1. JEE
  2. IIT JEE Study Material
  3. System Of Linear Equations Using Determinants
Top Banner

System of Linear Equations using Determinants

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.

Solving System Of Linear Equations Using Determinants

 

How To Solve a Linear Equation System Using Determinants?

1. System Of Linear Equations with Two Variables

Let the equations be a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0

The solution to a system of equations having 2 variables is given by:

x=Δ1Δ,y=Δ2Δ  or  x=b1c2−b2c1a1b2−a2b1;y=a2c1−a1c2a1b2−a2b1x=\frac{{{\Delta }_{1}}}{\Delta },y=\frac{{{\Delta }_{2}}}{\Delta } \;or \;x=\frac{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}};y=\frac{{{a}_{2}}{{c}_{1}}-{{a}_{1}}{{c}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}

Where Δ1=∣b1c1b2c2∣,  Δ2=∣c1a1c2a2∣  and  Δ=∣a1b1a2b2∣{{\Delta }_{1}}=\left| \begin{matrix} {{b}_{1}} & {{c}_{1}} \\ {{b}_{2}} & {{c}_{2}} \\ \end{matrix} \right|,\;{{\Delta }_{2}}=\left| \begin{matrix} {{c}_{1}} & {{a}_{1}} \\ {{c}_{2}} & {{a}_{2}} \\ \end{matrix} \right|\; and \;\Delta =\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{2}} & {{b}_{2}} \\ \end{matrix} \right|

2. System Of Linear Equations Involving Three Variables

a1x+b1y+c1z=d1    and    a2x+b2y+c2z=d2    and  a3x+b3y+c3z=d3{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z={{d}_{1}} \;\;and\;\;{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z={{d}_{2}} \;\;and\;{{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}z={{d}_{3}}

To solve this system we first define the following determinants

Δ=∣a1b1c1a2b2c2a3b3c3∣,Δ1=∣d1b1c1d2b2c2d3b3c3∣,Δ2=∣a1d1c1a2d2c2a3d3c3∣,Δ3=∣a1b1d1a2b2d2a3b3d3∣\Delta =\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|,{{\Delta }_{1}}=\left| \begin{matrix} {{d}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{d}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{d}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|,{{\Delta }_{2}}=\left| \begin{matrix} {{a}_{1}} & {{d}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{d}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{d}_{3}} & {{c}_{3}} \\ \end{matrix} \right|,{{\Delta }_{3}}=\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{d}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{d}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{d}_{3}} \\ \end{matrix} \right|

Now following algorithm is followed to solve the system (CRITERION FOR CONSISTENCY)

How to Solve a System Of Linear Equations Using Determinants

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\Delta =0 and Δ1=Δ2=Δ3=0,{{\Delta }_{1}}={{\Delta }_{2}}={{\Delta }_{3}}=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,{{d}_{1}}={{d}_{2}}={{d}_{3}}=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

A=[a1b1c1a2b2c2a3b3c3],X=[xyz]  and  B=[d1d2d3]A=\left[ \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right],X=\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]\;and\;B=\left[ \begin{matrix} {{d}_{1}} \\ {{d}_{2}} \\ {{d}_{3}} \\ \end{matrix} \right]

Now, solution set is obtained by solving X = A-1 B. Hence the solution set exists only if the inverse of A exists.

Some Important Results

Condition for the consistency of three simultaneous linear equations in 2 variables

The lines: a1x+b1y+c1=0{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0 … (i)

a2x+b2y+c2=0{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0 … (ii)

a3x+b3y+c3=0{{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}=0 … (iii)

are concurrent if, ∣a1b1c1a2b2c2a3b3c3∣=0\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|=0

(a) ax2+2hxy+by2+2gx+2fy+c=0a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0 represents a pair of straight lines if

abc+2fgh−af2−bg2−ch2=0=∣ahghbfgfc∣abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0=\left| \begin{matrix} a & h & g \\ h & b & f \\ g & f & c \\ \end{matrix} \right|

(b) Area of a triangle whose vertices are(xr,yr);  r=1,2,3    is:      D=12∣x1y11x2y21x3y31∣.\left( {{x}_{r}},{{y}_{r}} \right);\,\,r=1,2,3 \;\;is: \;\;\;D=\frac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\ \end{matrix} \right|.

If D = 0 then the three points are collinear.

(c) Equation of a straight line passing through (x1,y1)&(x2,y2)    is    ∣xy1x1y11x2y21∣=0.\left( {{x}_{1}},{{y}_{1}} \right)\And \left( {{x}_{2}},{{y}_{2}} \right) \;\;is \;\;\left| \begin{matrix} x & y & 1 \\ {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ \end{matrix} \right|=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.

E.g., ∣a1+xb1+yc1+za2b2c2a3b3c3∣=∣a1b1c1a2b2c2a3b3c3∣+∣xyza2b2c2a3b3c3∣\left| \begin{matrix} {{a}_{1}}+x & {{b}_{1}}+y & {{c}_{1}}+z \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|=\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|+\left| \begin{matrix} x & y & z \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|

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.x+y+z=9,2x+5y+7z=52,2x+y-z=0.

Solution:

Here in this problem define the determinants Δ,Δ1,Δ2  and  Δ3\Delta ,{{\Delta }_{1}},{{\Delta }_{2}} \;and\; {{\Delta }_{3}} 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=∣11125721−1∣  (Applying  C2→  C2−C1  and  C3→  C3−C1)\Delta =\left| \begin{matrix} 1 & 1 & 1 \\ 2 & 5 & 7 \\ 2 & 1 & -1 \\ \end{matrix} \right| \;(Applying\; {{C}_{2}}\to\; {{C}_{2}}-{{C}_{1}}\;and \;{{C}_{3}}\to\; {{C}_{3}}-{{C}_{1}})

Now,

Δ=∣1002352−1−3∣=1(−9+5)=−4\Delta =\left| \begin{matrix} 1 & 0 & 0 \\ 2 & 3 & 5 \\ 2 & -1 & -3 \\ \end{matrix} \right|=1\left( -9+5 \right)=-4

⇒ Δ1=∣911525701−1∣  (Applying  C2→C2+C3){\Delta }_{1}=\left| \begin{matrix} 9 & 1 & 1 \\ 52 & 5 & 7 \\ 0 & 1 & -1 \\ \end{matrix} \right| \;(Applying \;C_{2}\to C_{2}+C_{3})

⇒ Δ1=∣9215212700−1∣=−1(108−104)=−4;  Δ2=∣191252720−1∣  (Applying  C1→C1+2C3){{\Delta }_{1}}=\left| \begin{matrix} 9 & 2 & 1 \\ 52 & 12 & 7 \\ 0 & 0 & -1 \\ \end{matrix} \right|=-1\left( 108-104 \right)=-4;\,\,{{\Delta }_{2}}=\left| \begin{matrix} 1 & 9 & 1 \\ 2 & 52 & 7 \\ 2 & 0 & -1 \\ \end{matrix} \right|\; (Applying \;{{C}_{1}}\to {{C}_{1}}+2{{C}_{3}}),

⇒ Δ2=∣3911652700−1∣=−1(156−144)=−12  and  Δ3=∣1192552210∣  (applying  C1→C1−2C2){{\Delta }_{2}}=\left| \begin{matrix} 3 & 9 & 1 \\ 16 & 52 & 7 \\ 0 & 0 & -1 \\ \end{matrix} \right|=-1\left( 156-144 \right)=-12\;and\;{{\Delta }_{3}}=\left| \begin{matrix} 1 & 1 & 9 \\ 2 & 5 & 52 \\ 2 & 1 & 0 \\ \end{matrix} \right|\; (applying \;{{C}_{1}}\to {{C}_{1}}-2{{C}_{2}})

∴ Δ3=∣−119−8552010∣         (Applying  C1→C1−2C2)=−1(−52+72)=−20{{\Delta }_{3}}=\left| \begin{matrix} -1 & 1 & 9 \\ -8 & 5 & 52 \\ 0 & 1 & 0 \\ \end{matrix} \right|\ \;\;\;\; (Applying \;{{C}_{1}}\to {{C}_{1}}-2{{C}_{2}}) =-1\left( -52+72 \right)=-20

By Cramer’s rule x=Δ1Δ=−4−4=1,y=Δ2Δ=−12−4=3  and  z=Δ3Δ=−20−4=5x=\frac{{{\Delta }_{1}}}{\Delta }=\frac{-4}{-4}=1,y=\frac{{{\Delta }_{2}}}{\Delta }=\frac{-12}{-4}=3 \; and \;z=\frac{{{\Delta }_{3}}}{\Delta }=\frac{-20}{-4}=5

. x=1,y=3,z=5

Illustration: Solve the following linear equations: 4x+5+3y+7=−1  and  6x+5−6y+7=−5\frac{4}{x+5}+\frac{3}{y+7}=-1 \; and \; \frac{6}{x+5}-\frac{6}{y+7}=-5

Solution:

Here in this problem first put 1x+5=a\frac{1}{x+5}=a and 1y+7=b\frac{1}{y+7}=b and then define the determinants

Δ,Δ1  and    Δ2\Delta ,{{\Delta }_{1}}\;and \;\;{{\Delta }_{2}} .

Then by using Cramer’s rule we can get the values of x and y.

Let us put Δ,1x+5=a  and  1y+7=b\Delta ,\frac{1}{x+5}=a\; and \;\frac{1}{y+7}=b then the 2 linear equations become

4a+3b=-1 … (i)

And 6a-6b=-5 … (ii);

Using Cramer’s Rule, we get,

x∣−13−5−6∣=y∣4−16−5∣=1∣436−6∣⇒a6+15=b−20+6=1−24−18\frac{x}{\left| \begin{matrix} -1 & 3 \\ -5 & -6 \\ \end{matrix} \right|}=\frac{y}{\left| \begin{matrix} 4 & -1 \\ 6 & -5 \\ \end{matrix} \right|}=\frac{1}{\left| \begin{matrix} 4 & 3 \\ 6 & -6 \\ \end{matrix} \right|}\Rightarrow \frac{a}{6+15}=\frac{b}{-20+6}=\frac{1}{-24-18} a21=b−14=1−42    ⇒ a=−12  and    b=13\frac{a}{21}=\frac{b}{-14}=\frac{1}{-42} \;\; \Rightarrow \,a=\frac{-1}{2}\;and\;\;b=\frac{1}{3} a=−12   ⇒   1x+5=−12;⇒  2=−x−5⇒x=−7a=-\frac{1}{2}\,\,\, \Rightarrow \,\,\,\frac{1}{x+5}=-\frac{1}{2};\Rightarrow \,\,2=-x-5\Rightarrow x=-7

b=13   ⇒1y+7=13      ⇒   3=y+7   ⇒  y=−4b=\frac{1}{3}\,\,\,\Rightarrow \frac{1}{y+7}=\frac{1}{3} \;\;\;\Rightarrow \,\,\,3=y+7\,\,\,\Rightarrow \,\,y=-4

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.

For non-trivial solution, Δ= 0

⇒∣11−k3−1−21−12∣=0    ⇒∣20−k+220−41−12∣=0    [R1→R1+R3,R2→R2−R3]\Rightarrow \left| \begin{matrix} 1 & 1 & -k \\ 3 & -1 & -2 \\ 1 & -1 & 2 \\ \end{matrix} \right|=0\,\,\,\,\Rightarrow \left| \begin{matrix} 2 & 0 & -k+2 \\ 2 & 0 & -4 \\ 1 & -1 & 2 \\ \end{matrix} \right|=0 \;\;\left[ {{R}_{1}}\to {{R}_{1}}+{{R}_{3}},{{R}_{2}}\to {{R}_{2}}-{{R}_{3}} \right]

Expanding along C2.⇒−(−1)[−8−2(2−k)]=0   ⇒2k−12=0   ⇒k=6{{C}_{2}}.\Rightarrow -\left( -1 \right)\left[ -8-2\left( 2-k \right) \right]=0\,\,\,\Rightarrow 2k-12=0\,\,\,\Rightarrow 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\Rightarrow 4x-8z=0

∴  z=x2\,z=\frac{x}{2}

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=t2,x=t,y=2t,z=\frac{t}{2}, 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 [2122−1113−1][xyz]=[0105]\left[ \begin{matrix} 2 & 1 & 2 \\ 2 & -1 & 1 \\ 1 & 3 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} 0 \\ 10 \\ 5 \\ \end{matrix} \right]

∴ AX = D where A = [2122−1113−1],X=[xyz],D=[0105]\left[ \begin{matrix} 2 & 1 & 2 \\ 2 & -1 & 1 \\ 1 & 3 & -1 \\ \end{matrix} \right],X=\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right],D=\left[ \begin{matrix} 0 \\ 10 \\ 5 \\ \end{matrix} \right] ⇒   A−1(AX)=A−1D     ⇒X=A−1D\Rightarrow \,\,\,{{A}^{-1}}\left( AX \right)={{A}^{-1}}D\,\,\,\,\,\Rightarrow X={{A}^{-1}}D … (i)

Now A−1=adj A∣A∣;      ∣A∣=∣2122−1113−1∣=2(1−3)−1(−2−1)+2(6+1)=13{{A}^{-1}}=\frac{adj\,A}{|A|}; \;\;\; |A|=\left| \begin{matrix} 2 & 1 & 2 \\ 2 & -1 & 1 \\ 1 & 3 & -1 \\ \end{matrix} \right|=2\left( 1-3 \right)-1\left( -2-1 \right)+2\left( 6+1 \right)=13

The matrix of cofactors of |A| is [−2377−4−5324].    So,      adj A=[−2733−427−5−4];  A−1=13[−2733−427−5−4].\left[ \begin{matrix} -2 & 3 & 7 \\ 7 & -4 & -5 \\ 3 & 2 & 4 \\ \end{matrix} \right].\;\;So, \;\;\;adj\,A=\left[ \begin{matrix} -2 & 7 & 3 \\ 3 & -4 & 2 \\ 7 & -5 & -4 \\ \end{matrix} \right];\,\,{{A}^{-1}}=\frac{1}{3}\left[ \begin{matrix} -2 & 7 & 3 \\ 3 & -4 & 2 \\ 7 & -5 & -4 \\ \end{matrix} \right].

from (1), X=113[−2733−427−54][0105]=113[0+70+150−40+100−50−20]=[85/13−30/13−70/13];[xyz]=[85/13−30/13−70/13]X=\frac{1}{13}\left[ \begin{matrix} -2 & 7 & 3 \\ 3 & -4 & 2 \\ 7 & -5 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} 0 \\ 10 \\ 5 \\ \end{matrix} \right]=\frac{1}{13}\left[ \begin{matrix} 0+70+15 \\ 0-40+10 \\ 0-50-20 \\ \end{matrix} \right]=\left[ \begin{matrix} 85/13 \\ -30/13 \\ -70/13 \\ \end{matrix} \right];\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} 85/13 \\ -30/13 \\ -70/13 \\ \end{matrix} \right] ⇒     x=8513,y=−3013,z=−7013\Rightarrow \,\,\,\,\,x=\frac{85}{13},y=\frac{-30}{13},z=\frac{-70}{13}

Related Topics: