Question

Show that the following systems of linear equations is consistent and also find their solutions:
5x + 3y + 7z = 4
3x + 26y + 2z = 9
7x + 2y + 10z = 5

Answer 1

 Here,
\[5x + 3y + 7z = 4 . . . (1) \]
\[3x + 26y + 2z = 9 . . . (2)\]
\[7x + 2y + 10z = 5 . . . (3)\]
\[or , AX = B \]
where, 
\[ A = \begin{bmatrix}5 & 3 & 7 \\ 3 & 26 & 2 \\ 7 & 2 & 10\end{bmatrix}, X = \begin{bmatrix}x \\ y \\ z\end{bmatrix} \text{ and } B = \begin{bmatrix}4 \\ 9 \\ 5\end{bmatrix}\]
\[\begin{bmatrix}5 & 3 & 7 \\ 3 & 26 & 2 \\ 7 & 2 & 10\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}4 \\ 9 \\ 5\end{bmatrix}\]
\[\left| A \right| = \begin{vmatrix}5 & 3 & 7 \\ 3 & 26 & 2 \\ 7 & 2 & 10\end{vmatrix}\]
\[ = 5\left( 260 - 4 \right) - 3\left( 30 - 14 \right) + 7(6 - 182)\]
\[ = 1280 - 48 - 1232\]
\[ = 0\]
So, A is singular . Thus, the given system of equations is either inconsistent or it is consistent with 
\[\text{ infinitely many solutions because }\left( adj A \right)B \neq 0\text{ or }\left( adj A \right)B = 0 . \]
\[ {\text{ Let }C}_{ij} {\text{ be the co-factors of the elements a }}_{ij}\text{ in }A\left[ a_{ij} \right]. \text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}26 & 2 \\ 2 & 10\end{vmatrix} = 256 , C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}3 & 2 \\ 7 & 10\end{vmatrix} = - 16, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}3 & 26 \\ 7 & 2\end{vmatrix} = - 176\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}3 & 7 \\ 2 & 10\end{vmatrix} = - 16 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}5 & 7 \\ 7 & 10\end{vmatrix} = 1, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}5 & 3 \\ 7 & 2\end{vmatrix} = 11\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}3 & 7 \\ 26 & 2\end{vmatrix} = - 176, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}5 & 7 \\ 3 & 2\end{vmatrix} = 11, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}5 & 3 \\ 3 & 26\end{vmatrix} = 121\]
\[adj A = \begin{bmatrix}256 & - 16 & - 176 \\ - 16 & 1 & 11 \\ - 176 & 11 & 121\end{bmatrix}^T \]
\[ = \begin{bmatrix}256 & - 16 & - 176 \\ - 16 & 1 & 11 \\ - 176 & 11 & 121\end{bmatrix}\]
\[\left( adj A \right)B = \begin{bmatrix}256 & - 16 & - 176 \\ - 16 & 1 & 11 \\ - 176 & 11 & 121\end{bmatrix}\begin{bmatrix}4 \\ 9 \\ 5\end{bmatrix}\]
\[ = \begin{bmatrix}1024 - 144 - 880 \\ - 64 + 9 + 55 \\ - 704 + 99 + 605\end{bmatrix}\]
\[ = \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}\]
\[\text{ if }\left| A \right|=0\text{ and }\left( adjA \right)B=0, \text{ then the system is consistent and has infinitely many solutions.}\]
\[\text{ Thus, }AX=B \text{ has infinitely many solutions.}\]
 Substituting z=k in eq. (1) and eq. (2), we get
\[5x + 3y = 4 - 7k\text{ and }3x + 26y = 9 - 2k\]
\[\begin{bmatrix}5 & 3 \\ 3 & 26\end{bmatrix}\binom{x}{y} = \binom{4 - 7k}{9 - 2k}\]
Now,
\[\left| A \right| = \begin{vmatrix}5 & 3 \\ 3 & 26\end{vmatrix}\]
\[ = 130 - 9\]
\[ = 121 \neq 0\]
\[adj A = \begin{vmatrix}26 & - 3 \\ - 3 & 5\end{vmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{121}\begin{bmatrix}26 & - 3 \\ - 3 & 5\end{bmatrix}\]
\[ \therefore X = A^{- 1} B\]
\[ \Rightarrow \binom{x}{y} = \frac{1}{121}\begin{bmatrix}26 & - 3 \\ - 3 & 5\end{bmatrix}\binom{4 - 7k}{9 - 2k}\]
\[ \Rightarrow \binom{x}{y} = \frac{1}{121}\binom{104 - 182k - 27 + 6k}{ - 12 + 21k + 45 - 10k}\]
\[ \Rightarrow \binom{x}{y} = \binom{\frac{77 - 176k}{121}}{\frac{33 + 11k}{121}}\]
\[ \Rightarrow x = \frac{11\left( 7 - 16k \right)}{121}, y = \frac{11\left( 3 + k \right)}{121} and z = k\]
\[ \therefore x = \frac{7 - 16k}{11}, y = \frac{3 + k}{11}and z = k\]
These values of x, y and z also satisfy the third equation . 
\[\text{ Thus, }x = \frac{7 - 16k}{11}, y = \frac{3 + k}{11}\text{ and }z = k \left( \text{where k is a real number } \right)\text{ satisfy the given system of equations .}\]