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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, below:

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:

\(\begin{array}{l}x=\frac{{{\Delta }_{1}}}{\Delta },y=\frac{{{\Delta }_{2}}}{\Delta } \;or \;x=\frac{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}};y=\frac{{{a}_{2}}{{c}_{1}}-{{a}_{1}}{{c}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\end{array} \)

Where

\(\begin{array}{l}{{\Delta }_{1}}=\left| \begin{matrix} {{b}_{1}} & {{c}_{1}} \\ {{b}_{2}} & {{c}_{2}} \\ \end{matrix} \right|,\;{{\Delta }_{2}}=\left| \begin{matrix} {{c}_{1}} & {{a}_{1}} \\ {{c}_{2}} & {{a}_{2}} \\ \end{matrix} \right|\; and \;\Delta =\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{2}} & {{b}_{2}} \\ \end{matrix} \right|\end{array} \)

2. System of Linear Equations Involving Three Variables

\(\begin{array}{l}{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z={{d}_{1}} \;\;and\;\;{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z={{d}_{2}} \;\;and\;{{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}z={{d}_{3}}\end{array} \)

To solve this system, we need to first define the following determinants:

\(\begin{array}{l}\Delta =\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|,{{\Delta }_{1}}=\left| \begin{matrix} {{d}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{d}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{d}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|,{{\Delta }_{2}}=\left| \begin{matrix} {{a}_{1}} & {{d}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{d}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{d}_{3}} & {{c}_{3}} \\ \end{matrix} \right|,{{\Delta }_{3}}=\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{d}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{d}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{d}_{3}} \\ \end{matrix} \right|\end{array} \)

Now, the following algorithm is used 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 and Δ1 = Δ2 = Δ3 = 0, then the system of the 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, and 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 the 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, the 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

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

Now, the 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:

\(\begin{array}{l}{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0….(i)\end{array} \)
\(\begin{array}{l}{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0….(ii)\end{array} \)
\(\begin{array}{l}{{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}=0….(iii)\end{array} \)

are concurrent if,

\(\begin{array}{l}\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|=0\end{array} \)

 

(a)

\(\begin{array}{l}a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0\end{array} \)
represents a pair of straight lines if

\(\begin{array}{l}abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0=\left| \begin{matrix} a & h & g \\ h & b & f \\ g & f & c \\ \end{matrix} \right|\end{array} \)

(b) Area of a triangle whose vertices are

\(\begin{array}{l}\left( {{x}_{r}},{{y}_{r}} \right);\,\,r=1,2,3 \;\;is: \;\;\;D=\frac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\ \end{matrix} \right|.\end{array} \)

If D = 0, then the three points are collinear.

(c) Equation of a straight line passing through

\(\begin{array}{l}\left( {{x}_{1}},{{y}_{1}} \right)\And \left( {{x}_{2}},{{y}_{2}} \right) \;\;is \;\;\left| \begin{matrix} x & y & 1 \\ {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ \end{matrix} \right|=0.\end{array} \)

(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.,

\(\begin{array}{l}\left| \begin{matrix} {{a}_{1}}+x & {{b}_{1}}+y & {{c}_{1}}+z \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|=\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|+\left| \begin{matrix} x & y & z \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|\end{array} \)

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

\(\begin{array}{l}x+y+z=9,2x+5y+7z=52,2x+y-z=0.\end{array} \)

Solution:

Here, in this problem, define the determinants Δ1, Δ2, and Δ3 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

\(\begin{array}{l}\Delta =\left| \begin{matrix} 1 & 1 & 1 \\ 2 & 5 & 7 \\ 2 & 1 & -1 \\ \end{matrix} \right| \;(Applying\; {{C}_{2}}\to\; {{C}_{2}}-{{C}_{1}}\;and \;{{C}_{3}}\to\; {{C}_{3}}-{{C}_{1}})\end{array} \)

Now,

\(\begin{array}{l}\Delta =\left| \begin{matrix} 1 & 0 & 0 \\ 2 & 3 & 5 \\ 2 & -1 & -3 \\ \end{matrix} \right|=1\left( -9+5 \right)=-4\end{array} \)

⇒ 

\(\begin{array}{l}{\Delta }_{1}=\left| \begin{matrix} 9 & 1 & 1 \\ 52 & 5 & 7 \\ 0 & 1 & -1 \\ \end{matrix} \right| \;(Applying \;C_{2}\to C_{2}+C_{3})\end{array} \)

\(\begin{array}{l}{{\Delta }_{1}}=\left| \begin{matrix} 9 & 2 & 1 \\ 52 & 12 & 7 \\ 0 & 0 & -1 \\ \end{matrix} \right|=-1\left( 108-104 \right)=-4;\,\,{{\Delta }_{2}}=\left| \begin{matrix} 1 & 9 & 1 \\ 2 & 52 & 7 \\ 2 & 0 & -1 \\ \end{matrix} \right|\; (Applying \;{{C}_{1}}\to {{C}_{1}}+2{{C}_{3}})\end{array} \)
,

\(\begin{array}{l}{{\Delta }_{2}}=\left| \begin{matrix} 3 & 9 & 1 \\ 16 & 52 & 7 \\ 0 & 0 & -1 \\ \end{matrix} \right|=-1\left( 156-144 \right)=-12\;and\;{{\Delta }_{3}}=\left| \begin{matrix} 1 & 1 & 9 \\ 2 & 5 & 52 \\ 2 & 1 & 0 \\ \end{matrix} \right|\; (applying \;{{C}_{1}}\to {{C}_{1}}-2{{C}_{2}})\end{array} \)

\(\begin{array}{l}{{\Delta }_{3}}=\left| \begin{matrix} -1 & 1 & 9 \\ -8 & 5 & 52 \\ 0 & 1 & 0 \\ \end{matrix} \right|\ \;\;\;\; (Applying \;{{C}_{1}}\to {{C}_{1}}-2{{C}_{2}}) =-1\left( -52+72 \right)=-20\end{array} \)

By Cramer’s rule

\(\begin{array}{l}x=\frac{{{\Delta }_{1}}}{\Delta }=\frac{-4}{-4}=1,y=\frac{{{\Delta }_{2}}}{\Delta }=\frac{-12}{-4}=3 \; and \;z=\frac{{{\Delta }_{3}}}{\Delta }=\frac{-20}{-4}=5\end{array} \)

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

Illustration: Solve the following linear equations:

\(\begin{array}{l}\frac{4}{x+5}+\frac{3}{y+7}=-1 \; and \; \frac{6}{x+5}-\frac{6}{y+7}=-5\end{array} \)

Solution:

\(\begin{array}{l}\text{Here in this problem first put}\ \frac{1}{x+5}=a\ \text{and}\ \frac{1}{y+7}=b\ \text{and then define the determinants}\ \Delta ,{{\Delta }_{1}}\ \text{and}\ {{\Delta }_{2}}.\end{array} \)

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

Let us put

\(\begin{array}{l}\Delta ,\frac{1}{x+5}=a\; and \;\frac{1}{y+7}=b\end{array} \)
, then the 2 linear equations become

4a + 3b = -1 … (i)

And 6a – 6b = -5 … (ii);

Using Cramer’s rule, we get,

\(\begin{array}{l}\frac{x}{\left| \begin{matrix} -1 & 3 \\ -5 & -6 \\ \end{matrix} \right|}=\frac{y}{\left| \begin{matrix} 4 & -1 \\ 6 & -5 \\ \end{matrix} \right|}=\frac{1}{\left| \begin{matrix} 4 & 3 \\ 6 & -6 \\ \end{matrix} \right|}\\\Rightarrow \frac{a}{6+15}=\frac{b}{-20+6}=\frac{1}{-24-18}\end{array} \)
\(\begin{array}{l}\frac{a}{21}=\frac{b}{-14}=\frac{1}{-42} \;\; \Rightarrow \,a=\frac{-1}{2}\;and\;\;b=\frac{1}{3}\end{array} \)
\(\begin{array}{l}a=-\frac{1}{2}\,\,\, \Rightarrow \,\,\,\frac{1}{x+5}=-\frac{1}{2};\Rightarrow \,\,2=-x-5\Rightarrow x=-7\end{array} \)

\(\begin{array}{l}b=\frac{1}{3}\,\,\,\Rightarrow \frac{1}{y+7}=\frac{1}{3} \;\;\;\Rightarrow \,\,\,3=y+7\,\,\,\Rightarrow \,\,y=-4\end{array} \)

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:

In this problem, first, define Δ. As we know, for a 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

\(\begin{array}{l}\Rightarrow \left| \begin{matrix} 1 & 1 & -k \\ 3 & -1 & -2 \\ 1 & -1 & 2 \\ \end{matrix} \right|=0\,\,\,\,\Rightarrow \left| \begin{matrix} 2 & 0 & -k+2 \\ 2 & 0 & -4 \\ 1 & -1 & 2 \\ \end{matrix} \right|=0 \;\;\left[ {{R}_{1}}\to {{R}_{1}}+{{R}_{3}},{{R}_{2}}\to {{R}_{2}}-{{R}_{3}} \right]\end{array} \)

Expanding along

\(\begin{array}{l}{{C}_{2}}.\Rightarrow -\left( -1 \right)\left[ -8-2\left( 2-k \right) \right]=0\,\,\,\Rightarrow 2k-12=0\,\,\,\Rightarrow k=6\end{array} \)

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)

\(\begin{array}{l}\Rightarrow 4x-8z=0\end{array} \)

∴ z = x/2

Putting the value of z in (i), we get x+y-3x = 0

∴ y = 2x

Thus when k = 6, the solution of the given system of equations will be

\(\begin{array}{l}x=t,y=2t,z=\frac{t}{2},\end{array} \)
, 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 sides by A-1, we will get the required value of x, y and z.

In the matrix form, the equations can be written as

\(\begin{array}{l}\left[ \begin{matrix} 2 & 1 & 2 \\ 2 & -1 & 1 \\ 1 & 3 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} 0 \\ 10 \\ 5 \\ \end{matrix} \right]\end{array} \)

∴ AX = D where

\(\begin{array}{l}A = \left[ \begin{matrix} 2 & 1 & 2 \\ 2 & -1 & 1 \\ 1 & 3 & -1 \\ \end{matrix} \right],X=\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right],D=\left[ \begin{matrix} 0 \\ 10 \\ 5 \\ \end{matrix} \right]\end{array} \)
\(\begin{array}{l}\Rightarrow \,\,\,{{A}^{-1}}\left( AX \right)={{A}^{-1}}D\,\,\,\,\,\Rightarrow X={{A}^{-1}}D….(i)\end{array} \)

Now

\(\begin{array}{l}{{A}^{-1}}=\frac{adj\,A}{|A|}; \;\;\; |A|=\left| \begin{matrix} 2 & 1 & 2 \\ 2 & -1 & 1 \\ 1 & 3 & -1 \\ \end{matrix} \right|=2\left( 1-3 \right)-1\left( -2-1 \right)+2\left( 6+1 \right)=13\end{array} \)

The matrix of cofactors of |A| is

\(\begin{array}{l}\left[ \begin{matrix} -2 & 3 & 7 \\ 7 & -4 & -5 \\ 3 & 2 & 4 \\ \end{matrix} \right]\\ So, \;\;\;adj\,A=\left[ \begin{matrix} -2 & 7 & 3 \\ 3 & -4 & 2 \\ 7 & -5 & -4 \\ \end{matrix} \right];\,\,{{A}^{-1}}=\frac{1}{3}\left[ \begin{matrix} -2 & 7 & 3 \\ 3 & -4 & 2 \\ 7 & -5 & -4 \\ \end{matrix} \right].\end{array} \)

from (1),

\(\begin{array}{l}X=\frac{1}{13}\left[ \begin{matrix} -2 & 7 & 3 \\ 3 & -4 & 2 \\ 7 & -5 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} 0 \\ 10 \\ 5 \\ \end{matrix} \right]\\=\frac{1}{13}\left[ \begin{matrix} 0+70+15 \\ 0-40+10 \\ 0-50-20 \\ \end{matrix} \right]\\=\left[ \begin{matrix} 85/13 \\ -30/13 \\ -70/13 \\ \end{matrix} \right];\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]\\=\left[ \begin{matrix} 85/13 \\ -30/13 \\ -70/13 \\ \end{matrix} \right]\end{array} \)
\(\begin{array}{l}\Rightarrow \,\,\,\,\,x=\frac{85}{13},y=\frac{-30}{13},z=\frac{-70}{13}\end{array} \)

Related Topics:

Frequently Asked Questions

Q1

Give the condition that a system of linear equations has an infinite solution.

If a1/a2 = b1/b2 = c1/c2, then the system of equations has an infinite solution.

Q2

Give the condition that a system of linear equations has no solution.

If a1/a2 = b1/b2 ≠ c1/c2, then the system of equations has no solution.

Q3

How to check whether the 3 points are collinear?

Three points are collinear if the value of the area of the triangle formed by the three points is zero.

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