Linear System : A Linear System is group of linear equations in the variables.
Linear Equation :
A general linear equation has the form\[a_1x_1 + a_2x_2 + a_3x_3 + · · · + a_nx_n = b_1\] where \(a_1 , a_2, a_3, · · · ,a_n , b_1\) are real numbers.
Examples : \[2x = 1,\]
\[y = 3x – 1,\]
\[3x-4y = 1,\]
\[x+\frac{1}{2}y-z=\frac{5}{2},\]
\[0.1x-y+2z+3w = -1.5\]
are linear equations.
Solving Linear equations :
A value of the variable that satisfies linear equation is called solution.Example (1) : Substituting \(x = \frac{1}{2} \) in to left hand side
\[2x = 1\]
\[\Rightarrow 2*\frac{1}{2} =\frac{2}{2} = 1 \]
which is equals right hand side. Therefore \(x = \frac{1}{2} \) is a solution.
A value of the variable that satisfies linear equation is called solution.Example (2) : Substituting \(x = 1, y =2 \) in to
\[y = 3x – 1\]
and evaluating both sides
\[ \Rightarrow 2 =3*1- 1 \]
\[ \Rightarrow 2 = 2 \]
which is a true statement. Therefore \(x = 1, y = 2 \) is a solution.
Note : \(x =k, y =3k -1\) is also a solution, for any real number \(k\).
Example (3) : Substituting \(x = 3 , y = 2\) in left hand side
\[3x-4y = 1\]
\[\Rightarrow 3*3-4*2= 9-8 = 1\]
equals right hand side . Therefore tuple \((3, 2) \) is a solution.
Note : \(x =c, y = \frac{3c -1}{4}\) for any real constant \(c\), is also a solution.
System of Linear equations :
Example (1)
Solve the following system of linear equations \[2x – y – z = 0\]\[-x+2y-z=0\]\[ -x-y+2z=0\]
Solution:
Step 1: Writing system in matrix form \(AX=B\)
\[\left( {\begin{array}{*{20}{c}}2&{ – 1}&{ – 1}\\{ – 1}&2&{ – 1}\\{ – 1}&{ – 1}&2\end{array}} \right)\left( \begin{array}{l}x\\y\\z\end{array} \right) = \left( \begin{array}{l}0\\0\\0\end{array} \right)\]
Step 2: Writing Augmented matrix \([A, B]\)
\[[A , B] =\left( {\begin{array}{*{20}{c}}2&{ – 1}&{ – 1}&{0}\\{ – 1}&2&{ – 1}&{0}\\{ – 1}&{ – 1}&2&{0}\end{array}} \right)\]
Step 3: Echelon form of \([A, B]\)
performing row elementary operation \(R_2\rightarrow R_2+(\frac{1}{2})R_1, R_3\rightarrow R_3+(\frac{1}{2})R_1\)
\[\Rightarrow [A , B] \sim \left( {\begin{array}{*{20}{c}}2&{ – 1}&{ – 1}&0\\0&{\frac{3}{2}}&{ – \frac{3}{2}}&0\\0&{ – \frac{3}{2}}&{\frac{3}{2}}&0\end{array}} \right)\]
and \( R_3\rightarrow R_3+R_2\)
\[\Rightarrow [A , B] \sim \left( {\begin{array}{*{20}{c}}2&{ – 1}&{ – 1}&0\\0&{\frac{3}{2}}&{ – \frac{3}{2}}&0\\0&0&0&0\end{array}} \right)\]
and performing \( R_1\rightarrow (\frac{1}{2})R_1, R_2\rightarrow (\frac{2}{3})R_2\)
\[\Rightarrow [A , B] \sim \left( {\begin{array}{*{20}{c}}1&-{ \frac{1}{2}}&-{ \frac{1}{2}}&0\\0&1&-1&0\\0&0&0&0\end{array}} \right)\]
This is Echelon form of \([A, B]\).
\(\Rightarrow Rank([A, B]) = 2\) and \(Rank(A) = 2\)
Since \(Rank([A, B]) = Rank(A) \) hence system is consistent and has solution.
Now the reduced system can be written as
\[\left( {\begin{array}{*{20}{c}}1&-{ \frac{1}{2}}&-{ \frac{1}{2}}\\0&1&-1\\0&0&0\end{array}} \right)\left( \begin{array}{l}x\\y\\z\end{array} \right) = \left( \begin{array}{l}0\\0\\0\end{array} \right)\]
\[\Rightarrow x -\frac{1}{2} y – \frac{1}{2}z = 0 , y-z=0 \Rightarrow y = z\]
\[\Rightarrow x = y = z\]
To get a solution we need to substitute \(z = c_1\) an arbitrary number then \(x =c_1 , y =c_1 , z =c_1 \) is a solution of the system
Since \(c_1\) is arbitrary number hence system has infinitely solutions of the system.
Therefore option (D) is correct.