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. …
Coefficient determinant Δ=221\1−11\312=0. Since Δ=0, test Δx (replace the x-column with the constants): Δx=521\1−11\412=−6eq0. Because Δ=0 but Δxeq0, the system i …
Compute the coefficient determinant Δ by Cramer's rule; since Δ=0, test Δx (and confirm via elimination) to show the system is inconsistent with no solution.
Write the coefficient determinant: Δ=2132−11112.
Expand along column 1: Δ=2[(−1)(2)−(1)(1)]−1[(2)(2)−(1)(1)]+3[(2)(1)−(1)(−1)]=2(−3)−1(3)+3(3)=−6−3+9=0.
Since Δ=0, the system either has no solution or infinitely many; test Δx (replace the x-coefficients 2,1,3 with the RHS constants 5,1,4): Δx=5142−11112.
Expand along row 1: Δx=5[(−1)(2)−(1)(1)]−2[(1)(2)−(1)(4)]+1[(1)(1)−(−1)(4)]=5(−3)−2(−2)+1(5)=−15+4+5=−6. …