A system of linear equations having two and three variables can be easily solved using determinants. Here, the formulas and steps to find the solution of a system of linear equations are given along with practice problems. Cramer’s rule is well explained along with a diagram.
Now following algorithm is followed to solve the system (CRITERION FOR CONSISTENCY)
This method of finding a solution to a system of equations is called Cramer’s rule.
Conditions for Infinite and No Solutions
(a) If Δ=0 and Δ1=Δ2=Δ3=0, then system of equation may or may not be consistent:
(i) If the value of x, y and z in terms of t satisfy the third equation then the system is said to be consistent and will have infinite solutions.
(ii) If the values of x, y, z don’t satisfy the third equation, the system is said to be inconsistent and will have no solution.
(b) If d1=d2=d3=0, then system of linear equations is known as Homogeneous linear equations, which always possess at least one solution i.e. (0, 0, 0). This is called a trivial solution for homogeneous linear equations.
(c) If the system of homogeneous linear equations possesses non-zero/nontrivial solutions, and Δ = 0. In such a case given system has infinite solutions.
We can also solve these solutions using the matrix inversion method.
We can write the linear equations in the matrix form as AX = B where
(b) Area of a triangle whose vertices are(xr,yr);r=1,2,3is:D=21∣∣∣∣∣∣∣x1x2x3y1y2y3111∣∣∣∣∣∣∣.
If D = 0 then the three points are collinear.
(c) Equation of a straight line passing through (x1,y1)&(x2,y2)is∣∣∣∣∣∣∣xx1x2yy1y2111∣∣∣∣∣∣∣=0.
(d) If each element of any row (or column) can be expressed as a sum of two terms, then the determinant can be expressed as the sum of the determinants.
It should be noted that while applying operations on determinants at least one row (or column) must remain unchanged i.e.
The maximum number of simultaneous operations = order of determinant – 1.
Practice Problems On System Of Linear Equations Using Determinants
Illustration: Solve the following equations by Cramer’s rule x+y+z=9,2x+5y+7z=52,2x+y−z=0.
Solution:
Here in this problem define the determinants Δ,Δ1,Δ2andΔ3 and find out their value by using the invariance property and then by using Cramer’s rule, we can get the values of x, y and z.
Here Δ=∣∣∣∣∣∣∣12215117−1∣∣∣∣∣∣∣(ApplyingC2→C2−C1andC3→C3−C1)Δ=∣∣∣∣∣∣∣12203−105−3∣∣∣∣∣∣∣=1(−9+5)=−4
By Cramer’s rule x=ΔΔ1=−4−4=1,y=ΔΔ2=−4−12=3andz=ΔΔ3=−4−20=5
. x=1,y=3,z=5
Illustration: Solve the following linear equations: x+54+y+73=−1andx+56−y+76=−5
Solution:
Here in this problem first put x+51=aandy+71=bandthendefinethedeterminantsΔ,Δ1andΔ2 . Then by using Cramer’s rule we can get the values of x and y.
Let us put Δ,x+51=aandy+71=b then the 2 linear equations become
Illustration: For what value of k will the following system of equations possess nontrivial solutions. Also find all the solutions of the system for that value of k.
x+y-kz=0; 3x-y-2z=0; x-y+2z=0.
Solution:
Here in this problem first define Δ. As we know that, for non-trivial solution Δ= 0.
So by using the invariance property we can solve Δ= 0 and will get the value of k.
Expanding along C2.⇒−(−1)[−8−2(2−k)]=0⇒2k−12=0⇒k=6
Putting the value of k in the given equation, we get,
x+y-6z=0 … (i)
3x-y-2z=0 … (ii)
x-y+2z=0 … (iii)
(i) + (ii) ⇒4x−8z=0
∴ z=2x
Putting the value of z in (i), we get x+y-3x=0
∴ y = 2x
Thus when k = 6, solution of the given system of equations will be x=t,y=2t,z=2t, when t is an arbitrary number.
Illustration: Solve the following equations by matrix inversion.
2x + y + 2z = 0, 2x – y + z = 10, x + 3y – z = 5
Solution:
By writing the given equations into the form of AX = D and then multiplying both side by A-1 we will get the required value of x, y and z.
In the matrix form, the equations can be written as ⎣⎢⎡2211−1321−1⎦⎥⎤⎣⎢⎡xyz⎦⎥⎤=⎣⎢⎡0105⎦⎥⎤
∴ AX = D where A = ⎣⎢⎡2211−1321−1⎦⎥⎤,X=⎣⎢⎡xyz⎦⎥⎤,D=⎣⎢⎡0105⎦⎥⎤⇒A−1(AX)=A−1D⇒X=A−1D … (i)
Now A−1=∣A∣adjA;∣A∣=∣∣∣∣∣∣∣2211−1321−1∣∣∣∣∣∣∣=2(1−3)−1(−2−1)+2(6+1)=13
The matrix of cofactors of |A| is ⎣⎢⎡−2733−427−54⎦⎥⎤.So,adjA=⎣⎢⎡−2377−4−532−4⎦⎥⎤;A−1=31⎣⎢⎡−2377−4−532−4⎦⎥⎤.
from (1), X=131⎣⎢⎡−2377−4−5324⎦⎥⎤⎣⎢⎡0105⎦⎥⎤=131⎣⎢⎡0+70+150−40+100−50−20⎦⎥⎤=⎣⎢⎡85/13−30/13−70/13⎦⎥⎤;⎣⎢⎡xyz⎦⎥⎤=⎣⎢⎡85/13−30/13−70/13⎦⎥⎤⇒x=1385,y=13−30,z=13−70