Question

The sum of three numbers is 6. If we multiply the third number by 2 and add the first number to the result, we get 7. By adding second and the third numbers to the three times of the first number we get 12. Use matrices to find the number.

Answer 1

Hint : In this question, we need to determine the value of the three numbers such that their sum is 6 and if we multiply the third number by 2 and add the first number to the result, we get 7. Also, on adding second and the third numbers to the three times of the first number we get 12. For this, we will first establish the mathematical equation for all the three conditions and then frame the matrix equation to solve for the values of the numbers.

Complete step-by-step answer :
Let the three numbers are $x,y$ and $z$.
According to the question, the sum of the three numbers is 6. So, we can write
$x + y + z = 6 - - - - (i)$
Also, it has been given that if we multiply the third number by 2 and add the first number to the result, we get 7. So, we can write
$x + 2z = 7 - - - - (ii)$
Again, on adding second and the third numbers to the three times of the first number we get 12. So, we can write
$3x + y + z = 12 - - - - (iii)$
By equation (i), (ii) and (iii), we can frame the matrix equation as:
\left[ {\begin{array}{*{20}{c}} x&y&z \\\ x&0&{2z} \\\ {3x}&y&z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 6 \\\ 7 \\\ {12} \end{array}} \right]
\Rightarrow \left[ {\begin{array}{*{20}{c}} 1&1&1 \\\ 1&0&2 \\\ 3&1&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\\ y \\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 6 \\\ 7 \\\ {12} \end{array}} \right] - - - - (iv) \\\
The equation (iv) resembles to the matrix equation $AX = B$ where A = \left[ {\begin{array}{*{20}{c}} 1&1&1 \\\ 1&0&2 \\\ 3&1&1 \end{array}} \right];B = \left[ {\begin{array}{*{20}{c}} 6 \\\ 7 \\\ {12} \end{array}} \right] and X = \left[ {\begin{array}{*{20}{c}} x \\\ y \\\ z \end{array}} \right] .
The matrix equation can also be written as $X = {A^{ - 1}}B$ provided that the matrix A is consistent and its determinant value id not zero.
Now, we will calculate the determinant value of the matrix A as:

A = \left[ {\begin{array}{*{20}{c}} 1&1&1 \\\ 1&0&2 \\\ 3&1&1 \end{array}} \right] $$ $$ \Rightarrow \left| A \right| = 1(0 - 2) - 1(1 - 6) + 1(1 - 0) $$ $$ \Rightarrow \left| A \right| = - 2 + 5 + 1 $$ $$ \Rightarrow \left| A \right| = 4 $$ Here, the determinant value of A is 4 (not equals to zero) hence we can use the solution equation as $X = {A^{ - 1}}B$ to get the value of the three numbers. Now, evaluating the value of the inverse of the matrix A as: $$ A = \left[ {\begin{array}{*{20}{c}} 1&1&1 \\\ 1&0&2 \\\ 3&1&1 \end{array}} \right] $$ $$ \Rightarrow {A^{ - 1}} = \dfrac{{adjA}}{{\left| A \right|}} $$ $$ \Rightarrow {A^{ - 1}} = {\dfrac{{\left[ {\begin{array}{*{20}{c}} {(0 - 2)}&{ - (1 - 6)}&{(1 - 0)} \\\ { - (1 - 1)}&{(1 - 3)}&{ - (1 - 3)} \\\ {(2 - 0)}&{ - (2 - 1)}&{(0 - 1)} \end{array}} \right]}}{4}^T} $$ $$ \Rightarrow {A^{ - 1}} = \dfrac{{{{\left[ {\begin{array}{*{20}{c}} { - 2}&5&1 \\\ 0&{ - 2}&2 \\\ 2&{ - 1}&{ - 1} \end{array}} \right]}^T}}}{4} $$ $$ \Rightarrow {A^{ - 1}} = \dfrac{{\left[ {\begin{array}{*{20}{c}} { - 2}&0&2 \\\ 5&{ - 2}&{ - 1} \\\ 1&2&{ - 1} \end{array}} \right]}}{4} - - - - (v) $$ Substituting the value of the inverse of matrix A and the value of matrix B in the equation $X = {A^{ - 1}}B$ to evaluate the value of the variables. $$ X = {A^{ - 1}}B $$ $$ \Rightarrow \left[ {\begin{array}{*{20}{c}} x \\\ y \\\ z \end{array}} \right] = \dfrac{{\left[ {\begin{array}{*{20}{c}} { - 2}&0&2 \\\ 5&{ - 2}&{ - 1} \\\ 1&2&{ - 1} \end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}} 6 \\\ 7 \\\ {12} \end{array}} \right]}}{4} $$ $$ \Rightarrow \left[ {\begin{array}{*{20}{c}} x \\\ y \\\ z \end{array}} \right] = \dfrac{{\left[ {\begin{array}{*{20}{c}} {\left( { - 2 \times 6} \right) + 0 + \left( {2 \times 12} \right)} \\\ {\left( {5 \times 6} \right) - \left( {2 \times 7} \right) - \left( {1 \times 12} \right)} \\\ {\left( {1 \times 6} \right) + \left( {2 \times 7} \right) - \left( {1 \times 12} \right)} \end{array}} \right]}}{4} $$ $$ \Rightarrow \left[ {\begin{array}{*{20}{c}} x \\\ y \\\ z \end{array}} \right] = \dfrac{{\left[ {\begin{array}{*{20}{c}} { - 12 + 24} \\\ {30 - 14 - 12} \\\ {6 + 14 - 12} \end{array}} \right]}}{4} $$ $$ \Rightarrow \left[ {\begin{array}{*{20}{c}} x \\\ y \\\ z \end{array}} \right] = \dfrac{{\left[ {\begin{array}{*{20}{c}} {12} \\\ 4 \\\ 8 \end{array}} \right]}}{4} $$ $$ \Rightarrow \left[ {\begin{array}{*{20}{c}} x \\\ y \\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 3 \\\ 1 \\\ 2 \end{array}} \right] $$ Hence, the values of the three numbers which are all satisfying the given conditions of the problem is 3, 1 and 2. **So, the correct answer is “3, 1 ,2”.** **Note** : It is very important to note here that, before proceeding with the use of the solution of the matrix equation, it must be made sure that the matrix A must be a consistent, unique solution and its determinant value should not be equals to zero. Moreover, care should be taken while calculating the value of the inverse of the matrix as it includes many calculations. Many times, students forgets to do the transpose during the calculation of the inverse of the matrix which led to the wrong result.