Question

Ramesh buys a half dozen pencils, two erasers, and two sharpeners in a shop and pays 14 rupees. Suresh buys 15 pencils, three erasers, 5 erasers, and 5 sharpeners and pays 34 rupees where his friend who accompanied them buys 1 pencil, 1 eraser, and 1 sharpener and pays 3 rupees. Find the price of a pencil, sharpener, and eraser using matrices. \[\]

Answer 1

Hint: We assume the price of one pencil as $x$, the price of one eraser as $y$, and the price of one sharpener as $z$. We use the given information and form a $3\times 3$ linear system of equations $AX=B$ in matrices. We find the solution $X={{A}^{-1}}B.$\[\]

Complete step-by-step solution
We know the $3\times 3$ linear system of equations with three unknowns $x,y,z$ represented in matrix form as
\[\begin{align}
  & \left[ \begin{matrix}
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
   {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right]\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   {{d}_{1}} \\
   {{d}_{2}} \\
   {{d}_{3}} \\
\end{matrix} \right] \\
 & \Rightarrow AX=B \\
\end{align}\]
The solution of the above system of equation is given by
\[X={{A}^{-1}}B=\left( \dfrac{1}{\det \left( A \right)}\text{adj}A \right)B\]
Let us assume the price of one pencil as$x$, the price of one eraser as $y$ and the price of one sharpener as $y$. We are given the question that Rajesh buys a half dozen pencils, 2 erasers and 2 sharpeners in a shop and pays 14 rupees. So we have
\[6x+2y+2z=14...(1)\]
Suresh buys 15 pencils, three erasers, 5 erasers and 5 sharpeners and pays 34 rupees. So we have
\[15x+5y+3z=34....\left( 2 \right)\]
The accompanying friend buys 1 pencil, 1 eraser and 1 sharpener and pays 3 rupees. So we have,
\[x+y+z=3......\left( 3 \right)\]
We write equation (1), (2), (3) in matrix from as,
\[\begin{align}
  & \left[ \begin{matrix}
   6 & 2 & 2 \\
   15 & 5 & 3 \\
   1 & 1 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   14 \\
   34 \\
   3 \\
\end{matrix} \right] \\
 & \Rightarrow AX=B \\
\end{align}\]
We know that the adjoint of the $A$ is the transpose of the cofactor matrix of $A$. The cofactor matrix is matrix where each element ${{a}_{ij}}$ is equal to the determinant value of the matrix formed by rows and columns excluding${{a}_{ij}}$multiplied by ${{\left( -1 \right)}^{i+j}}$for example co-factor of ${{a}_{11}}=6$ is the determinant excluding row-1 first column -1 that is
\[C\left( 6 \right)={{\left( -1 \right)}^{11}}\left| \begin{matrix}
   5 & 3 \\
   1 & 1 \\
\end{matrix} \right|=1\left( 5\times 1-3\times 1 \right)=2\]
Summarily we find the cofactors of the elements in the first row as
\[\begin{align}
  & C\left( {{a}_{12}}=2 \right)={{\left( -1 \right)}^{1+2}}\left| \begin{matrix}
   15 & 3 \\
   1 & 1 \\
\end{matrix} \right|=-1\left( 15-3 \right)=-12 \\
 & C\left( {{a}_{13}}=2 \right)={{\left( -1 \right)}^{1+3}}\left| \begin{matrix}
   15 & 5 \\
   1 & 1 \\
\end{matrix} \right|=1\left( 15-5 \right)=10 \\
\end{align}\]
The cofactors of the elements in the second row are
\[\begin{align}
  & C\left( {{a}_{21}}=15 \right)={{\left( -1 \right)}^{2+1}}\left| \begin{matrix}
   2 & 2 \\
   1 & 1 \\
\end{matrix} \right|=-1\left( 2-2 \right)=0 \\
 & C\left( {{a}_{22}}=5 \right)={{\left( -1 \right)}^{2+2}}\left| \begin{matrix}
   6 & 2 \\
   1 & 1 \\
\end{matrix} \right|=1\left( 6-2 \right)=4 \\
 & C\left( {{a}_{23}}=3 \right)={{\left( -1 \right)}^{2+3}}\left| \begin{matrix}
   6 & 2 \\
   1 & 1 \\
\end{matrix} \right|=-1\left( 6-2 \right)=-4 \\
\end{align}\]
The cofactors of the elements in the third row are

\[\begin{align}
  & C\left( {{a}_{31}}=1 \right)={{\left( -1 \right)}^{3+1}}\left| \begin{matrix}
   2 & 2 \\
   5 & 3 \\
\end{matrix} \right|=1\left( 6-10 \right)=-4 \\
 & C\left( {{a}_{32}}=1 \right)={{\left( -1 \right)}^{3+2}}\left| \begin{matrix}
   6 & 2 \\
   15 & 3 \\
\end{matrix} \right|=-1\left( 18-30 \right)=12 \\
 & C\left( {{a}_{33}}=1 \right)={{\left( -1 \right)}^{3+3}}\left| \begin{matrix}
   6 & 2 \\
   15 & 5 \\
\end{matrix} \right|=1\left( 30-30 \right)=0 \\
\end{align}\]
 The cofactor matrix of $A$ is
\[\text{coeff }\left( A \right)=\left[ \begin{matrix}
   C\left( 6 \right) & C\left( 2 \right) & C\left( 2 \right) \\
   C\left( 15 \right) & C\left( 5 \right) & C\left( 3 \right) \\
   C\left( 1 \right) & C\left( 1 \right) & C\left( 1 \right) \\
\end{matrix} \right]=\left[ \begin{matrix}
   2 & -12 & 10 \\
   0 & 4 & -4 \\
   -4 & 12 & 0 \\
\end{matrix} \right]\]
So the adjoint matrix of $A$ is
\[\text{adj}.A={{\left( \text{coeff}.A \right)}^{T}}=\left[ \begin{matrix}
   2 & 0 & -4 \\
   -12 & 4 & 12 \\
   10 & -4 & 0 \\
\end{matrix} \right]\]
We find the determinant value of matrix $A$ by expanding in third column. We have
\[\det \left( A \right)=\left| \begin{matrix}
   6 & 2 & 2 \\
   15 & 5 & 3 \\
   1 & 1 & 1 \\
\end{matrix} \right|=1\left( 6-10 \right)-1\left( 18-30 \right)+1\left( 30-30 \right)=8\]
So the solutions are
\[X={{A}^{-1}}B=\left( \dfrac{adj.A}{\det \left( A \right)} \right)B=\dfrac{1}{8}\left[ \begin{matrix}
   2 & 0 & -4 \\
   -12 & 4 & 12 \\
   10 & -4 & 0 \\
\end{matrix} \right]\left[ \begin{matrix}
   14 \\
   34 \\
   3 \\
\end{matrix} \right]=\dfrac{1}{8}\left[ \begin{matrix}
   16 \\
   4 \\
   4 \\
\end{matrix} \right]=\left[ \begin{matrix}
   2 \\
   0.5 \\
   0.5 \\
\end{matrix} \right]\]
So the price of one pencil is 2 rupees one eraser is 0.5 and one sharpeners is 0.5 rupees.\[\]

Note: We can alternatively solve the problem by using Gauss-Jordan elimination of the augmented matrix$\left( A,B \right)$. We can use the same elimination to find inverse in $\left( A,{{A}^{-1}} \right)$.We note that we cannot find a unique solution when$\det \left( A \right)=0$, infinite solution if $\det \left( A \right)=0,\text{adj}A\cdot B=0$ and no solution if $\det \left( A \right)=0,\text{adj}A\cdot B\ne 0$/