Question

The sum of three numbers is 2. If twice the second number is added in the sum of first and third, we get 1. On adding the sum of the second and third number to five times the first number, we get 6. Find the three numbers using Cramer’s rule.

Answer 1

Hint: We will first make linear equations in three variables by the above given conditions. Now, we have three variables and three linear equations related to them. We will express this set in matrix form, which will have a coefficient matrix, a variable matrix and the constant matrix. The set of linear equations can be solved by Cramer’s rule, i.e.
${\text{x = }}\dfrac{{\left| {{{\text{D}}_{\text{x}}}} \right|}}{{\left| {\text{D}} \right|}}$,
${\text{y = }}\dfrac{{\left| {{{\text{D}}_{\text{y}}}} \right|}}{{\left| {\text{D}} \right|}}$,
${\text{z = }}\dfrac{{\left| {{{\text{D}}_{\text{z}}}} \right|}}{{\left| {\text{D}} \right|}}$
Where, ${{\text{D}}_{\text{x}}}{\text{,}}{{\text{D}}_{\text{y}}}{\text{,}}{{\text{D}}_{\text{z}}}$can be obtained by replacing first, second and third column of coefficient matrix by constant matrix respectively.

Complete step by step solution: Let the required three numbers be ${\text{x,y,z}}$
Now, according to the first condition, the sum of three numbers is 2
${\text{x + y + z = 2}}$ …(i)
According to the second condition, twice the second number is added in the sum of first and third, we get 1
${\text{2y + }}\left( {{\text{x + z}}} \right){\text{ = 1}}$
${\text{x + 2y + z = 1}}$ …(ii)
According to the third condition, on adding the sum of second and third number to five times the first number, we get 6
$\left( {{\text{y + z}}} \right){\text{ + 5x = 6}}$
${\text{5x + y + z = 6}}$ …(iii)
From (i), (ii) and (iii), we get:
${\text{x + y + z = 2}}$
${\text{x + 2y + z = 1}}$
${\text{5x + y + z = 6}}$
Now, we will represent the above equations in matrix form:
\[\left[ {\begin{array}{*{20}{c}}
  {\text{1}}&{\text{1}}&{\text{1}} \\
  {\text{1}}&{\text{2}}&{\text{1}} \\
  {\text{5}}&{\text{1}}&{\text{1}}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  {\text{x}} \\
  {\text{y}} \\
  {\text{z}}
\end{array}} \right]{\text{ = }}\left[ {\begin{array}{*{20}{c}}
  {\text{2}} \\
  {\text{1}} \\
  {\text{6}}
\end{array}} \right]\]
${\text{AX = B}}$
Where A is the coefficient matrix, X is the variable matrix and B is the constant matrix
Now, calculating determinant of coefficient matrix,
$
  \left| {\text{A}} \right|{\text{ = }}\left| {\begin{array}{*{20}{c}}
  \Rightarrow{\text{1}}&{\text{1}}&{\text{1}} \\
  \Rightarrow{\text{1}}&{\text{2}}&{\text{1}} \\
  \Rightarrow{\text{5}}&{\text{1}}&{\text{1}}
\end{array}} \right| \\
 \Rightarrow \left| {\text{A}} \right|{\text{ = 1 - 1}}\left( {{\text{ - 4}}} \right){\text{ + 1 - 10}} \\
  \Rightarrow \left| {\text{A}} \right|{\text{ = 1 + 4 + 1 - 10}} \\
  \Rightarrow \left| {\text{A}} \right|{\text{ = - 4}} \\
 $
So,
Now, the Cramer’s rule for solving set of linear equations is
${\text{x = }}\dfrac{{\left| {{{\text{D}}_{\text{x}}}} \right|}}{{\left| {\text{D}} \right|}}$,
${\text{y = }}\dfrac{{\left| {{{\text{D}}_{\text{y}}}} \right|}}{{\left| {\text{D}} \right|}}$,
${\text{z = }}\dfrac{{\left| {{{\text{D}}_{\text{z}}}} \right|}}{{\left| {\text{D}} \right|}}$
Where, ${{\text{D}}_{\text{x}}}{\text{,}}{{\text{D}}_{\text{y}}}{\text{,}}{{\text{D}}_{\text{z}}}$can be obtained by replacing first, second and third column of coefficient matrix by constant matrix respectively.
Now,
\[
  {{\text{D}}_{\text{x}}}{\text{ = }}\left[ {\begin{array}{*{20}{c}}
  {\text{2}}&{\text{1}}&{\text{1}} \\
  {\text{1}}&{\text{2}}&{\text{1}} \\
  {\text{6}}&{\text{1}}&{\text{1}}
\end{array}} \right]{\text{,}} \\
  {{\text{D}}_{\text{y}}}{\text{ = }}\left[ {\begin{array}{*{20}{c}}
  {\text{1}}&{\text{2}}&{\text{1}} \\
  {\text{1}}&{\text{1}}&{\text{1}} \\
  {\text{5}}&{\text{6}}&{\text{1}}
\end{array}} \right]{\text{,}} \\
  {{\text{D}}_{\text{z}}}{\text{ = }}\left[ {\begin{array}{*{20}{c}}
  {\text{1}}&{\text{1}}&{\text{2}} \\
  {\text{1}}&{\text{2}}&{\text{1}} \\
  {\text{5}}&{\text{1}}&{\text{6}}
\end{array}} \right] \\
 \]
Now, calculating determinant of above matrices,
\[
  \left| {{{\text{D}}_{\text{x}}}} \right|{\text{ = }}\left| {\begin{array}{*{20}{c}}
  {\text{2}}&{\text{1}}&{\text{1}} \\
  {\text{1}}&{\text{2}}&{\text{1}} \\
  {\text{6}}&{\text{1}}&{\text{1}}
\end{array}} \right|{\text{ = 2 + 5 - 11 = - 4}} \\
  \left| {{{\text{D}}_{\text{y}}}} \right|{\text{ = }}\left| {\begin{array}{*{20}{c}}
  {\text{1}}&{\text{2}}&{\text{1}} \\
  {\text{1}}&{\text{1}}&{\text{1}} \\
  {\text{5}}&{\text{6}}&{\text{1}}
\end{array}} \right|{\text{ = - 5 + 8 + 1 = 4}} \\
  \left| {{{\text{D}}_{\text{z}}}} \right|{\text{ = }}\left| {\begin{array}{*{20}{c}}
  {\text{1}}&{\text{1}}&{\text{2}} \\
  {\text{1}}&{\text{2}}&{\text{1}} \\
  {\text{5}}&{\text{1}}&{\text{6}}
\end{array}} \right|{\text{ = 11 - 1 - 18 = - 8}} \\
 \]
${\text{x = }}\dfrac{{\left| {{{\text{D}}_{\text{x}}}} \right|}}{{\left| {\text{D}} \right|}}{\text{ = }}\dfrac{{{\text{ - 4}}}}{{{\text{ - 4}}}}{\text{ = 1}}$,
${\text{y = }}\dfrac{{\left| {{{\text{D}}_{\text{y}}}} \right|}}{{\left| {\text{D}} \right|}}{\text{ = }}\dfrac{{\text{4}}}{{{\text{ - 4}}}}{\text{ = - 1}}$,
${\text{z = }}\dfrac{{\left| {{{\text{D}}_{\text{z}}}} \right|}}{{\left| {\text{D}} \right|}}{\text{ = }}\dfrac{{{\text{ - 8}}}}{{{\text{ - 4}}}}{\text{ = 2}}$
Thus, solution is ${\text{x = 1,y = - 1,z = 2}}$
Hence, the required numbers are, 1, -1, 2

Note: A different approach to the question could be solving the linear equations simply by eliminating variables, if it was not mentioned to solve by Cramer’s rule. As we have
${\text{x + y + z = 2}}$ …(i)
${\text{x + 2y + z = 1}}$ …(ii)
${\text{5x + y + z = 6}}$ …(iii)
Subtracting (ii) from (i)
${\text{x + 2y + z - }}\left( {{\text{x + y + z}}} \right){\text{ = 1 - 2}}$
$\Rightarrow{\text{y = - 1}}$
Now, subtracting, (ii) from (iii)
${\text{5x + y + z - }}\left( {{\text{x + 2y + z}}} \right){\text{ = 6 - 1}}$
$\Rightarrow{\text{4x - y = 5}}$
Substituting, ${\text{y = - 1}}$
${\text{4x + 1 = 5}}$
$\Rightarrow{\text{4x = 4}}$
$\Rightarrow{\text{x = 1}}$
Now, we have
${\text{x + y + z = 2}}$
$\Rightarrow{\text{1 + }}\left( {{\text{ - 1}}} \right){\text{ + z = 2}}$
$\Rightarrow{\text{z = 2}}$
Thus, solution is ${\text{x = 1,y = - 1,z = 2}}$
Hence, this method is easy and less time consuming.