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Show that the following system of linear equations is consistent and also find their solution:
\[x + y + z = 6\]
\[x + 2y + 3z = 14\]
\[x + 4y + 7z = 30\]

seo-qna
Last updated date: 13th Jun 2024
Total views: 371.4k
Views today: 10.71k
Answer
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Hint:
This question can be solved by using matrices and determinants. We will use matrix formulas and then try to find whether the given equations are consistent or not. Then, we will use the matrix method further, to find the required solution. Also, if the determinant is 0 then either the equations are inconsistent or they have infinitely many solutions.

Complete step by step solution:
In the given question, we have,
\[x + y + z = 6\]
\[x + 2y + 3z = 14\]
\[x + 4y + 7z = 30\]
Now, writing these 3 equations as \[AX = B\], here, \[A = \left[ {\begin{array}{*{20}{l}}1&1&1\\1&2&3\\1&4&7\end{array}} \right]\] , \[X = \left[ \begin{array}{l}x\\y\\z\end{array} \right]\] and \[B = \left[ \begin{array}{l}6\\14\\30\end{array} \right]\].
Hence, \[AX = B\] can be written as:
\[\left[ {\begin{array}{*{20}{l}}1&1&1\\1&2&3\\1&4&7\end{array}} \right].\left[ \begin{array}{l}x\\y\\z\end{array} \right] = \left[ \begin{array}{l}6\\14\\30\end{array} \right]\]
Now, first of all, we are required to show whether these linear equations are consistent or not.
Hence, we will find the determinant,
\[\left| A \right| = \left| {\begin{array}{*{20}{l}}1&1&1\\1&2&3\\1&4&7\end{array}} \right|\]
Now, solving the determinant, we get,
\[\left| A \right| = \left( {2 \times 7 - 4 \times 3} \right) - 1\left( {1 \times 7 - 1 \times 3} \right) + 1\left( {1 \times 4 - 1 \times 2} \right)\]
Simplifying the expression, we get
\[ \Rightarrow \left| A \right| = \left( {14 - 12} \right) - \left( {7 - 3} \right) + \left( {4 - 2} \right)\]
\[ \Rightarrow \left| A \right| = 2 - 4 + 2\]
Adding and subtracting the terms, we get
\[ \Rightarrow \left| A \right| = 4 - 4 = 0\]
Since, the determinant of A is equal to 0
Hence, either the equations are inconsistent or they have infinitely many solutions.
Now, we will find the Co-factors of the elements \[{a_{ij}}\] in the matrix A.
\[{A_{11}} = {\left( { - 1} \right)^{1 + 1}}\left| \begin{array}{l}2{\rm{ }}3\\4{\rm{ 7}}\end{array} \right| = 14 - 12 = 2\]
\[{A_{12}} = {\left( { - 1} \right)^{1 + 2}}\left| \begin{array}{l}{\rm{1 }}3\\{\rm{1 7}}\end{array} \right| = - 4\]
\[{A_{13}} = {\left( { - 1} \right)^{1 + 3}}\left| \begin{array}{l}{\rm{1 2}}\\{\rm{1 4}}\end{array} \right| = 4 - 2 = 2\]
\[{A_{21}} = {\left( { - 1} \right)^{2 + 1}}\left| \begin{array}{l}{\rm{1 1}}\\{\rm{4 7}}\end{array} \right| = - 3\]
\[{A_{22}} = {\left( { - 1} \right)^{2 + 2}}\left| \begin{array}{l}{\rm{1 1}}\\{\rm{1 7}}\end{array} \right| = 6\]
\[{A_{23}} = {\left( { - 1} \right)^{2 + 3}}\left| \begin{array}{l}{\rm{1 1}}\\{\rm{1 4}}\end{array} \right| = - 3\]
\[{A_{31}} = {\left( { - 1} \right)^{3 + 1}}\left| \begin{array}{l}{\rm{1 1}}\\{\rm{2 3}}\end{array} \right| = 1\]
\[{A_{32}} = {\left( { - 1} \right)^{3 + 2}}\left| \begin{array}{l}{\rm{1 1}}\\{\rm{1 3}}\end{array} \right| = - 2\]
\[{A_{33}} = {\left( { - 1} \right)^{3 + 3}}\left| \begin{array}{l}{\rm{1 1}}\\{\rm{1 2}}\end{array} \right| = 1\]
Now,
\[adj{\rm{A}} = {\left[ {{\rm{cofactor}}} \right]^{\rm{T}}}\]
\[ \Rightarrow adj{\rm{A}} = {\left[ {\begin{array}{*{20}{l}}2&{ - 4}&2\\{ - 3}&6&{ - 3}\\1&{ - 2}&1\end{array}} \right]^{\rm{T}}}\]
\[ \Rightarrow adj{\rm{A}} = \left[ {\begin{array}{*{20}{l}}2&{ - 3}&1\\{ - 4}&6&{ - 2}\\2&{ - 3}&1\end{array}} \right]\]
Now, we will find \[\left( {adj{\rm{A}}} \right){\rm{B}}\].
\[\left( {adj{\rm{A}}} \right){\rm{B}} = \left[ {\begin{array}{*{20}{l}}2&{ - 3}&1\\{ - 4}&6&{ - 2}\\2&{ - 3}&1\end{array}} \right].\left[ \begin{array}{l}6\\14\\30\end{array} \right]\]
Solving the matrix further,
\[ \Rightarrow \left( {adj{\rm{A}}} \right){\rm{B}} = \left[ \begin{array}{l}12 - 18 + 6\\ - 24 + 84 - 60\\12 - 42 + 30\end{array} \right] = \left[ \begin{array}{l}0\\0\\0\end{array} \right]\]
Since, \[\left( {adj{\rm{A}}} \right){\rm{B}}\] is equal to 0, hence, the linear equations are inconsistent and have infinitely many solutions.
Now, we will substitute the value of \[z = k\] in the \[x + y + z = 6\] and \[x + 2y + 3z = 14\]. Therefore, we get
\[x + y + k = 6\]
\[ \Rightarrow x + y = 6 - k\]……………………….\[\left( 1 \right)\]
\[x + 2y + 3k = 14\]
\[ \Rightarrow x + 2y = 14 - 3k\]…………………..\[\left( 2 \right)\]
Now, writing equation \[\left( 1 \right)\] and \[\left( 2 \right)\] in the form of \[AX = B\]
Where, \[A = \left[ \begin{array}{l}{\rm{1 1}}\\{\rm{1 2}}\end{array} \right]\], \[X = \left[ \begin{array}{l}x\\y\end{array} \right]\] and \[B = \left[ \begin{array}{l}6 - k\\14 - 3k\end{array} \right]\]
Now, we will find the determinant of A. Therefore, we get
\[\left| A \right| = \left| \begin{array}{l}{\rm{1 1}}\\{\rm{1 2}}\end{array} \right|\]
Simplifying the determinant, we get
\[ \Rightarrow \left| A \right| = \left( {2 - 1} \right) = 1\]
Hence, the determinant is not equal to 0.
Therefore, it is invertible.
Now, Cofactors:
\[{C_{11}} = {\left( { - 1} \right)^{1 + 1}} \cdot \left( 2 \right) = 2\]
\[{C_{12}} = {\left( { - 1} \right)^{1 + 2}} \cdot \left( 1 \right) = - 1\]
\[{C_{21}} = {\left( { - 1} \right)^{2 + 1}} \cdot \left( 1 \right) = - 1\]
\[{C_{22}} = {\left( { - 1} \right)^{2 + 2}} \cdot \left( 1 \right) = 1\]
Now,
\[adj{\rm{A}} = {\left[ {\begin{array}{*{20}{l}}2&{ - 1}\\{ - 1}&1\end{array}} \right]^T}\]
\[ \Rightarrow adjA = \left[ {\begin{array}{*{20}{l}}2&{ - 1}\\{ - 1}&1\end{array}} \right]\]
Now,
\[{{\rm{A}}^{{ - 1}}} = \dfrac{1}{{|{\rm{A|}}}}adj{\rm{A}}\]
\[ \Rightarrow {A^{ - 1}} = \dfrac{1}{1}\left[ {\begin{array}{*{20}{l}}2&{ - 1}\\{ - 1}&1\end{array}} \right]\]
Now, as we know,\[X = {A^{ - 1}}B\], so substituting the values, we get
\[X = \left[ {\begin{array}{*{20}{l}}2&{ - 1}\\{ - 1}&1\end{array}} \right]\left[ \begin{array}{l}6 - k\\14 - 3k\end{array} \right]\]
Multiplying the matrix, we get
\[ \Rightarrow X = \left[ \begin{array}{l}12 - 2k - 14 + 3k\\ - 6 + k + 14 - 3k\end{array} \right]\]
Hence,
\[\left[ \begin{array}{l}x\\y\end{array} \right] = \left[ \begin{array}{l}k - 2\\8 - 2k\end{array} \right]\]
Hence, \[x = k - 2\], \[y = 8 - 2k\] and \[z = k\]
Now, putting these values in the third equation, \[x + 4y + 7z = 30\]
\[ \Rightarrow k - 2 + 4\left( {8 - 2k} \right) + 7k = 30\]
\[ \Rightarrow k - 2 + 32 - 8k + 7k = 30\]
\[ \Rightarrow 30 = 30\]
Hence, LHS \[ = \] RHS
Therefore, the values \[x = k - 2\], \[y = 8 - 2k\] and \[z = k\] satisfies all the three equations, where \[k\] is any real number.

Hence, we can find infinitely many solutions using these values.

Note:
This question is of a special type because it is very rare to see that equations have infinitely many solutions. Hence, one must understand this concept very carefully as we have applied the matrix method twice and expressed our answer in terms of a constant \[k\]. Also, as we have to find cofactors, we must choose the plus/minus signs carefully as there are high possibilities of interchanging them and making our answer completely wrong.