Question

The sum of three numbers is 6. Thrice the third number when added to the first number gives 7. On adding the sum of second and third number to three times the first number, we get 12. Find the three numbers using determinants.

Answer 1

Hint: In this question, we have to find three numbers for whom conditions are given. For this, we will first suppose these numbers as some variables and then form three equations using given conditions. We have to find these numbers using determinant, so we will write the equations in the form of AX = B where A is a coefficient matrix, X represents a matrix of variables and B represents a matrix of constant terms of equations. Using AX = B, we will find X by taking A to the other side and get $X={{A}^{-1}}B$. Hence, at last we will find ${{A}^{-1}}$ using ${{A}^{-1}}=\dfrac{1}{\left| A \right|}adj\left( A \right)$ and solve ${{A}^{-1}}B$ to find the value of X.

Complete step-by-step answer:
Let us suppose three numbers as x, y and z. We are given that, sum of three numbers is 6. Therefore,
\[x+y+z=6\cdots \cdots \cdots \cdots \cdots \left( 1 \right)\]
Also, according to question, thrice the third number when added to the first number gives 7. Therefore,
\[x+3z=7\cdots \cdots \cdots \cdots \cdots \left( 2 \right)\]
Also, on adding sum of second and third number to three times the first number, we get 12. Therefore,
\[3x+y+z=12\cdots \cdots \cdots \cdots \cdots \left( 3 \right)\]
Hence, we have obtained three equations (1), (2) and (3). Let us write these equations in the form AX = B where A is coefficient matrix, X is variable matrix and B is constant value matrix, we get AX = B as,
\[\left[ \begin{matrix}
   1 & 1 & 1 \\
   1 & 0 & 3 \\
   3 & 1 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   6 \\
   7 \\
   12 \\
\end{matrix} \right]\]
AX = B can be written as $X={{A}^{-1}}B$. Hence, we need to find ${{A}^{-1}}$ for calculating X.
As we know, ${{A}^{-1}}=\dfrac{1}{\left| A \right|}adj\left( A \right)$. So, let us calculate $\left| A \right|$ (determinant of A) first,
\[\begin{align}
  & \left| A \right|=\left| \begin{matrix}
   1 & 1 & 1 \\
   1 & 0 & 3 \\
   3 & 1 & 1 \\
\end{matrix} \right| \\
 & \Rightarrow 1\left( -3 \right)-1\left( 1-9 \right)+1\left( 1 \right) \\
 & \Rightarrow -3+8+1 \\
 & \Rightarrow 6 \\
\end{align}\]
Since $\left| A \right|=6\ne 0$ hence, ${{A}^{-1}}$ exists.
\[\left| A \right|=6\cdots \cdots \cdots \cdots \cdots \left( 4 \right)\]
Now, we need to calculate adjoin of A. As adj(A) is transpose of cofactor matrix. So, let us find cofactors of matrix A.
\[\begin{align}
  & {{A}_{11}}=+1\left| \begin{matrix}
   0 & 3 \\
   1 & 1 \\
\end{matrix} \right|=1\left( -3 \right)=-3 \\
 & {{A}_{12}}=-1\left| \begin{matrix}
   1 & 3 \\
   3 & 1 \\
\end{matrix} \right|=-1\left( 1-9 \right)=8 \\
 & {{A}_{13}}=+1\left| \begin{matrix}
   1 & 0 \\
   3 & 1 \\
\end{matrix} \right|=1\left( 1-0 \right)=1 \\
 & {{A}_{21}}=-1\left| \begin{matrix}
   1 & 1 \\
   1 & 1 \\
\end{matrix} \right|=-1\left( 1-1 \right)=0 \\
\end{align}\]
\[\begin{align}
  & {{A}_{22}}=+1\left| \begin{matrix}
   1 & 1 \\
   3 & 1 \\
\end{matrix} \right|=1\left( 1-3 \right)=-2 \\
 & {{A}_{23}}=-1\left| \begin{matrix}
   1 & 1 \\
   3 & 1 \\
\end{matrix} \right|=-1\left( 1-3 \right)=2 \\
 & {{A}_{31}}=1\left| \begin{matrix}
   1 & 1 \\
   0 & 3 \\
\end{matrix} \right|=1\left( 3-0 \right)=3 \\
 & {{A}_{32}}=-1\left| \begin{matrix}
   1 & 1 \\
   1 & 3 \\
\end{matrix} \right|=-1\left( 3-1 \right)=-2 \\
 & {{A}_{33}}=1\left| \begin{matrix}
   1 & 1 \\
   1 & 0 \\
\end{matrix} \right|=1\left( 0-1 \right)=-1 \\
\end{align}\]
Hence,
\[\left[ \begin{matrix}
   {{A}_{11}} & {{A}_{12}} & {{A}_{13}} \\
   {{A}_{21}} & {{A}_{22}} & {{A}_{23}} \\
   {{A}_{31}} & {{A}_{32}} & {{A}_{33}} \\
\end{matrix} \right]=\left[ \begin{matrix}
   -3 & 8 & 1 \\
   0 & -2 & 2 \\
   3 & -2 & -1 \\
\end{matrix} \right]\]
Where ${{A}_{ij}}$ are cofactors, ${{a}_{ij}}$ are elements of coefficient matrix for i, j = 1, 2, 3.
\[\begin{align}
  & adjA=\left[ \begin{matrix}
   {{A}_{11}} & {{A}_{12}} & {{A}_{13}} \\
   {{A}_{21}} & {{A}_{22}} & {{A}_{23}} \\
   {{A}_{31}} & {{A}_{32}} & {{A}_{33}} \\
\end{matrix} \right]=\left[ \begin{matrix}
   {{A}_{11}} & {{A}_{21}} & {{A}_{31}} \\
   {{A}_{12}} & {{A}_{22}} & {{A}_{32}} \\
   {{A}_{13}} & {{A}_{23}} & {{A}_{33}} \\
\end{matrix} \right] \\
 & \Rightarrow \left[ \begin{matrix}
   -3 & 0 & 3 \\
   8 & -2 & -2 \\
   1 & 2 & -1 \\
\end{matrix} \right] \\
\end{align}\]
\[{{A}^{-1}}=\dfrac{1}{\left| A \right|}adj\left( A \right)\]
Putting value of $\left| A \right|$ from (6) we get:
\[{{A}^{-1}}=\dfrac{1}{6}\left[ \begin{matrix}
   -3 & 0 & 3 \\
   8 & -2 & -2 \\
   1 & 2 & -1 \\
\end{matrix} \right]\]
Let us multiply ${{A}^{-1}}$ by B, we get:
\[\begin{align}
  & {{A}^{-1}}B=\dfrac{1}{6}\left[ \begin{matrix}
   -3 & 0 & 3 \\
   8 & -2 & -2 \\
   1 & 2 & -1 \\
\end{matrix} \right]\left[ \begin{matrix}
   6 \\
   7 \\
   12 \\
\end{matrix} \right] \\
 & \Rightarrow \dfrac{1}{6}\left[ \begin{matrix}
   -18+36 \\
   48-14-24 \\
   6+14-12 \\
\end{matrix} \right] \\
 & \Rightarrow \dfrac{1}{6}\left[ \begin{matrix}
   18 \\
   10 \\
   8 \\
\end{matrix} \right] \\
 & \Rightarrow \left[ \begin{matrix}
   3 \\
   \dfrac{5}{3} \\
   \dfrac{4}{3} \\
\end{matrix} \right] \\
\end{align}\]
But $X={{A}^{-1}}B$. Therefore,
\[X=\left[ \begin{matrix}
   3 \\
   \dfrac{5}{3} \\
   \dfrac{4}{3} \\
\end{matrix} \right]\]
Hence, \[\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   3 \\
   \dfrac{5}{3} \\
   \dfrac{4}{3} \\
\end{matrix} \right]\]
By comparing $x=3,y=\dfrac{5}{3},z=\dfrac{4}{3}$.
Hence, required numbers are $3\dfrac{5}{3},\dfrac{4}{3}.$

Note: Students should be careful of the signs while calculating cofactors of the elements. Don't forget to take transpose of the cofactor matrix to find adjoin of matrix. Students can make mistakes while multiplying two matrices, so take care of signs while doing that. At last, you can check your answer by putting them into equation.