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 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: We start solving the above problem by taking the values of Honesty, Regularity and Hard work as $x,y$ and $z$respectively. Then, we represent the above given statements in algebraic forms. We get a system of three linear equations. We represent them in the matrix form $AX=B$. We have to find the solution that is, matrix $X={{A}^{-1}}B$. Before finding the solution, we make sure that the matrix ${{A}^{-1}}$ exists by checking whether the determinant of matrix $A$ is zero or not. After making sure that the determinant of matrix $A$ is not equal to zero, we find the matrix ${{A}^{-1}}$ using the formula $\dfrac{1}{\left| A \right|}Adj\left( A \right)$. Then we obtain the final result $X$ using the formula $X={{A}^{-1}}B$.

Complete step by step answer:
We were given that the total cash award for the values of Honesty, Regularity and Hard work is Rs 6,000. So, we get,
$\Rightarrow x+y+z=6000............\left( 1 \right)$
And we were also given that three times the award money for Hard work added to that given for honesty amounts to Rs 11,000. So, we get,
$\Rightarrow x+3z=11000.............\left( 2 \right)$
We were also given that the award money given for Honesty and Hard work together is double the one given for Regularity. So, we get,
$\begin{align}
  & \Rightarrow x+z=2y \\
 & \Rightarrow x-2y+z=0...............\left( 3 \right) \\
\end{align}$
Now, we represent this system of linear equations in matrix form that is in form of, $AX=B$ as
$A=\left[ \begin{matrix}
   1 & 1 & 1 \\
   1 & 0 & 3 \\
   1 & -2 & 1 \\
\end{matrix} \right]$ , $X=\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]$, $B=\left[ \begin{matrix}
   6000 \\
   11000 \\
   0 \\
\end{matrix} \right]$
So, let us check whether the determinant is equal to zero or not equal to zero.
Now, let us consider the formula,
$\left| \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right|={{a}_{11}}\left( {{a}_{22}}{{a}_{33}}-{{a}_{23}}{{a}_{32}} \right)-{{a}_{21}}\left( {{a}_{12}}{{a}_{33}}-{{a}_{13}}{{a}_{32}} \right)+{{a}_{31}}\left( {{a}_{12}}{{a}_{23}}-{{a}_{13}}{{a}_{22}} \right)$
By using the above formula,
$\begin{align}
  & \left| A \right|=\left| \begin{matrix}
   1 & 1 & 1 \\
   1 & 0 & 3 \\
   1 & -2 & 1 \\
\end{matrix} \right|=1\left( 0+6 \right)-1\left( 1+2 \right)+1\left( 3-0 \right) \\
 & \left| A \right|=1\left( 6 \right)-1\left( 3 \right)+1\left( 3 \right) \\
 & \left| A \right|=6-3+3 \\
 & \left| A \right|=6 \\
\end{align}$
So, we got
$\left| A \right|\ne 0$
So, we can say that ${{A}^{-1}}$exists.
Let us consider the formula,
${{A}^{-1}}=\dfrac{1}{\left| A \right|}AdjA$
So now, we find the matrix $AdjA$
To find the adjoint of a matrix, first we find the cofactor matrix of the given matrix. Then we find the transpose of the cofactor matrix.
We have, $A=\left[ \begin{matrix}
   1 & 1 & 1 \\
   1 & 0 & 3 \\
   1 & -2 & 1 \\
\end{matrix} \right]$
Let us find the cofactor matrix for A first.
Cofactor of 1 = $\left| \begin{matrix}
   0 & 3 \\
   -2 & 1 \\
\end{matrix} \right|=0\left( 1 \right)-3\left( -2 \right)=6$
Cofactor of 1 = $-\left| \begin{matrix}
   1 & 3 \\
   1 & 1 \\
\end{matrix} \right|=-\left( 1\left( 1 \right)-3\left( 1 \right) \right)=2$
Cofactor of 1 = $\left| \begin{matrix}
   1 & 0 \\
   1 & -2 \\
\end{matrix} \right|=1\left( -2 \right)-0\left( 1 \right)=-2$
Cofactor of 1 = $-\left| \begin{matrix}
   1 & 1 \\
   -2 & 1 \\
\end{matrix} \right|=-\left( 1\left( 1 \right)-1\left( -2 \right) \right)=-3$
Cofactor of 0 = $\left| \begin{matrix}
   1 & 1 \\
   1 & 1 \\
\end{matrix} \right|=1\left( 1 \right)-1\left( 1 \right)=0$
Cofactor of 3 = $-\left| \begin{matrix}
   1 & 1 \\
   1 & -2 \\
\end{matrix} \right|=-\left( 1\left( -2 \right)-1\left( 1 \right) \right)=3$
Cofactor of 1 = $\left| \begin{matrix}
   1 & 1 \\
   0 & 3 \\
\end{matrix} \right|=1\left( 3 \right)-1\left( 0 \right)=3$
Cofactor of -2 = $-\left| \begin{matrix}
   1 & 1 \\
   1 & 3 \\
\end{matrix} \right|=-\left( 1\left( 3 \right)-1\left( 1 \right) \right)=-2$
Cofactor of 1 = $\left| \begin{matrix}
   1 & 1 \\
   1 & 0 \\
\end{matrix} \right|=1\left( 0 \right)-1\left( 1 \right)=-1$
So, we get that cofactor matrix as
$\left[ \begin{matrix}
   6 & 2 & -2 \\
   -3 & 0 & 3 \\
   3 & -2 & -1 \\
\end{matrix} \right]$
Now, we need to find $AdjA$, by taking the transpose of cofactor matrix.
$AdjA=\left[ \begin{matrix}
   6 & -3 & 3 \\
   2 & 0 & -2 \\
   -2 & 3 & -1 \\
\end{matrix} \right]$
Now, let us consider the formula, ${{A}^{-1}}=\dfrac{1}{\left| A \right|}AdjA$
By using the above formula, we get
${{A}^{-1}}=\dfrac{1}{6}\left[ \begin{matrix}
   6 & -3 & 3 \\
   2 & 0 & -2 \\
   -2 & 3 & -1 \\
\end{matrix} \right]$
Let us consider the formula,$X={{A}^{-1}}B$
By using the above formula, we get
$\begin{align}
  & X=\dfrac{1}{6}\left[ \begin{matrix}
   6 & -3 & 3 \\
   2 & 0 & -2 \\
   -2 & 3 & -1 \\
\end{matrix} \right]\left[ \begin{matrix}
   6000 \\
   11000 \\
   0 \\
\end{matrix} \right] \\
 & \\
 & X=\dfrac{1}{6}\left[ \begin{matrix}
   6\left( 6000 \right)-3\left( 11000 \right)+3\left( 0 \right) \\
   2\left( 6000 \right)+0\left( 11000 \right)-2\left( 0 \right) \\
   -2\left( 6000 \right)+3\left( 11000 \right)-1\left( 0 \right) \\
\end{matrix} \right] \\
 & \\
 & X=\dfrac{1}{6}\left[ \begin{matrix}
   36000-33000+0 \\
   12000+0-0 \\
   -12000+33000-0 \\
\end{matrix} \right] \\
 & \\
 & X=\dfrac{1}{6}\left[ \begin{matrix}
   3000 \\
   12000 \\
   21000 \\
\end{matrix} \right] \\
 & \\
 & X=\left[ \begin{matrix}
   \dfrac{3000}{6} \\
   \dfrac{12000}{6} \\
   \dfrac{21000}{6} \\
\end{matrix} \right] \\
 & \\
 & X=\left[ \begin{matrix}
   500 \\
   2000 \\
   3500 \\
\end{matrix} \right] \\
 & \\
\end{align}$
Therefore, we get $x=500$, $y=2000$ and $z=3500$.

Hence, the values of Honesty, Regularity and Hard work are 500, 2000 and 3500 respectively.
Apart from the Honesty, Regularity and Hard work, Punctuality is one more value which the school must include for awards.

Note: The possible mistakes that can done in this type of questions is while calculating the cofactor matrix, one might forget to multiply the determinant of ${{a}_{ij}}$with negative sign if $i+j$ is an odd number. After solving the problem one can also substitute the obtained values of x, y and z in the equations for verifying the answer.
Substituting in equation (1),
$\begin{align}
  & \Rightarrow x+y+z=6000 \\
 & \Rightarrow 500+2000+3500=6000 \\
 & \Rightarrow 6000=6000 \\
\end{align}$
Substituting in equation (2),
$\begin{align}
  & \Rightarrow x+3z=11000 \\
 & \Rightarrow 500+3\left( 3500 \right)=11000 \\
 & \Rightarrow 500+10500=11000 \\
 & \Rightarrow 11000=11000 \\
\end{align}$
Substituting in equation (3),
$\begin{align}
  & \Rightarrow x-2y+z=0 \\
 & \Rightarrow 500-2\left( 2000 \right)+3500=0 \\
 & \Rightarrow 500-4000+3500=0 \\
 & \Rightarrow 0=0 \\
\end{align}$
So, they satisfy the equation given the conditions, so, the obtained answer is correct.