inverse of matrix that is $$${{A}^{-1}}=\dfrac{1}{\left| A \right|}\left( adj\left( A \right) \right)$where$Ais a square matrix and $$adj\left( A \right)$$ represents adjoint of matrixA$and get the values of$x,y$and$z$.

Complete step by step answer:
In the question it is given that $x,y$ and $z$ is the prize amount per student for the values of Discipline, Politeness, and Punctuality respectively.
Now, as is school P is awarding a total amount of Rs 1000 to its 3,2 and 1 students for three respective values, we can also write it in equation form as
$3x+2y+z=1000................(1)$
Similarly, School Q wants to spend Rs 1500 to award its 4,1 and 3 students on the respective values, so it can also be written in equation form as,
$4x+y+3z=1500...............(2)$
And finally, the total amount of awards for one prize on each value is Rs 600, it can also be represented in equation form as,
$x+y+z=600.............(3)$
Now representing these three equations in matrix form, we get
$\therefore \left( \begin{matrix} 3 & 2 & 1 \\\ 4 & 1 & 3 \\\ 1 & 1 & 1 \\\ \end{matrix} \right)\left( \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right)=\left( \begin{matrix} 1000 \\\ 1500 \\\ 600 \\\ \end{matrix} \right)$
Now, represent the above equation as $AX=B$ where $A=\left( \begin{matrix} 3 & 2 & 1 \\\ 4 & 1 & 3 \\\ 1 & 1 & 1 \\\ \end{matrix} \right),X=\left( \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right)$ and $B=\left( \begin{matrix} 1000 \\\ 1500 \\\ 600 \\\ \end{matrix} \right)$.
So, we will find the determinant as $\left| A \right|=3(1-3)-2(4-3)+1(4-1)=-6-2+3=-5$
As, $\left| A \right|\ne 0$, so we get that the matrix is invertible and ${{A}^{-1}}$ exists.
Now the solution of the above equation $AX=B$ can be written as $X={{A}^{-1}}B$.
We also know that ${{A}^{-1}}=\dfrac{1}{\left| A \right|}\left( adj\left( A \right) \right)$
Now adjoint of a matrix is calculated by using cofactors which is represented as ${{C}_{ij}}$. We can find it as shown below,

{{C}_{11}} & {{C}_{12}} & {{C}_{13}} \\\ {{C}_{21}} & {{C}_{22}} & {{C}_{23}} \\\ {{C}_{31}} & {{C}_{32}} & {{C}_{33}} \\\ \end{matrix} \right)}^{T}}$$ ${{C}_{11}}={{\left( -1 \right)}^{1+1}}\left| \begin{matrix} 1 & 3 \\\ 1 & 1 \\\ \end{matrix} \right|=1-3=-2$ ${{C}_{12}}={{\left( -1 \right)}^{1+2}}\left| \begin{matrix} 4 & 3 \\\ 1 & 1 \\\ \end{matrix} \right|=-(4-3)=-1$ $${{C}_{13}}={{\left( -1 \right)}^{1+3}}\left| \begin{matrix} 4 & 1 \\\ 1 & 1 \\\ \end{matrix} \right|=4-1=3$$ ${{C}_{21}}={{\left( -1 \right)}^{2+1}}\left| \begin{matrix} 2 & 1 \\\ 1 & 1 \\\ \end{matrix} \right|=-(2-1)=-1$ ${{C}_{22}}={{\left( -1 \right)}^{2+2}}\left| \begin{matrix} 3 & 1 \\\ 1 & 1 \\\ \end{matrix} \right|=3-1=2$ ${{C}_{23}}={{\left( -1 \right)}^{2+3}}\left| \begin{matrix} 3 & 2 \\\ 1 & 1 \\\ \end{matrix} \right|=-\left( 3-2 \right)=-1$ ${{C}_{31}}={{\left( -1 \right)}^{3+1}}\left| \begin{matrix} 2 & 1 \\\ 1 & 3 \\\ \end{matrix} \right|=6-1=5$ ${{C}_{32}}={{\left( -1 \right)}^{3+2}}\left| \begin{matrix} 3 & 1 \\\ 4 & 3 \\\ \end{matrix} \right|=-\left( 9-4 \right)=-5$ ${{C}_{33}}={{\left( -1 \right)}^{3+3}}\left| \begin{matrix} 3 & 2 \\\ 4 & 1 \\\ \end{matrix} \right|=3-8=-5$ Now, $adj(A)={{\left( \begin{matrix} -2 & -1 & 3 \\\ -1 & 2 & -1 \\\ 5 & -5 & -5 \\\ \end{matrix} \right)}^{T}}=\left( \begin{matrix} -2 & -1 & 5 \\\ -1 & 2 & -5 \\\ 3 & -1 & -5 \\\ \end{matrix} \right)$ As, we have $X={{A}^{-1}}B=\dfrac{1}{\left| A \right|}\left( adj\left( A \right) \right)B$ We can now write the matrix as $\Rightarrow X=\dfrac{-1}{5}\left( \begin{matrix} -2 & -1 & 5 \\\ -1 & 2 & -5 \\\ 3 & -1 & -5 \\\ \end{matrix} \right)\left( \begin{matrix} 1000 \\\ 1500 \\\ 600 \\\ \end{matrix} \right)$ $\Rightarrow \left( \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right)=\dfrac{-1}{5}\left( \begin{matrix} -2000-1500+3000 \\\ -1000+3000-3000 \\\ 3000-1500-3000 \\\ \end{matrix} \right)$ $\Rightarrow \left( \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right)=\dfrac{-1}{5}\left( \begin{matrix} -500 \\\ -1000 \\\ -1500 \\\ \end{matrix} \right)=\left( \begin{matrix} 100 \\\ 200 \\\ 300 \\\ \end{matrix} \right)$ Hence, $x=100,y=200$ and $z=300$ rupees is the prize amount per student for the values of Discipline, Politeness, and Punctuality respectively. Other values like Honesty, Kindness can be considered for award. **Note:** This question checks your understanding of matrices, it’s inverse and ability to form equations from word problems. In this problem we have to formulate equations based on the statements given in the question and after representing them in matrix form, solve it and get the desired value. Forming the right equations is an integral part of the solution. So, always confirm your equations before solving them. Students should carefully take care of signs while calculating cofactors of the matrices as they often make silly mistakes in it. Apart from that there is nothing tricky, calculate the values carefully and get the right answer. A final check can be done after obtaining the values and re-substituting them in the equations.