Question

Show that the following systems of linear equations is consistent and also find their solutions:
x + y + z = 6
x + 2y + 3z = 14
x + 4y + 7z = 30

Answer 1

Here,
\[x + y + z = 6 . . . (1)\]
\[x + 2y + 3z = 14 . . . (2)\]
\[x + 4y + 7z = 30 . . . (3)\]
\[or, AX = B \]
where, 
\[ A = \begin{bmatrix}1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7\end{bmatrix}, X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}\text{ and }B = \begin{bmatrix}6 \\ 14 \\ 30\end{bmatrix}\]
\[\begin{bmatrix}1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}6 \\ 14 \\ 30\end{bmatrix}\]
\[\left| A \right| = \begin{vmatrix}1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7\end{vmatrix}\]
\[ = 1\left( 14 - 12 \right) - 1\left( 7 - 3 \right) + 1(4 - 2)\]
\[ = 2 - 4 + 2\]
\[ = 0\]
So, A is singular . Thus, the given system of equations is either inconsistent or it is consistent with
\[\text{ infinitely many solutions because }\left( adj A \right)B \neq 0\text{ or }\left( adj A \right) = 0 . \]
\[ {\text{ Let }C}_{ij} {\text{ be the co-factors of the elements a }}_{ij}\text{ in }A\left[ a_{ij} \right]. \text{Then, }\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}2 & 3 \\ 4 & 7\end{vmatrix} = 2, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}1 & 3 \\ 1 & 7\end{vmatrix} = - 4 , C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}1 & 2 \\ 1 & 4\end{vmatrix} = 2\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & 1 \\ 4 & 7\end{vmatrix} = - 3, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 1 \\ 1 & 7\end{vmatrix} = 6, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 1 \\ 1 & 4\end{vmatrix} = - 3\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & 1 \\ 2 & 3\end{vmatrix} = 1 , C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 1 \\ 1 & 3\end{vmatrix} = - 2 , C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 1 \\ 1 & 2\end{vmatrix} = 1\]
\[adj A = \begin{bmatrix}2 & - 4 & 2 \\ - 3 & 6 & - 3 \\ 1 & - 2 & 1\end{bmatrix}^T \]
\[ = \begin{bmatrix}2 & - 3 & 1 \\ - 4 & 6 & - 2 \\ 2 & - 3 & 1\end{bmatrix}\]
\[\left( adj A \right)B = \begin{bmatrix}2 & - 3 & 1 \\ - 4 & 6 & - 2 \\ 2 & - 3 & 1\end{bmatrix}\begin{bmatrix}6 \\ 14 \\ 30\end{bmatrix}\]
\[ = \begin{bmatrix}12 - 42 + 30 \\ - 24 + 84 - 60 \\ 12 - 42 + 30\end{bmatrix}\]
\[ = \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}\]
\[\text{ If }\left| A \right|=0\text{ and }\left( adjA \right)B=0, \text{ then the system is consistent and has infinitely many solutions.}\]
\[\text{ Thus, } AX=B \text{ has infinitely many solutions}.\]
\[\text{ Substituting z=k in eq. (1) and eq. (2), we get}\]
\[x + y = 6 - k \text{ and }x + 2y = 14 - 3k\]
\[\begin{bmatrix}1 & 1 \\ 1 & 2\end{bmatrix}\binom{x}{y} = \binom{6 - k}{14 + 3k}\]
Now,
\[\left| A \right| = \begin{vmatrix}1 & 1 \\ 1 & 2\end{vmatrix}\]
\[ = 2 - 1 = 1 \neq 0\]
\[adj A = \begin{vmatrix}2 & - 1 \\ - 1 & 1\end{vmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{1}\begin{bmatrix}2 & - 1 \\ - 1 & 1\end{bmatrix}\]
\[ \therefore X = A^{- 1} B\]
\[ \Rightarrow \binom{x}{y} = \frac{1}{1}\begin{bmatrix}2 & - 1 \\ - 1 & 1\end{bmatrix}\binom{6 - k}{14 - 3k}\]
\[ \Rightarrow \binom{x}{y} = \frac{1}{1}\binom{12 - 2k - 14 + 3k}{ - 6 + k + 14 - 3k}\]
\[ \Rightarrow \binom{x}{y} = \binom{\frac{k - 2}{1}}{\frac{8 - 2k}{1}}\]
\[ \therefore x = k - 2, y = 8 - 2k\text{ and }z = k\]
These values of x, y and z also satisfy the third equation .
\[\text{ Thus, }x = k - 2, y = 8 - 2k\text{ and }z = k \left(\text{ where k is a real number }\right)\text{ satisfy the given system of equations }.\]