For a system of 3 linear equations in 3 unknowns ai1x1+ai2x2+ai3x3=bi (i=1,2,3) whose coefficient determinant Δ=a11a21a31a12a22a32a13a23a33 is non-zero, Cramer's rule gives each unknown directly as a ratio of two determinants:
x1=ΔΔ1,x2=ΔΔ2,x3=ΔΔ3,
where Δk is Δ with its kth column replaced by the constants column (b1,b2,b3)T, everything else unchanged. (The same pattern extends to 2 equations in 2 unknowns: Δ=a11a21a12a22, x=Δ1/Δ, y=Δ2/Δ.)
Why it works. Multiplying Δ by x1 and using the linearity-in-a-column property of determinants (splitting the first column ai1x1 into the sum ai1x1+ai2x2+ai3x3 using the original equations, then subtracting off the x2,x3 multiples of the identical columns 2 and 3, which vanish) collapses the first column to exactly the constants bi -- giving x1Δ=Δ1, and dividing by Δ=0 gives the rule.
Worked illustration. For x+y=3,2x−y=0: Δ=121−1=−3, Δ1=301−1=−3, Δ2=1230=−6. So x=Δ1/Δ=1, y=Δ2/Δ=2 -- check: 1+2=3 and 2(1)−2=0, correct.
Word problems that produce equations like y=ax2+bx+c through three given points, or rate/mixture/scoring problems, translate to a 3×3 system in the unknown constants exactly as for matrix inversion, then Cramer's rule reads off each unknown independently -- convenient when only one or two of the unknowns are actually needed. A system with fractional unknowns like xa+by=c is first turned linear by the substitution u=x1 (or y1, z1), solved for u,v,(w) by Cramer's rule, and only inverted back to x,y,(z) at the very last step. …
Forming the coefficient determinant D and the three determinants Dx,Dy,Dz (replacing one column at a time with the constants) and applying Cramer's rule $x= …
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
CBSE 2018Set ANNUAL1 markMCQ
Q.In a system of 3 linear non-homogeneous equations with three unknowns, if Δ=0; and Δx=0; Δy=0; Δz=0 then the system has :
(a) infinitely many solutions
(b) unique solution
(c) no solution
(d) two solutions
›Reveal solutionSolution
With Δ=0 but Δy=0, the standard consistency test for a 3×3 linear system shows it is inconsistent, i.e. it has no solution.
For a system of 3 linear equations in 3 unknowns with determinant of coefficients Δ and determinants Δx,Δy,Δz (formed by replacing the respective column with the constants), the consistency rules are: if Δ=0, unique solution; if Δ=0 and Δx=Δy=Δz=0, either no solution or infinitely many (needs rank check); if Δ=0 and at least one of Δx,Δy,Δz=0, the system is inconsistent. …
Q.If aex+bey=c; pex+qey=d and Δ1=apbq; Δ2=cdbq; Δ3=apcd then the value of (x,y) is :
(a) (Δ1Δ2,Δ1Δ3)
(b) (logΔ1Δ2,logΔ1Δ3)
(c) (logΔ3Δ1,logΔ2Δ1)
(d) (logΔ2Δ1,logΔ3Δ1)
›Reveal solutionSolution
Substituting X=ex,Y=ey turns the exponential system into a linear system solvable by Cramer's rule; solving for X,Y and then taking logarithms recovers x,y in terms of the given determinants.
Given: aex+bey=c and pex+qey=d.
Substitute X=ex, Y=ey to linearize: aX+bY=c and pX+qY=d.
This is a standard 2×2 linear system in X,Y. By Cramer's rule, with coefficient determinant Δ1=apbq:
X=Δ1cdbq=Δ1Δ2,Y=Δ1apcd=Δ1Δ3 …