Q.Solve the following linear equations by using Cramer's Rule: x+z=1, y+z=1, x+y=4
Concept understanding — Cramer's Rule
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.
If Δ=0, Cramer's rule simply cannot be applied -- it says nothing about whether the system has no solution or infinitely many; that question needs the rank method (§ Consistency by Rank Method). A closely related determinant trick solves xayb=em, xcyd=en: taking logs turns this into a linear system in lnx,lny, so Cramer's rule on that linear system gives lnx,lny as ratios of 2×2 determinants, and x,y follow by exponentiating.
"Cramer's rule formula for 3 variables" and "Cramer's rule vs matrix method" are frequent search terms among Class 12 students, since this method for solving linear systems is an important topic for JEE Main and several state board and CET-level exams that build on the NCERT Class 12 Determinants and Matrices curriculum. Being able to switch fluently between Cramer's rule and the matrix-inversion method taught in NCERT is a common requirement in exam questions that ask for a specific unknown without solving the full system.
Insert zero coefficients for missing variables, then use Cramer's Rule.
x=2, y=2, z=-1
Rewrite as x+0y+z=1, 0x+y+z=1, x+y+0z=4.
D=101011110=1(0−1)−0(0−1)+1(0−1)=−1+0−1=−2.
Dx=114011110=1(0−1)−0(0−4)+1(1−4)=−1+0−3=−4.
Dy=101114110=1(0−4)−1(0−1)+1(0−1)=−4+1−1=−4.
Dz=101011114=1(4−1)−0(0−1)+1(0−1)=3−0−1=2.
x=−2−4=2, y=−2−4=2, z=−22=−1. Check: x+z=2−1=1 ✓, y+z=2−1=1 ✓, x+y=2+2=4 ✓.
x=2, y=2, z=-1
Rewrite each equation with all three variables present (zero coefficient where a variable is missing), then apply Cramer's Rule.
- Forgetting to insert the 0 coefficient for the variable missing from an equation
- Sign errors while expanding determinants that contain zeros
- Not verifying the solution satisfies all three original equations
- CBSE 2018Set ANNUAL1 markMCQQ.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.
- Here Δ=0, Δx=0, Δz=0, but Δy=0.
- Since not all of Δx,Δy,Δz vanish while Δ=0, the system falls into the inconsistent case.
- Hence the system has no solution.
✓Final answerThe system is inconsistent — it has no solution — option (c).
- CBSE 2017Set ANNUAL1 markMCQQ.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 (matching exactly the Δ2 and Δ3 defined in the question, since Δ2 replaces column 1 of Δ1 with (c,d), and Δ3 replaces column 2 of Δ1 with (c,d)).
- Since X=ex, we get x=lnX=lnΔ1Δ2.
- Since Y=ey, we get y=lnY=lnΔ1Δ3.
- So (x,y)=(logΔ1Δ2, logΔ1Δ3).
- This matches option (b).
✓Final answer(x,y)=(logΔ1Δ2, logΔ1Δ3).
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