Q27
3 marksShort AnswerSection C

Using Cramer's rule, show that the following system of linear equations is consistent and hence solve it :

2x−3y+5z=112x - 3y + 5z = 11

3x+2y−4z=−53x + 2y - 4z = -5

x+y−2z=−3x + y - 2z = -3

Determinants
Cramer's Rule for a $3\times3$ System
Official Answer

By Cramer's rule the system is consistent because the coefficient determinant D is non-zero.


Coefficient determinant:

  • D=∣2−3532−411−2∣=2(0)+3(−2)+5(1)=−1D = \begin{vmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{vmatrix} = 2(0) + 3(-2) + 5(1) = -1.
  • Since D=−1D = -1 (not equal to 0), the system is consistent with a unique solution.

Replacement determinants:

  • Dx=∣11−35−52−4−31−2∣=−1D_x = \begin{vmatrix} 11 & -3 & 5 \\ -5 & 2 & -4 \\ -3 & 1 & -2 \end{vmatrix} = -1.
  • Dy=∣21153−5−41−3−2∣=−2D_y = \begin{vmatrix} 2 & 11 & 5 \\ 3 & -5 & -4 \\ 1 & -3 & -2 \end{vmatrix} = -2.
  • Dz=∣2−31132−511−3∣=−3D_z = \begin{vmatrix} 2 & -3 & 11 \\ 3 & 2 & -5 \\ 1 & 1 & -3 \end{vmatrix} = -3.

Solution:

  • x=Dx/D=−1/−1=1x = D_x/D = -1/-1 = 1.
  • y=Dy/D=−2/−1=2y = D_y/D = -2/-1 = 2.
  • z=Dz/D=−3/−1=3z = D_z/D = -3/-1 = 3.

Final answer: x=1,y=2,z=3x = 1, y = 2, z = 3.

Cramer's ruledeterminantconsistent systemcoefficient determinantunique solutionD not equal to zerox=1 y=2 z=3linear equations

Marking Scheme

  • 11 mark: correct coefficient determinant D=−1D = -1 and statement that D≠0D \ne 0 implies the system is consistent.
  • 21.5 marks: correct evaluation of Dx=−1D_x = -1, Dy=−2D_y = -2, Dz=−3D_z = -3 (0.5 mark each).
  • 30.5 mark: correct solution x=1,y=2,z=3x = 1, y = 2, z = 3 (award only if divisions are shown).

Hint

First evaluate the coefficient determinant D. If D is non-zero the system is consistent; then find Dx,Dy,DzD_x, D_y, D_z and use x=Dx/D,y=Dy/D,z=Dz/Dx = D_x/D, y = D_y/D, z = D_z/D.

Quick Oral Answer

I compute the coefficient determinant D; it equals -1, which is non-zero, so the system is consistent with a unique solution, and Dx/D,Dy/D,Dz/DD_x/D, D_y/D, D_z/D give x=1,y=2,z=3x=1, y=2, z=3.

Analysis & Explanation

Cramer's rule turns solving simultaneous equations into a determinant computation, and the sign of D immediately settles consistency.


Concept:

  • For a system AX=BAX = B, if D=det⁡(A)D = \det(A) is non-zero the system has a unique solution and is therefore consistent. Each variable equals the ratio of a modified determinant (formed by replacing the relevant coefficient column with B) to D.

Consistency check:

  • Here D=−1D = -1, which is non-zero, so we can guarantee consistency before even finding x, y, z. This is what the phrase 'show that the system is consistent' demands.

Verification (always do this):

  • Substitute x=1,y=2,z=3x=1, y=2, z=3: 2−6+15=112-6+15 = 11, 3+4−12=−53+4-12 = -5, 1+2−6=−31+2-6 = -3. All three original equations hold.

Exam trap:

  • Sign errors while expanding 3×33 \times 3 determinants are the biggest score-killers. Expanding consistently along the same row and rechecking the cofactor signs (+−+)(+ - +) avoids them.

Real-world link:

  • Cramer's rule and determinant methods underpin solving equilibrium equations in economics (supply-demand across markets) and circuit analysis in engineering.

Common Mistakes

  1. 1Sign errors in expanding the 3×33 \times 3 determinants, especially forgetting the alternating cofactor signs (+−+)(+ - +).
  2. 2Replacing the wrong column when forming Dx,DyD_x, D_y or DzD_z.
  3. 3Concluding inconsistency because D is negative; consistency depends only on D being non-zero, not on its sign.

Interesting Facts

Cramer's rule is named after Swiss mathematician Gabriel Cramer, who published it in 1750, though Colin Maclaurin had described the idea earlier around 1729.

For large systems Cramer's rule is computationally expensive (it needs many determinants), which is why computers use Gaussian elimination instead, but for 2×22\times2 and 3×33\times3 systems it is fast and exam-friendly.

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Frequently Asked Questions

How does Cramer's rule prove a system is consistent?

If the coefficient determinant D is not equal to zero, the matrix is invertible, so the system has exactly one solution and is therefore consistent. Here D=−1D = -1 which is non-zero, guaranteeing a unique consistent solution before you even compute x, y and z.

Should I verify the answer after using Cramer's rule?

Yes. Substituting x=1,y=2,z=3x = 1, y = 2, z = 3 back into all three equations gives 11, -5 and -3 exactly, confirming the solution. Verification catches sign errors in determinant expansion and often earns confidence marks.