Question

The sum of the three numbers is 6, twice the third number when added to the first number gives 7. On adding the sum of the second and third to thrice the first number, we get 12. Find the numbers, using the adjoint method.

Answer 1

Hint: We will first find the three equation with three variables using the given conditions and then use the equation \[AX = B\], where \[X = \left[ {\begin{array}{*{20}{c}}
  a \\
  b \\
  c
\end{array}} \right]\]. Then we will use formula of finding the inverse using the adjoint method is \[{A^{ - 1}} = \dfrac{{adjA}}{{\left| A \right|}}\] in the equation. Apply this formula, and then use the given conditions to find the required values.

Complete step by step answer:

Given that the sum of the three numbers is 6, twice the third number when added to the first number gives 7 and on adding the sum of the second and third to thrice the first number, we get 12.

Let us assume that the first number is \[a\], the second number is \[b\] and the third number is\[c\].

We will now find the equation when the sum of the three numbers is 6.

\[a + b + c = 6{\text{ ......}}\left( 1 \right)\]

Now we will find the equation when twice the third number, when added to the first number, gives 7.

\[a + 2c = 7{\text{ ......}}\left( 2 \right)\]

We will now find the equation when adding the sum of the second and third to thrice the first number is equal to 12.

\[3a + b + c = 12{\text{ ......}}\left( 3 \right)\]

We know that the equation of matrix using determinant property for finding the value of \[a\], \[b\], \[c\], is \[AX = B\] where \[X = \left[ {\begin{array}{*{20}{c}}
  a \\
  b \\
  c
\end{array}} \right]\].

Multiplying the \[{A^{ - 1}}\]in the above equation of matrix \[AX = B\] on both sides , we get

\[
  {A^{ - 1}}AX = {A^{ - 1}}B \\
   \Rightarrow X = {A^{ - 1}}B \\
\]

Using the equation \[\left( 1 \right)\], equation \[\left( 2 \right)\] and equation \[\left( 3 \right)\] to find the value of the matrices \[A\] and \[B\], we get

\[A = \left[ {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&0&2 \\
  3&1&1
\end{array}} \right]\]

\[B = \left[ {\begin{array}{*{20}{c}}
  6 \\
  7 \\
  {12}
\end{array}} \right]\]

Substituting the values of \[A\] and \[B\] in equation \[AX = B\], we get

\[\left[ {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&0&2 \\
  3&1&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  a \\
  b \\
  c
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  6 \\
  7 \\
  {12}
\end{array}} \right]\]

We know that the formula to find the inverse of a matrix \[A\] using the adjoint method is \[{A^{ - 1}} = \dfrac{{adjA}}{{\left| A \right|}}\].

First, we will find the determinant of \[A\].

\[
  \left| A \right| = \left| {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&0&2 \\
  3&1&1
\end{array}} \right| \\
   = 1\left( {0 - 2} \right) - 1\left( {1 - 6} \right) + 1\left( 1 \right) \\
   = - 2 + 5 + 1 \\
   = 4 \\
\]

We will now find the value of \[adjA\] by finding the cofactors of the given elements of the matrix, \[Co{f_{i,j}} = {\left( { - 1} \right)^{i + j}}Det\left( {S{M_i}} \right)\], where \[S{M_i}\] are the sub-matrices of the given matrix.

First, we will take \[i = 1\] and \[j = 1\] in the formula of finding the cofactors.

\[
  Co{f_{1,1}} = {\left( { - 1} \right)^{1 + 1}}\left| {\begin{array}{*{20}{c}}
  0&2 \\
  1&1
\end{array}} \right| \\
   = 0 - 2 \\
   = - 2 \\
\]

Now taking \[i = 1\] and \[j = 2\] in the above formula of cofactors.

\[
  Co{f_{1,2}} = {\left( { - 1} \right)^{1 + 2}}\left| {\begin{array}{*{20}{c}}
  1&2 \\
  3&1
\end{array}} \right| \\
   = - 1\left( {1 - 6} \right) \\
   = 5 \\
\]

We will now take \[i = 1\] and \[j = 3\] in the above formula of cofactors.

\[
  Co{f_{1,3}} = {\left( { - 1} \right)^{1 + 3}}\left| {\begin{array}{*{20}{c}}
  1&0 \\
  3&1
\end{array}} \right| \\
   = 1 - 0 \\
   = 1 \\
\]

Now taking \[i = 2\] and \[j = 1\] in the above formula of cofactors.

\[
  Co{f_{2,1}} = {\left( { - 1} \right)^{2 + 1}}\left| {\begin{array}{*{20}{c}}
  1&1 \\
  1&1
\end{array}} \right| \\
   = - 1\left( {1 - 1} \right) \\
   = 0 \\
\]

We will now take \[i = 2\] and \[j = 2\] in the above formula of cofactors.

\[
  Co{f_{2,2}} = {\left( { - 1} \right)^{2 + 2}}\left| {\begin{array}{*{20}{c}}
  1&1 \\
  3&1
\end{array}} \right| \\
   = 1 - 3 \\
   = - 2 \\
\]

We will now take \[i = 2\] and \[j = 3\] in the above formula of cofactors.

\[
  Co{f_{2,3}} = {\left( { - 1} \right)^{2 + 3}}\left| {\begin{array}{*{20}{c}}
  1&1 \\
  3&1
\end{array}} \right| \\
   = - 1\left( {3 - 1} \right) \\
   = - 2 \\
\]

Now taking \[i = 3\] and \[j = 1\] in the above formula of cofactors.

\[
  Co{f_{3,1}} = {\left( { - 1} \right)^{3 + 1}}\left| {\begin{array}{*{20}{c}}
  1&1 \\
  0&2
\end{array}} \right| \\
   = 2 - 0 \\
   = 2 \\
\]

We will now take \[i = 3\] and \[j = 2\] in the above formula of cofactors.

\[
  Co{f_{3,2}} = {\left( { - 1} \right)^{3 + 2}}\left| {\begin{array}{*{20}{c}}
  1&1 \\
  0&2
\end{array}} \right| \\
   = - 1\left( {2 - 0} \right) \\
   = - 2 \\
\]

We will now take \[i = 3\] and \[j = 3\] in the above formula of cofactors.

\[
  Co{f_{3,3}} = {\left( { - 1} \right)^{3 + 3}}\left| {\begin{array}{*{20}{c}}
  1&1 \\
  1&0
\end{array}} \right| \\
   = 0 - 1 \\
   = - 1 \\
\]

We will use these values to form a matrix and then find the transpose to calculate the \[adjA\].

\[
  adjA = {\left[ {\begin{array}{*{20}{c}}
  { - 2}&5&1 \\
  0&{ - 2}&2 \\
  2&{ - 1}&{ - 1}
\end{array}} \right]^T} \\
   = \left[ {\begin{array}{*{20}{c}}
  { - 2}&0&2 \\
  5&{ - 2}&{ - 1} \\
  1&2&{ - 1}
\end{array}} \right] \\
\]

Dividing the value of \[adjA\] by \[\left| A \right|\] to find the value of \[{A^{ - 1}}\], we get

\[\dfrac{1}{4}\left[ {\begin{array}{*{20}{c}}
  { - 2}&0&2 \\
  5&{ - 2}&{ - 1} \\
  1&2&{ - 1}
\end{array}} \right]\]

Substituting the values of \[X\], \[{A^{ - 1}}\] and \[B\] in the equation \[X = {A^{ - 1}}B\], we get

\[
  \left[ {\begin{array}{*{20}{c}}
  a \\
  b \\
  c
\end{array}} \right] = \dfrac{1}{4}\left[ {\begin{array}{*{20}{c}}
  { - 2}&0&2 \\
  5&{ - 2}&{ - 1} \\
  1&2&{ - 1}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  6 \\
  7 \\
  {12}
\end{array}} \right] \\
   = \dfrac{1}{4}\left[ {\begin{array}{*{20}{c}}
  { - 12 + 0 + 24} \\
  {30 - 14 - 12} \\
  {6 + 14 - 12}
\end{array}} \right] \\
   = \dfrac{1}{4}\left[ {\begin{array}{*{20}{c}}
  {12} \\
  4 \\
  8
\end{array}} \right] \\
   = \left[ {\begin{array}{*{20}{c}}
  {\dfrac{{12}}{4}} \\
  {\dfrac{4}{4}} \\
  {\dfrac{8}{4}}
\end{array}} \right] \\
   = \left[ {\begin{array}{*{20}{c}}
  3 \\
  1 \\
  2
\end{array}} \right] \\
\]

Thus, the values are \[a = 3\], \[b = 1\] and \[c = 2\].

Note: In solving these types of questions, you should be familiar with the formulas of finding the inverse using the adjoint method, \[{A^{ - 1}} = \dfrac{{adjA}}{{\left| A \right|}}\], where \[\left| A \right|\] is determinant of the matrix \[A\]. Then use the given conditions and values given in the question, and substitute in the formula, to find the required values. Also, we are supposed to write the values properly to avoid any miscalculation.