1. Algebra — Matrices and Determinants

Consistency of system of linear equations

Q: FAQ 1: Can a homogeneous system ever be inconsistent?

No. A homogeneous system $AX = 0$ always possesses the zero solution ($X = 0$), known as the trivial solution. Hence, a homogeneous system is always consistent.

Q: FAQ 2: What is the most common reason students lose marks in consistency questions?

Students often assume that $\det(A) = 0$ automatically implies infinitely many solutions without checking the augmented matrix rank or Cramer determinants ($\Delta_x, \Delta_y, \Delta_z$). When $\det(A) = 0$ and $\Delta_i \neq 0$, the system has no solution.

Q: FAQ 3: How can I solve parameter-based consistency problems?

Immediately evaluate $\det(A) = 0$ to find critical parameter values. Then test whether those values yield identical equations (infinite solutions) or contradictory equations (no solution) by quick substitution.

The consistency of a system of linear equations determines whether a solution exists by analyzing the ranks and determinants of the coefficient and augmented matrices.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Rouché-Capelli Theorem (Kronecker-Capelli Theorem)

A system of $m$ linear equations in $n$ unknowns represented matrix-wise as $AX = B$ is consistent (i.e., possesses at least one solution) if and only if the rank of the coefficient matrix $A$ is equal to the rank of the augmented matrix $[A \mid B]$, that is, $\operatorname{rank}(A) = \operatorname{rank}([A \mid B])$. Furthermore, the system has a unique solution if $\operatorname{rank}(A) = \operatorname{rank}([A \mid B]) = n$, and infinitely many solutions if $\operatorname{rank}(A) = \operatorname{rank}([A \mid B]) < n$.

📐 Geometric Interpretation

In a system of linear equations, each equation represents a hyperplane in $\mathbb{R}^n$. Consistency implies that these hyperplanes intersect in at least one common point, line, or subspace. Inconsistent systems correspond to parallel, non-intersecting hyperplanes.

⚙️ Algebraic & Structural Meaning

It translates geometric intersection questions into purely algebraic rank-computation tasks using elementary row operations, Gauss-Jordan elimination, or determinant criteria.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Field of Scalars	`Field $\mathbb{R}$ or $\mathbb{C}$`	Strictly Mandatory	Coefficients must belong to a defined field; otherwise, matrix operations and rank definitions break down.
Square vs. Rectangular Matrix	`$m \times n$ system`	Essential for method selection	If $m \neq n$, standard determinant tests via Cramer's rule cannot be applied directly, requiring rank analysis.
Non-zero Augmented Column for Inhomogeneity	$B \neq 0$	Defines non-homogeneous behavior	If $B = 0$, the system is homogeneous and is always consistent (trivial solution always exists).

Useful results and shortcuts

Use these identities and results when their conditions apply:

Cramer's Rule for Unique Solution

$$ \Delta = \det(A) \neq 0 $$

Rule: If the determinant of the coefficient matrix $A$ is non-zero, the system is consistent and possesses a unique solution given by $x_i = \frac{\Delta_i}{\Delta}$.

Application: Directly tested in UP PGT to find parameter values for unique solutions.

Infinite Solutions Condition for Homogeneous System

$$ \det(A) = 0 $$

Rule: A homogeneous system $AX = 0$ of $n$ equations in $n$ variables has non-trivial (infinitely many) solutions if and only if $\det(A) = 0$.

Application: Extremely frequent in TGT/PGT for finding values of $k$ where non-trivial solutions exist.

Inconsistency Condition

$$ \operatorname{rank}(A) \neq \operatorname{rank}([A \mid B]) $$

Rule: If $\det(A) = 0$ but at least one Cramer determinant $\Delta_i \neq 0$, the system is strictly inconsistent (no solution).

Application: Used to eliminate options in parameter-based consistency questions.

Three Variable Plane Intersections

$$ \text{Check } \Delta, \Delta_x, \Delta_y, \Delta_z $$

Rule: For 3 planes: $\Delta \neq 0 \implies$ unique intersection point. $\Delta = 0$ and all $\Delta_i = 0$ with $\operatorname{rank}(A)=2 \implies$ line of intersection (infinite solutions). $\Delta = 0$ and any $\Delta_i \neq 0 \implies$ parallel/no intersection.

Application: Standard 3-variable system questions in UP PGT Mathematics.

Homogeneous Overdetermined / Underdetermined Systems

$$ m < n \implies \text{Infinitely many solutions} $$

Rule: A homogeneous system with fewer equations than variables ($m < n$) always has infinitely many solutions.

Application: Quick conceptual check for verbal or theory-based questions.

Rapid Substitution of Options

$$ \text{Test options for } k $$

Rule: When asked for a value of a parameter $k$ making the system inconsistent, plug the option values into $\det(A)$ and check for vanishing determinants followed by non-zero augmented rank.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Assuming $\det(A) = 0$ always means Infinite Solutions

Common Pitfall: Candidates jump to 'infinite solutions' whenever $\det(A) = 0$ for a non-homogeneous system.

Counterexample: $x + y = 1, \; x + y = 2$

Why it fails: Here $\det(A) = 0$, but the system is inconsistent (parallel lines, no solution) because $\operatorname{rank}(A) = 1 \neq \operatorname{rank}([A \mid B]) = 2$.

🛡️ Exam Defense: Always check $\Delta_x, \Delta_y, \Delta_z$ when $\det(A) = 0$ before concluding infinite solutions.

⚠️ COMMON ERROR #02

Ignoring Homogeneous System Consistency

Common Pitfall: Thinking a homogeneous system can be inconsistent.

Counterexample: $AX = 0$

Why it fails: The zero vector $X = 0$ is always a solution (trivial solution). Homogeneous systems are ALWAYS consistent.

🛡️ Exam Defense: Never mark 'inconsistent' for any homogeneous system of linear equations.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Unique Solution	$\det(A) \neq 0$ or $\operatorname{rank}(A) = \operatorname{rank}([A \mid B]) = n$	$x_i = \frac{\Delta_i}{\Delta}$	Intersecting hyperplanes meeting at a single unique point	Non-zero determinant of coefficient matrix
Infinite Solutions	$\det(A) = 0$ and $\Delta_x = \Delta_y = \Delta_z = 0$ (with $\operatorname{rank} < n$)	Parametric solution set	Hyperplanes intersecting along a common line or plane	Dependent equations / proportional rows
No Solution (Inconsistent)	$\det(A) = 0$ and at least one $\Delta_i \neq 0$	$\operatorname{rank}(A) \neq \operatorname{rank}([A \mid B])$	Parallel or skew hyperplanes with no common intersection	Contradictory row results like $0 = 5$
Homogeneous System	$B = 0$ ($m$ equations, $n$ variables)	Always consistent; non-trivial if $\det(A)=0$	All hyperplanes pass through the origin	Right-hand side is identically zero

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

For what value of $k$ does the system of equations $x + y + z = 1$, $x + 2y + 3z = 2$, $x + 2y + kz = 4$ have no solution?

(A) $k = 3$(B) $k \neq 3$(C) $k = 2$(D) $k = 0$

📚 Method A: Formal Textbook Derivation

Step 1: Write coefficient matrix $A$ and compute $\det(A) = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & k \end{vmatrix}$. Step 2: Expand to get $\det(A) = 1(2k - 6) - 1(k - 3) + 1(0) = k - 3$. Step 3: For no solution, $\det(A) = 0 \implies k = 3$. Step 4: Substitute $k = 3$ into the augmented matrix to verify $\operatorname{rank}(A) \neq \operatorname{rank}([A \mid B])$.

⚡ Method B: 25-Second Exam Shortcut

Set $\det(A) = 0 \implies k = 3$. Since substituting $k=3$ creates a contradiction in the third row, $k=3$ yields no solution.

💡 Tip: Check if setting $\det(A)=0$ gives a unique critical value; if so, that is almost always the answer for 'no solution' or 'infinite solution' questions.

CANONICAL PROBLEM 02 Dual-Method Breakdown

The system of linear equations $kx + y + z = 1$, $x + ky + z = 1$, $x + y + kz = 1$ has no solution if $k$ equals:

(A) $1$(B) $-2$(C) $2$(D) $0$

📚 Method A: Formal Textbook Derivation

Step 1: Calculate $\det(A) = \begin{vmatrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{vmatrix} = (k-1)^2(k+2)$. Step 2: For no solution or infinite solutions, $\det(A) = 0 \implies k = 1$ or $k = -2$. Step 3: Test $k = 1$: all three equations become $x + y + z = 1$, which represents a single plane (infinitely many solutions). Step 4: Test $k = -2$: the sum of the left-hand sides is $-(-2+1+1)x\dots$ leading to a contradiction with the right-hand sides, confirming no solution at $k = -2$.

⚡ Method B: 25-Second Exam Shortcut

Use the determinant factor formula $\det(A) = (k+2)(k-1)^2$. $k=1$ gives identical equations (infinite solutions), hence $k=-2$ must give no solution.

💡 Tip: Remember that $k=1$ collapses the system to a single equation, yielding infinite solutions, leaving $k=-2$ as the inconsistency condition.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: If the system $x + y + z = 0$, $2x + 3y + 2z = 0$, $x + 2y + az = 0$ has a non-trivial solution, then the value of $a$ is:

A $1$

B $2$

C $3$

D $0$

Derivation & Analysis:

For a homogeneous system to have a non-trivial solution, $\det(A) = 0$. $\begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & 2 \\ 1 & 2 & a \end{vmatrix} = 0 \implies 1(3a - 4) - 1(2a - 2) + 1(4 - 3) = 0 \implies a - 1 = 0 \implies a = 1$.

MCQ 2: A system of $n$ homogeneous linear equations in $n$ variables $AX = 0$ has a non-trivial solution if:

A $\det(A) \neq 0$

B $\det(A) = 0$

C $\operatorname{rank}(A) = n$

D $A$ is a null matrix

Derivation & Analysis:

A homogeneous system has a non-trivial (non-zero) solution if and only if the coefficient matrix is singular, i.e., $\det(A) = 0$.

MCQ 3: The system $x - 2y = 3$, $2x - 4y = 6$ has:

A A unique solution

B No solution

C Infinitely many solutions

D Two solutions

Derivation & Analysis:

The second equation is exactly twice the first ($2x - 4y = 6$). They represent the same line, hence the system has infinitely many solutions.

MCQ 4: If $\operatorname{rank}(A) \neq \operatorname{rank}([A \mid B])$ for the linear system $AX = B$, then the system is:

A Consistent with unique solution

B Consistent with infinite solutions

C Inconsistent (no solution)

D Trivially consistent

Derivation & Analysis:

By the Rouché-Capelli theorem, inequality of ranks ($\operatorname{rank}(A) \neq \operatorname{rank}([A \mid B])$) is the exact condition for system inconsistency.

MCQ 5: For what value of $\lambda$ does the system $x + y + z = 1$, $x + 2y + 4z = \lambda$, $x + 4y + 10z = \lambda^2$ be consistent?

A $\lambda = 1, 2$

B $\lambda = 2$

C $\lambda = 1$ only

D Any real number

Derivation & Analysis:

Applying consistency condition via determinant of augmented matrix or row reduction yields consistency when $\lambda = 1$ or $\lambda = 2$.

MCQ 6: If $A$ is a $3 \times 3$ matrix and $\det(A) = 0$, then the system $AX = D$ ($D \neq 0$) can never have:

A No solution

B Infinite solutions

C Unique solution

D Consistent solution

Derivation & Analysis:

Since $\det(A) = 0$, $\operatorname{rank}(A) < 3$, meaning a unique solution is impossible. It can either have no solution or infinitely many solutions.

MCQ 7: The number of solutions of the system $2x + y = 5$, $x - y = 1$, $3x + 2y = 8$ is:

A 0

B 1

C 2

D Infinite

Derivation & Analysis:

Solving the first two equations gives $x = 2, y = 1$. Substituting into the third: $3(2) + 2(1) = 8$, which satisfies the third equation. Thus, the unique solution is $(2, 1)$.

MCQ 8: Assertion (A): A homogeneous system of 4 linear equations in 3 variables always has a non-trivial solution. Reason (R): Such a system always has fewer equations than variables.

A Both (A) and (R) are true and (R) is the correct explanation of (A)

B Both (A) and (R) are true but (R) is not the correct explanation of (A)

C (A) is true but (R) is false

D (A) is false but (R) is true

Derivation & Analysis:

A homogeneous system with 4 equations and 3 variables has $m = 4, n = 3$, so $m > n$ (more equations than variables). Thus Reason (R) is false. However, (A) is true because any homogeneous system with $\operatorname{rank}(A) < 3$ (which happens when $\det = 0$ on submatrices) has non-trivial solutions.

MCQ 9: If the system $kx - y = 2$ and $6x - 2y = 3$ has no solution, then $k =$

A 3

B -3

C 2

D 0

Derivation & Analysis:

For no solution in two variables, $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$. Here $\frac{k}{6} = \frac{-1}{-2} = \frac{1}{2} \implies k = 3$.

Q10

MCQ 10: If $A$ is a non-singular square matrix of order $n$, then the system $AX = B$ has:

A No solution

B Infinitely many solutions

C A unique solution $X = A^{-1}B$

D Dependent solutions

Derivation & Analysis:

If $A$ is non-singular ($\det(A) \neq 0$), $A^{-1}$ exists, and multiplying $AX = B$ by $A^{-1}$ yields the unique solution $X = A^{-1}B$.

Common questions

No. A homogeneous system $AX = 0$ always possesses the zero solution ($X = 0$), known as the trivial solution. Hence, a homogeneous system is always consistent.

Students often assume that $\det(A) = 0$ automatically implies infinitely many solutions without checking the augmented matrix rank or Cramer determinants ($\Delta_x, \Delta_y, \Delta_z$). When $\det(A) = 0$ and $\Delta_i \neq 0$, the system has no solution.

Immediately evaluate $\det(A) = 0$ to find critical parameter values. Then test whether those values yield identical equations (infinite solutions) or contradictory equations (no solution) by quick substitution.

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