System Of Linear Equations Using Determinants
The solution of a system of linear equations with two and three variables can be easily found using determinants. Cramer’s rule is also explained with a diagram, along with formulas and steps to help solve practice problems.
How To Solve a Linear Equation System Using Determinants?
- Define the linear equation system.
- Calculate the determinant of the coefficient matrix.
- If the determinant is not 0, solve the system by using Cramer’s rule.
- If the determinant is 0, then the system has either no solution or an infinite number of solutions.
1. System of Linear Equations with Two Variables
Let the equations be
$$a_1x + b_1y + c_1 = 0$$
$$a_2x + b_2y + c_2 = 0$$
The Solution to a System of Equations with Two Variables is Given by:
(\begin{array}{l}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 $\Delta_1 = \left| \begin{matrix} b_1 & c_1 \ b_2 & c_2 \end{matrix} \right|$, $\Delta_2 = \left| \begin{matrix} c_1 & a_1 \ c_2 & a_2 \end{matrix} \right|$ and $\Delta = \left| \begin{matrix} a_1 & b_1 \ a_2 & b_2 \end{matrix} \right|$
2. System of Linear Equations Involving Three Variables
(\begin{array}{l}{{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}\end{array})
#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|)
CRITERION FOR CONSISTENCY: Now following algorithm is followed to solve the system
Cramer’s rule is a method of finding a solution to a system of equations.
Conditions for Infinite Solutions and No Solutions
(a) If Δ = 0 and Δ1 = Δ2 = Δ3 = 0, then the system of equations may or may not be consistent.
(i) If the value of x, y and z in terms of t satisfies the third equation, the system is said to be consistent and will have infinite solutions.
If the values of x, y, and z do not satisfy the third equation, the system is considered inconsistent and will have no solution.
If d1 = d2 = d3 = 0, then the system of linear equations is known as a Homogeneous Linear Equations, which always has at least one solution, i.e. (0, 0, 0), called a Trivial Solution for Homogeneous Linear Equations.
If the system of homogeneous linear equations has non-zero/nontrivial solutions and Δ = 0, then 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 A X = B, where
$A = \begin{bmatrix} a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \ a_3 & b_3 & c_3 \end{bmatrix}, ; X = \begin{bmatrix} x \ y \ z \end{bmatrix}, ; B = \begin{bmatrix} d_1 \ d_2 \ d_3 \end{bmatrix}$
The solution set can only exist if the inverse of A, $A^{-1}$, exists, which can be obtained by solving $X = A^{-1}B$.
Some Important Results
The Consistency of Three Simultaneous Linear Equations in Two Variables
(\begin{array}{l}{a_1}x + {b_1}y + {c_1} = 0 \dots (i)\end{array})
(\begin{array}{l}{{a}^{2}}x + {{b}^{2}}y + {{c}^{2}} = 0 \ldots (ii)\end{array})
(\begin{array}{l}{a}{3}x + {b}{3}y + {c}_{3} = 0 \dots \text{(iii)}\end{array})
$\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \ \end{matrix} \right|=0$
(a) (\begin{array}{l}a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0\end{array} ) represents a pair of straight lines if $a\neq b$ and $h^2-ab\neq 0$.
(\begin{vmatrix} a & h & g \ h & b & f \ g & f & c \ \end{vmatrix} + 2fgh - af^2 - bg^2 - ch^2 = 0 )
(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) The equation of a straight line passing through $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is given by $\left| \begin{matrix} x & y & 1 \ {{x}_{1}} & {{y}_{1}} & 1 \ {{x}_{2}} & {{y}_{2}} & 1 \ \end{matrix} \right|=0$.
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 of those two terms.
(\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| = \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|)
It should be noted that, when applying operations on determinants, at least one row (or column) must remain unchanged; that is,
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:
$$\begin{array}{l} x+y+z=9 \ 2x+5y+7z=52 \ 2x+y-z=0 \end{array}$$
Given:
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Solution:
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We can define the determinants Δ1, Δ2, and Δ3 in this problem and use the invariance property to find their values. After that, we can use Cramer’s rule to get the values of x, y, and z.
Here (\begin{array}{l}\Delta =\left| \begin{matrix} 1 & 0 & 0 \ 0 & 4 & 6 \ 0 & -1 & -2 \ \end{matrix} \right|\end{array} ) (Applying; {{C}_{2}}\to; {{C}_{2}}-2{{C}_{1}};and ;{{C}_{3}}\to; {{C}_{3}}-{{C}_{1}}) he said
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(\Delta = \left| \begin{matrix} 1 & 0 & 0 \ 2 & 3 & 5 \ 2 & -1 & -3 \ \end{matrix} \right| = 1 \cdot \left(-9 + 5 \right) = -4 )
⇒ (\begin{array}{l}{\Delta }_{1}=\left| \begin{matrix} 9 & 1 & 0 \ 52 & 5 & 8 \ 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|;\textrm{(Applying } {{C}_{1}}\textrm{ to } {{C}_{1}}+2{{C}_{3}})\end{array} )
⇒ $$\Delta_{2}=\left| \begin{matrix} 3 & 9 & 1 \ 16 & 52 & 7 \ 0 & 0 & -1 \ \end{matrix} \right|=-1\left( 156-144 \right)=-12 ;\text{and}; \Delta_{3}=\left| \begin{matrix} 1 & 1 & 9 \ 0 & 3 & 36 \ 2 & 1 & 0 \ \end{matrix} \right|; (\text{applying }{{C}_{1}}\to {{C}_{1}}-2{{C}_{2}})$$
∴ ${\Delta}_{3} = \left| \begin{matrix} -1 & 1 & 9 \ -8 & 5 & 52 \ 0 & 1 & 0 \ \end{matrix} \right| ;;;; (Applying ;{{C}_{1}}\to {{C}_{1}}-2{{C}_{2}}) = -1 \left(-52 + 72 \right) = -20$
By Cramer’s rule: $$ \begin{array}{l} x=\frac{{\Delta }{1}}{\Delta }=\frac{-4}{-4}=1, \ y=\frac{{\Delta }{2}}{\Delta }=\frac{-12}{-4}=3 ; and ; \ z=\frac{{\Delta }_{3}}{\Delta }=\frac{-20}{-4}=5 \end{array} $$
x = 1, y = 3, z = 5
Answer:
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})
Given:
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Solution:
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(\begin{array}{l}\text{Let}\ \frac{1}{x+5}=a\ \text{and}\ \frac{1}{y+7}=b.\ \text{Define the determinants}\ \Delta =ab,\ {{\Delta }_{1}}=a+b\ \text{and}\ {{\Delta }_{2}}=\frac{1}{a+b}.\end{array} )
Then, using Cramer’s rule, we can calculate 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:
(\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}{15-6}=\frac{b}{6-20}=\frac{1}{-24-18}\end{array} )
(\begin{array}{l}a=-\frac{1}{2};and;;b=\frac{1}{3};; \Rightarrow ,\frac{a}{21}=\frac{b}{-14}=\frac{1}{-42} \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}\frac{1}{y+7}=\frac{1}{3} ;;;\Rightarrow ,,,y+7=3 ;;;\Rightarrow ,,,y=-4\end{array} )
Question: 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
Given:
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Solution:
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Definition of Δ: Δ is a symbol used to represent the difference between two values.
Statement Rewrite: In this problem, we first define Δ as the difference between two values. It is known that, for a non-trivial solution, Δ must equal 0.
Using the invariance property, we can solve for $k$ when $\Delta = 0$.
For a non-trivial solution, Δ = 0
(\begin{array}{l}\Rightarrow \left| \begin{matrix} 2 & 0 & -k+2 \ 0 & 0 & -2 \ 1 & -1 & 2 \ \end{matrix} \right|=0 ;;\left[ {{R}_{2}}\to {{R}_{2}}-{{R}_{1}},{{R}_{3}}\to {{R}_{3}}+{{R}_{1}} \right]\end{array} )
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)
(iii) x - y + 2z = 0
(i) (4x - 8z = 0)
(ii) (\frac{4x}{8} = \frac{z}{-1})
z = \frac{x}{2}
Putting the value of $z$ in $(i)$, we get $x+y-3x=0$
y = 2x
When k = 6, the solution of the given system of equations is (x=t, y=2t, z=\frac{t}{2}), where t is an arbitrary number.
Illustration: Use matrix inversion to solve the following equations.
2x + y + 2z = 0
2x - y + z = 10
x + 3y - z = 5
Given:
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Solution:
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By writing the given equations into the form of $AX = D$, and then multiplying both sides by $A^{-1}$, we will get the required values of $x$, $y$, and $z$.
The equations can be expressed in matrix form as $$\begin{bmatrix} 2 & 1 & 2 \ 2 & -1 & 1 \ 1 & 3 & -1 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix}
\begin{bmatrix} 0 \ 10 \ 5 \end{bmatrix}$$
$$AX = D$$ where $$A = \left[ \begin{matrix} 2 & 1 & 2 \ 2 & -1 & 1 \ 1 & 3 & -1 \ \end{matrix} \right],\ X=\left[ \begin{matrix} x \ y \ z \ \end{matrix} \right],\ D=\left[ \begin{matrix} 0 \ 10 \ 5 \ \end{matrix} \right]$$
$$\Rightarrow ,,,{{A}^{-1}}AX = {{A}^{-1}}D,,,,,\Rightarrow X={{A}^{-1}}D….(i)$$
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]\end{array}); 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), $$\begin{array}{l}X=\frac{1}{13}\left[ \begin{matrix} -2 & 7 & 3 \ 3 & -4 & 2 \ 7 & -5 & 4 \ \end{matrix} \right]\left[ \begin{matrix} 0 \ 10 \ 5 \ \end{matrix} \right]\=\frac{1}{13}\left[ \begin{matrix} 0+70+15 \ 0-40+10 \ 0-50-20 \ \end{matrix} \right]\=\left[ \begin{matrix} 85/13 \ -30/13 \ -70/13 \ \end{matrix} \right];\left[ \begin{matrix} x \ y \ z \ \end{matrix} \right]\=\left[ \begin{matrix} 85/13 \ -30/13 \ -70/13 \ \end{matrix} \right]\end{array} \quad \Rightarrow \quad \left[ \begin{matrix} x \ y \ z \ \end{matrix} \right] = \left[ \begin{matrix} 85/13 \ -30/13 \ -70/13 \ \end{matrix} \right]\end{array}$$
(\begin{array}{l}x=\frac{85}{13},\ y=\frac{-30}{13},\ z=\frac{-70}{13}\end{array})
Related Topics:
Differentiation and Integration of Determinants
Frequently Asked Questions
A system of linear equations has an infinite solution when the equations are linearly dependent.
If $$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$, then the system of equations has infinite solutions.
The condition that a system of linear equations has no solution is when the equations are inconsistent.
If $$\frac{a_1}{a_2} \neq \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$$, then the system of equations has no solution.
Are Three Points Collinear?
Three points are collinear if the area of the triangle formed by the three points is equal to zero.
