NDA Maths · Matrices & Determinants

Linear Systems: Consistency & Cramer's Rule

A square linear system AX = B is solved and classified by one number — the coefficient determinant: nonzero gives a unique Cramer's-rule solution, zero gives either no solution or infinitely many.

Why this matters

Eight PYQs — small but a reliable scorer and the payoff of the whole chapter: determinants and inverses applied to solving equations. Questions test consistency from the coefficient determinant, Cramer's rule, homogeneous systems, and finding the parameter k that makes a system consistent or not. Four concepts cover it.

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Concept 1 of 4

Consistency from the coefficient determinant

Intuition

For a square system AX=BAX = B, the coefficient determinant A|A| tells you everything about the solution count BEFORE you solve: A0|A| \neq 0 means exactly one solution; A=0|A| = 0 means either no solution or infinitely many.

Definition

For AX=BAX = B (square): if A0|A| \neq 0unique solution. If A=0|A| = 0: the system is inconsistent (no solution) or has infinitely many solutions, decided by whether the equations are genuinely contradictory or just dependent (e.g. two equations the same scaling but different constants → no solution).

Worked example

Classify the system x+y+z=6, xy+z=2, 2x+yz=1x+y+z=6,\ x-y+z=2,\ 2x+y-z=1.
  1. Form the coefficient determinant A=111111211|A| = \begin{vmatrix}1&1&1\\1&-1&1\\2&1&-1\end{vmatrix}.
  2. Expand along row 1: 1(11)1(12)+1(1+2)=0+3+3=61(1-1) - 1(-1-2) + 1(1+2) = 0 + 3 + 3 = 6.
  3. A=60|A| = 6 \neq 0, so the coefficient matrix is non-singular.
Answer:Unique solution — consistent and independent (A=60|A| = 6 \neq 0).
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Matrices & DeterminantsEASY
The system of linear equations x+2y+z=4x+2y+z=4, 2x+4y+2z=82x+4y+2z=8 and 3x+6y+3z=103x+6y+3z=10 has

[Q8 · Apr · 2023]

Concept 2 of 4

Cramer's rule

Intuition

When A0|A| \neq 0, each unknown is a ratio of determinants: replace the column of coefficients for that variable with the constants column, take the determinant, and divide by A|A|.

Definition

For AX=BAX = B with A=Δ0|A| = \Delta \neq 0: x=Δx/Δx = \Delta_x/\Delta, y=Δy/Δy = \Delta_y/\Delta, z=Δz/Δz = \Delta_z/\Delta, where Δx\Delta_x is Δ\Delta with the xx-column replaced by BB (and similarly for Δy,Δz\Delta_y, \Delta_z).

Cramer's rule

x=ΔxΔ,y=ΔyΔ,z=ΔzΔx = \frac{\Delta_x}{\Delta}, \quad y = \frac{\Delta_y}{\Delta}, \quad z = \frac{\Delta_z}{\Delta}

Worked example

Solve x+y=3, xy=1x + y = 3,\ x - y = 1 by Cramer's rule.
  1. Δ=1111=2\Delta = \begin{vmatrix}1&1\\1&-1\end{vmatrix} = -2.
  2. Δx=3111=4\Delta_x = \begin{vmatrix}3&1\\1&-1\end{vmatrix} = -4; Δy=1311=2\Delta_y = \begin{vmatrix}1&3\\1&1\end{vmatrix} = -2.
  3. x=42=2, y=22=1x = \tfrac{-4}{-2} = 2,\ y = \tfrac{-2}{-2} = 1.
Answer:x=2, y=1x = 2,\ y = 1.
Practice this concept4 quick reps

From the bank · past-year question

Example 2Matrices & DeterminantsMODERATE
In obtaining the solution of the system of equations x+y+z=7x+y+z=7, x+2y+3z=16x+2y+3z=16 and x+3y+4z=22x+3y+4z=22 by Cramer's rule, the value of yy is obtained by dividing DD by D2D_2, where D=111123134D=\begin{vmatrix}1&1&1\\1&2&3\\1&3&4\end{vmatrix}. What is the value of the determinant D2D_2?

[Q14 · Sep · 2025]

Concept 3 of 4

Homogeneous systems and the solution space

Intuition

A homogeneous system AX=0AX = 0 always has the trivial solution X=0X = 0. It has extra (non-trivial) solutions exactly when A=0|A| = 0. And if a non-homogeneous system has two distinct solutions, it must have infinitely many.

Definition

AX=0AX = 0 has a non-trivial solution iff A=0|A| = 0 (otherwise only X=0X = 0). For AX=BAX = B: if X1X2X_1 \neq X_2 are both solutions, then X1+t(X2X1)X_1 + t(X_2 - X_1) is a solution for every ttinfinitely many (the solution set is never exactly two).

Worked example

If AX=BAX = B has two distinct solutions X1X_1 and X2X_2, how many solutions does it have?
  1. X1X_1 and X2X_2 solve it, so A(X1X2)=0A(X_1 - X_2) = 0 — a non-trivial solution of the homogeneous system.
  2. Then X1+t(X2X1)X_1 + t(X_2 - X_1) solves AX=BAX = B for every scalar tt.
  3. That's an infinite family.
Answer:Infinitely many solutions.
Practice this concept4 quick reps

From the bank · past-year question

Example 3Matrices & DeterminantsMODERATE
Let AX=BAX=B be a system of 3 linear equations with 3-unknowns. Let X1X_1 and X2X_2 be its two distinct solutions. If the combination aX1+bX2aX_1+bX_2 is a solution of AX=BAX=B; where aa, bb are real numbers, then which one of the following is correct?

[Q9 · Apr · 2023]

Concept 4 of 4

Finding the parameter for consistency

Intuition

When a coefficient (or constant) contains a parameter kk, the borderline cases live where A=0|A| = 0. Set the coefficient determinant to 0 to find the candidate kk values, then check each to see if it gives no solution or infinitely many.

Definition

Put the parameter into A|A| and solve A=0|A| = 0 for the critical kk. For each such kk, substitute back: dependent-and-consistent → infinitely many; contradictory → no solution. Away from those kk, the solution is unique.

Worked example

For what kk does kx+y+z=1, x+ky+z=k, x+y+kz=k2kx + y + z = 1,\ x + ky + z = k,\ x + y + kz = k^2 have no solution?
  1. Coefficient determinant =(k1)2(k+2)= (k-1)^2(k+2) (standard symmetric system).
  2. It is 0 at k=1k = 1 and k=2k = -2.
  3. At k=1k = 1 all three equations coincide → infinitely many. At k=2k = -2 they're contradictory → no solution.
Answer:k=2k = -2 (at k=1k = 1 there are infinitely many instead).
Practice this concept4 quick reps

From the bank · past-year question

Example 4Matrices & DeterminantsHARD
The system of equations kx+y+z=1kx + y + z = 1, x+ky+z=kx + ky + z = k and x+y+kz=k2x + y + kz = k^2 has no solution if kk equals

[Q27 · Sep · 2017]

Summary — formulas & gotchas at a glance

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Formulas (1)

  • Cramer's rule

    Cramer's rule

    x=ΔxΔ,y=ΔyΔ,z=ΔzΔx = \frac{\Delta_x}{\Delta}, \quad y = \frac{\Delta_y}{\Delta}, \quad z = \frac{\Delta_z}{\Delta}

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