# 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.

All Topics in Determinants

## How To Solve a Linear Equation System Using Determinants?

### 1. System Of Linear Equations Involving Two Variables Using Determinants

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

$x=\frac{{{\Delta }_{1}}}{\Delta },y=\frac{{{\Delta }_{2}}}{\Delta } \;or \;x=\frac{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}};y=\frac{{{a}_{2}}{{c}_{1}}-{{a}_{1}}{{c}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}$

Where ${{\Delta }_{1}}=\left| \begin{matrix} {{b}_{1}} & {{c}_{1}} \\ {{b}_{2}} & {{c}_{2}} \\ \end{matrix} \right|,\;{{\Delta }_{2}}=\left| \begin{matrix} {{c}_{1}} & {{a}_{1}} \\ {{c}_{2}} & {{a}_{2}} \\ \end{matrix} \right|\; and \;\Delta =\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{2}} & {{b}_{2}} \\ \end{matrix} \right|$

### 2. System Of Linear Equations Involving Three Variables Using Determinants

${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z={{d}_{1}} \;\;\;\;{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z={{d}_{2}} \;\;\;{{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}z={{d}_{3}}$

To solve this system we first define the following determinants

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

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

This method of finding a solution to a system of equations is called Cramer’s rule.

### Conditions for Infinite and No Solutions

(a) If $\Delta =0$ and ${{\Delta }_{1}}={{\Delta }_{2}}={{\Delta }_{3}}=0,$ 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 ${{d}_{1}}={{d}_{2}}={{d}_{3}}=0,$ 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

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

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: ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0$ … (i)

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

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

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

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

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

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

If D = 0 then the three points are collinear.

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

(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., $\left| \begin{matrix} {{a}_{1}}+x & {{b}_{1}}+y & {{c}_{1}}+z \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|=\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|+\left| \begin{matrix} x & y & z \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|$

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

Solution:

Here in this problem define the determinants $\Delta ,{{\Delta }_{1}},{{\Delta }_{2}} \;and\; {{\Delta }_{3}}$ 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 $\Delta =\left| \begin{matrix} 1 & 1 & 1 \\ 2 & 5 & 7 \\ 2 & 1 & -1 \\ \end{matrix} \right| \;(Applying\; {{C}_{2}}\to\; {{C}_{2}}-{{C}_{1}}\;and \;{{C}_{3}}\to\; {{C}_{3}}-{{C}_{1}})$ $\Delta =\left| \begin{matrix} 1 & 0 & 0 \\ 2 & 3 & 5 \\ 2 & -1 & -3 \\ \end{matrix} \right|=1\left( -9+5 \right)=-4$

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

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

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

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

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

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

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

Solution:

Here in this problem first put $\frac{1}{x+5}=a \;and\; \frac{1}{y+7}=b\;and\; then \;define \;the\; determinants \Delta ,{{\Delta }_{1}}\;and \;\;{{\Delta }_{2}}$ . Then by using Cramer’s rule we can get the values of x and y.

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

4a+3b=-1 … (i)

And 6a-6b=-5 … (ii);

Using Cramer’s Rule, we get,

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

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

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

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

Expanding along ${{C}_{2}}.\Rightarrow -\left( -1 \right)\left[ -8-2\left( 2-k \right) \right]=0\,\,\,\Rightarrow 2k-12=0\,\,\,\Rightarrow k=6$

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) $\Rightarrow 4x-8z=0$

∴ $\,z=\frac{x}{2}$

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 $x=t,y=2t,z=\frac{t}{2},$ 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 $\left[ \begin{matrix} 2 & 1 & 2 \\ 2 & -1 & 1 \\ 1 & 3 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} 0 \\ 10 \\ 5 \\ \end{matrix} \right]$

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

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

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

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