System of Linear Equations

System of Linear Equation

A system of linear equations are group of $n$ linear equations containing $n$ number of variables.

1. System of 2 Linear Equations:

It is a pair of linear equations in two variables. It is usually of the form

$a_1x +b_1y + c_1 = 0$

$a_2x +b_2y + c_2 = 0$

Finding a solution for this system means finding the values of $x$ and $y$ that satisfy both equations.

2. System of 3 Linear Equations:

It is a group of 3 linear equations in three variables. It is usually of the form

$a_1x +b_1y + +c_1z + d_1 = 0$

$a_2x +b_2y + +c_2z + d_2 = 0$

$a_3x +b_3y + +c_3z + d_3 = 0$

Finding a solution for this system means finding the values of $x, y$, and $z$ that satisfy all three equations.

The system of equations is broadly classified into two types:

Consistent system of equations:

A system of equations is said to be consistent if it has at least one solution. Let the given system of equations is

$
\begin{aligned}
&\text{The system of linear equations} \\
&a_1x + b_1y = c_1 \\
&a_2x + b_2y = c_2 \\
&\text{has exactly one solution if} \\
&\frac{a_1}{a_2} \neq \frac{b_1}{b_2}.
\end{aligned}
$

$
\begin{aligned}
& \text { E.g., } x+y=2 \\
& \qquad x-y=6 \text { is consistent because it has a solution } x=4 \text { and } y=-2 .
\end{aligned}
$
Given lines are non-parallel, hence lines will have one point of intersection.

$
\text{It has infinite solutions if} \\
\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}.
$

In this case, two lines represented by these lines coincide, so there are infinite pair of values of $x$ and $y$ that satisfy both the equations. This case is also counted as consistent as there is at least one solution.

Inconsistent equation:

A system of equations is said to be inconsistent if it has no solution.

$\\\mathrm{Let \; a_1x +b_1y + c_1 = 0\; and \; a_2x + b_2y + c_2 = 0, then} \\\mathrm{equation\; are \; inconsistent \; and\; has \; no\; solution\; if} \\\mathrm{\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}}$

For example, $x+y=5$ and $2 x+2 y=5$ are inconsistent as it has no solution, just by seeing the equation, we get that it is the equation of two different parallel lines which never intersect. These lines are non-intersecting, hence there is no solution to this system.

Homogenous System of Linear Equation

A linear equation with a constant value of zero is called a homogeneous equation.

$\\\mathrm{Let,} \\\mathrm{a_1x+b_1y +c_1z=0\;\;\; ...(i)} \\\mathrm{a_2x+b_2y +c_2z=0\;\;\; ...(ii)} \\\mathrm{a_3x+b_3y +c_3z=0\;\;\; ...(iii)} \\\mathrm{be \; three\; homogeneous\; equations} \\\\\mathrm{and \; let\; \Delta = \begin{vmatrix} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \end{vmatrix}}$

Note that $x=y=z=0$ will always satisfy this system of equations. So system of homogeneous equations will always have at least one solution.

Also, the solution $x=0, y=0$, and $z=0$ is called a trivial solution, and other solutions are called non-trivial solutions.

• If $\Delta \neq 0$, then $x=0, y=0, z=0$ is the only solution of the above system. This solution is also known as a trivial solution.
• If $\Delta=0$, at least one of $x, y$, and $z$ are non-zero. In this case, we will have non-trivial solutions as well. Also, there would be infinite solutions of such a system of equations.

Non Homogenous System of Linear Equations

A linear equation with a constant value not equal to zero is called a homogeneous equation.

Characteristics of non Homogenous Linear Equations

• If $|\mathrm{A}| \neq 0$, then the system of equations is consistent and has a unique solution $\mathrm{X}=\mathrm{A}-1 \mathrm{~B}$
• If $|A|=0$ and $(\operatorname{adj} A) \cdot B \neq 0$, then the system of equations is inconsistent and has no solution.
• If $|A|=0$ and $(\operatorname{adj} A) \cdot B=0$, then the system of equations is consistent and has an infinite number of solutions.

Methods to solve Systems of Linear Equations in two variables

We use the following method to solve a System of linear equations in two variables

• Graphical method
• Elimination method
• Substitution method
• Matrix method
• Cross Multiplication method

Methods to solve Systems of Linear Equations in three variables

We use the following method to solve a System of linear equations in three variables

• Cramers Rule
• Inverse method
• Gaussian elimination method
• Gaussian Jordan method
• LU Decomposition method

Recommended Video Based on System of Linear Equations:

Solved Examples Based on System of Linear Equations

Example 1:

Solution

$\begin{aligned} & -x+2 y-9 z=7-(1) \\ & -x+3 y-7 z=9-(2) \\ & -2 x+y+5 z=8-(3) \\ & (2)-(1) \\ & y+16 z=2(4) \\ & (3)-2 x(1) \\ & -3 y+23 z=-6-(5) \\ & 3 x(4)+(5) \\ & 71 z=0 \Rightarrow z=0 \\ & y=2 \\ & x=-3 \\ & (-3,2,0)-(a, \beta, \gamma) \\ & \text { Put in }-3 x+y+13 z=1 \\ & \lambda=9+2=11 \\ & d=|-6-4-11|=? \\ & 3 \end{aligned}$

Hence, the answer is 3.

Example 2: Let N denote the number that turns up when a fair die is rolled. If the probability that the system of equations

$\begin{aligned} & x+y+z=1 \\ & 2 x+\mathrm{Ny}+2 z=2 \\ & 3 x+3 y+\mathrm{N} z=3 \end{aligned}$

has a unique solution is $\frac{k}{6}$ then the sum of the value of k and all possible values of N is [JEE MAINS 2023]

Solution

For unique solution

$\begin{aligned} & \Delta \neq 0 \\ & \left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 & \mathrm{~N} & 2 \\ 3 & 3 & \mathrm{~N} \end{array}\right| \neq 0 \\ & \Rightarrow\left(\mathrm{N}^2-6\right)-(2 \mathrm{~N}-6)+(6-3 \mathrm{~N}) \neq 0 \\ & \Rightarrow \mathrm{N}^2-5 \mathrm{~N}+6 \neq 0 \end{aligned}$

$\Rightarrow \mathrm{N} \neq 3 \quad \& \quad \mathrm{~N} \neq 2$

$\text { Hence } N \text { can be }\{1,4,5,6\} \text { Fav case : } \frac{4}{6}=\frac{K}{6} \Rightarrow k=4$

$\operatorname{sum}=20$

Hence, the answer is 20.

Example 3:

Solution

$\begin{aligned} & \Delta=0 \\ & \Rightarrow\left|\begin{array}{ccc} 2 & 1 & -1 \\ 2 & -5 & \lambda \\ 1 & 2 & -5 \end{array}\right|=0 \\ & \Rightarrow 2(25-2 \lambda)-1(-10-\lambda)-1(4+5)=0 \\ & \Rightarrow 51-3 x=0 \\ & \Rightarrow \lambda=17 \\ & \Delta_{\mathrm{x}}=0 \\ & \left|\begin{array}{ccc} 5 & 1 & -1 \\ \mu & -5 & 17 \\ 7 & 2 & -5 \end{array}\right|=0 \end{aligned}$

$\begin{aligned} & \Rightarrow 5(25-34)-1(-5 \mu-119)-1(2 \mu+35)=0 \\ & \Rightarrow-45+5 \mu+119-2 \mu-35=0 \\ & \Rightarrow 39+3 \mu=0 \Rightarrow \mu=-13 \\ & (\lambda+\mu)^2+(\lambda-\mu)^2=4^2+(30)^2 \\ & =916 \end{aligned}$

Hence, the answer is 916.

Example 4 :

Solution

Given system of equations is inconsistent

$ \begin{aligned} & \Rightarrow \Delta=0 \\ & \left|\begin{array}{lll} \lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda \end{array}\right|=0 \\ & \Rightarrow \lambda^3-3 \lambda+2=0 \\ & \Rightarrow(\lambda-1)^2(\lambda+2)=0 \\ & \Rightarrow \lambda=1,-2 \\ & \end{aligned}$
But for $\lambda=1$ all planes are the same
Then $\lambda=-2$

$ \sum_{\lambda \in s}\left(|\lambda|^2+|\lambda|\right)=4+2=6$

Hence, the answer is 6.

Example 5: If the system of linear equations

$7 x+11 \mathrm{y}+\alpha z=13$

$5 x+4 y+7 z=\beta$

$175 x+194 y+57 z=361$

has infinitely many solutions, then $\alpha+B+2$ is equal to :

[JEE MAINS 2023]

Solution

$7 x+11 y+\alpha z=13$

$5 x+4 y+7 z=\beta$
$175 x+194 y+57 z=361$
Condition of Infinite Many solutions

$\Delta=0$ & $\Delta \mathrm{x}, \Delta \mathrm{y}, \Delta \mathrm{z}=0$
check.
After solving we get $\alpha+13+2=4$

Hence, the answer is 4.