Matrices and Determinants
9.0 Types of Linear Equations
9.0 Types of Linear Equations
(a) Consistent Equations: If Rank of $A$ =Rank of $C$ .
- Unique Solution: If Rank of $A$ =Rank of $C$=$n$. where $n$ is number of unknowns .
- Infinite Solution: If Rank of $A$ =Rank of $C$=$r$,where $r$<$n$.
(b) Inconsistent Equations: If Rank of $A$ $ \ne $ Rank of $C$,it has no solution.
Question 16.
Solve the following system of equations by Rank Method
$\begin{gathered} 5x + 3y + 7z = 4 \hspace{1em} \\ 3x + 26y + 2z = 9 \hspace{1em} \\ 7x + 2y + 10z = 5 \hspace{1em} \\ \end{gathered} $
Solution:
Consider $\left[ {\begin{array}{c} 5&3&7 \\ 3&{26}&2 \\ 7&2&{10} \end{array}} \right]\left[ {\begin{array}{c} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{c} 4 \\ 9 \\ 5 \end{array}} \right]$
which is in the form $AX = B$ ...(1)
Where $A = \left[ {\begin{array}{c} 5&3&7 \\ 3&{26}&2 \\ 7&2&{10} \end{array}} \right]$ and Augumented Matrix= $K = \left[ {\begin{array}{c} 5&3&7&4 \\ 3&{26}&2&9 \\ 7&2&{10}&5 \end{array}} \right]$
= $K = \left[ {\begin{array}{c} 1&{3/5}&{7/5}&{4/5} \\ 3&{26}&2&9 \\ 7&2&{10}&5 \end{array}} \right]$ $\left[ {{R_1} \to 1/5{R_1}} \right]$
= $\left[ {\begin{array}{c} 1&{3/5}&{7/5}&{4/5} \\ 0&{121/5}&{ - 11/5}&{33/5} \\ 0&{ - 11/5}&{1/5}&{ - 3/5} \end{array}} \right]$ $\left[ {{R_2} \to {R_2} - 3{R_1},{R_3} \to {R_3} - 7{R_1}} \right]$
= $\left[ {\begin{array}{c} 1&{3/5}&{7/5}&{4/5} \\ 0&{121/5}&{ - 11/5}&{33/5} \\ 0&0&0&0 \end{array}} \right]$ $\left[ {{R_3} \to {R_3} + 1/11{R_2}} \right]$
Here, Rank ($A$) = Rank ($K$) $=2<3$
It implies that system is consistent and has infinite number of solutions.
The equation (1) reduces to $\left[ {\begin{array}{c} 1&{3/5}&{7/5}&{} \\ 0&{121/5}&{ - 11/5}&{} \\ 0&0&0&{} \end{array}} \right]\left[ {\begin{array}{c} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{c} {4/5} \\ {33/5} \\ 0 \end{array}} \right]$
or
$\begin{gathered} x + 3/5y + 7/5z = 4/5 \hspace{1em} \\ \hspace{1em} \\ 121/5y - 11/5z = 33/5 \hspace{1em} \\ \hspace{1em} \\ 11y - z = 3 \hspace{1em} \\ \end{gathered} $
Let $Z$ = $k$, then
$\begin{gathered} y = 3/11 + k/11 \hspace{1em} \\ \hspace{1em} \\ \ and \ x = - 16/11k + 7/11 \hspace{1em} \\ \end{gathered} $
Since $k$ is arbitrary , hence the number of solutions is infinite.
