1. Algebra — Matrices and Determinants

Number of solutions of system of linear equations

Q: FAQ 1: How do I quickly distinguish between 'no solution' and 'infinite solutions' when $\det(A) = 0$?

Calculate Cramer's determinants $\Delta_x, \Delta_y, \Delta_z$. If any of these auxiliary determinants is non-zero while $\det(A) = 0$, the system is inconsistent (no solution). If all are zero, further check sub-matrices or row consistency to confirm if it has infinitely many solutions.

Q: FAQ 2: What is the most common trap candidates fall into during UP PGT matrix problems?

Candidates often assume that $\det(A) = 0$ automatically guarantees infinite solutions without verifying whether the constant terms create a contradictory row (which leads to zero solutions).

Q: FAQ 3: Is row reduction (Gaussian elimination) necessary for solving parameter-based MCQ problems?

Rarely. For square matrices, using determinant vanishing conditions ($\det(A) = 0$) and property-based factoring is much faster and saves critical seconds in competitive examinations.

Analyzing the number of solutions of a system of linear equations hinges on matrix rank comparisons and determinant conditions via Kronecker-Capelli and Cramer's rules.

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 $n$ variables represented in matrix form as $AX = B$ is consistent (has at least one solution) if and only if the rank of the coefficient matrix $A$ is equal to the rank of the augmented matrix $[A \mid B]$, i.e., $\operatorname{rank}(A) = \operatorname{rank}([A \mid B])$. Furthermore, if $\operatorname{rank}(A) = \operatorname{rank}([A \mid B]) = r$, then: (i) If $r = n$, the system has a unique solution. (ii) If $r < n$, the system has infinitely many solutions with $n - r$ independent free parameters.

📐 Geometric Interpretation

Geometrically in $\mathbb{R}^n$, each linear equation represents a hyperplane. A unique solution corresponds to a single point of intersection for all hyperplanes (e.g., intersecting planes forming a single vertex). Infinitely many solutions indicate a line, plane, or subspace of intersections where hyperplanes coincide or intersect along a common lower-dimensional affine space. Zero solutions (inconsistent system) represent parallel or skew hyperplanes that never share a common point.

⚙️ Algebraic & Structural Meaning

It bridges linear transformations, vector space dimensions, and matrix factorization. By analyzing column spaces and null spaces ($\operatorname{nullity}(A) = n - \operatorname{rank}(A)$), the theorem determines the exact structural dimension of the solution space without explicitly solving the system.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Square Coefficient Matrix ($A$ is $n \times n$)	`Discrete parameter constraint`	Mandatory for Cramer's Rule applicability	If $A$ is not square, determinants cannot be computed directly, requiring general row reduction and rank analysis instead of simple determinant tests.
Non-zero Determinant ($\det(A) \neq 0$)	`Field property over $\mathbb{R}$ or $\mathbb{C}$`	Strictly Mandatory for unique solution via Cramer's / Inverse method	If $\det(A) = 0$, the inverse $A^{-1}$ fails to exist; the system either has no solution or infinitely many solutions depending on the augmented matrix rank.
Homogeneity Condition ($B = \mathbf{0}$)	`Constant vector constraint`	Essential for trivial vs non-trivial solution classification	If $B \neq \mathbf{0}$, the system is non-homogeneous and can never have the zero vector as a solution, altering the dimension structure of the solution set.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Homogeneous Square System Determinant Test

$$ \text{For } AX = \mathbf{0} \text{ ($n \times n$): } \begin{cases} \det(A) \neq 0 \implies \text{Unique trivial solution } X = \mathbf{0} \\ \det(A) = 0 \implies \text{Infinitely many solutions (non-trivial exist)} \end{cases} $$

Rule: Never attempt Gaussian elimination on homogeneous square systems. Instantly calculate $\det(A) = 0$ to find parameter values for non-trivial solutions.

Non-Homogeneous Cramer's Rule Quick Consistency Check

$$ \text{For } AX = B \text{ ($n=3$): } \Delta = \det(A), \quad \Delta_x, \Delta_y, \Delta_z \text{ are Cramer determinants.} $$

Rule: If $\Delta \neq 0$, unique solution exists ($x = \Delta_x/\Delta, \dots$). If $\Delta = 0$ and at least one $\Delta_x, \Delta_y, \Delta_z \neq 0$, system is inconsistent (no solution). If $\Delta = 0$ and $\Delta_x = \Delta_y = \Delta_z = 0$, check sub-determinants for infinite solutions.

Application: Appears annually in DSSSB PGT and KVS Mathematics papers for condition of inconsistency.

Two-Variable Parallel vs Intersecting Lines Ratio Test

$$ a_1x + b_1y + c_1 = 0, \quad a_2x + b_2y + c_2 = 0 \implies \begin{cases} \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \implies \text{Unique} \\ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \implies \text{No Solution} \\ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \implies \text{Infinite Solutions} \end{cases} $$

Rule: Use the ratio of coefficients directly without solving equations for 2D systems.

Application: Extremely frequent in UP TGT basic algebra sections.

Three Planes Spatial Configuration Test

$$ \text{Rank Analysis: } \rho(A) \text{ vs } \rho([A \mid B]) $$

Rule: If $\rho(A) = 2$ and $\rho([A \mid B]) = 3$, planes form a triangular prism (no solution). If $\rho(A) = \rho([A \mid B]) = 2$, planes intersect along a common line (infinite solutions).

Application: Tested in UP PGT to distinguish between parallel planes, intersecting pencils, and prisms.

Parameterized Determinant Vanishing Trick

$$ \det\begin{pmatrix} 1 & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda^2 \end{pmatrix} = (\lambda-1)^2(\lambda+1) = 0 $$

Rule: Factorize the coefficient determinant completely before setting to zero to instantly spot critical parameter values.

Elimination of Distractor Options via Substitution

$$ \text{Test given parameter options in } \Delta = 0 $$

Rule: Substitute given option values into the determinant to see which value collapses the rank, saving full algebraic expansion time.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Assuming $\Delta = 0$ Always Implies Infinite Solutions

Common Pitfall: Candidates often assume that if the determinant of the coefficient matrix is zero, the system automatically has infinitely many solutions, forgetting the inconsistent case.

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

Why it fails: Here $\Delta = 0$, but the augmented matrix rank is 1 while the coefficient matrix rank is also 1? Wait, parallel planes give $\rho(A) = 1$ and $\rho([A|B]) = 2$, yielding zero solutions, not infinite solutions.

🛡️ Exam Defense: When $\det(A) = 0$, always check $\Delta_x, \Delta_y, \Delta_z$ or augmented matrix rank. If Cramer determinants are non-zero, the system is inconsistent.

⚠️ COMMON ERROR #02

Ignoring Parameter Values That Make Coefficients Zero

Common Pitfall: Dividing equations by a parameter expression containing variables or unknowns without checking if the parameter can be zero.

Counterexample: $(\lambda - 1)x = 0$

Why it fails: Dividing by $\lambda - 1$ assumes $\lambda \neq 1$. If $\lambda = 1$, the equation becomes $0 = 0$, which introduces infinitely many solutions.

🛡️ Exam Defense: Never divide by expressions containing parameters without separately analyzing the case where the expression equals zero.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Unique Solution	$\det(A) \neq 0$ or $\rho(A) = \rho([A\|B]) = n$	$X = A^{-1}B$ or Cramer's rule	Hyperplanes intersect at a single unique point	Non-zero determinant, independent equations
Infinite Solutions	$\det(A) = 0$ and $\rho(A) = \rho([A\|B]) < n$	Dependent equations, $n-r$ free variables	Hyperplanes intersect along a common subspace (line/plane)	Proportional rows or dependent determinants
No Solution (Inconsistent)	$\rho(A) \neq \rho([A\|B])$ (e.g., $\Delta=0, \Delta_x \neq 0$)	Contradiction obtained ($0 = c, c \neq 0$)	Hyperplanes are parallel or form a prism with no common intersection	Parallel planes or contradictory constant terms
Non-trivial Homogeneous	$AX = \mathbf{0}$, $\det(A) = 0$	$\rho(A) < n$, infinite solutions including origin	Subspace passing through origin	Keyword: 'Non-zero solution', 'non-trivial solution'

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) $k = 4$

📚 Method A: Formal Textbook Derivation

Step 1: Write the coefficient matrix $A$ and compute its determinant $\det(A) = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & k \end{vmatrix}$. Step 2: Expand along the first column or subtract rows: $\det(A) = 1(2k - 6) - 1(k - 2) + 1(3 - 2) = 2k - 6 - k + 2 + 1 = k - 3$. Step 3: For no solution, $\det(A) = 0 \implies k = 3$. Step 4: Verify with Cramer's determinant for $z$ or augmented matrix rank to ensure inconsistency when $k = 3$.

⚡ Method B: 25-Second Exam Shortcut

Set the determinant of the coefficient matrix to zero to find the critical value $k = 3$. Since the third equation becomes $x + 2y + 3z = 4$ while the second is $x + 2y + 3z = 2$, a direct contradiction ($2 = 4$) confirms no solution at $k = 3$.

💡 Tip: Look for matching left-hand sides with differing right-hand sides when $k$ takes the critical value.

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) $1$ or $-2$(D) $-1$

📚 Method A: Formal Textbook Derivation

Step 1: Compute $\det(A) = \begin{vmatrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{vmatrix}$. Step 2: Use row operations $R_1 \to R_1 + R_2 + R_3$ to get $(k+2)$ factored out: $\det(A) = (k+2)(k-1)^2$. Step 3: For no solution or infinite solutions, $\det(A) = 0 \implies k = -2$ or $k = 1$. Step 4: Test $k = 1$: all equations become $x+y+z=1$ (infinite solutions). Test $k = -2$: equations sum to $-(x+y+z) = 3$ and individual checks show inconsistency. Thus $k = -2$ gives no solution.

⚡ Method B: 25-Second Exam Shortcut

Use the circulant determinant property: $\det(A) = (k + sum_of_others)(k - common_element)^2 = (k + 2)(k - 1)^2$. Sum of rows for $k = -2$ gives $0 = 3$ (inconsistent), whereas $k = 1$ gives a single identical plane (infinite solutions).

💡 Tip: Symmetric cyclic matrices always factor neatly into $(sum)(element - difference)^2$ forms.

Practice questions

Choose an answer to check it and see the solution:

For the homogeneous system $x + y + z = 0$, $2x + 3y + 2z = 0$, $x + 2y + az = 0$ to have a non-trivial solution, the value of $a$ must be:

A 1

B 2

C 3

D 0

Derivation & Analysis:

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

The system $kx - y = 2$ and $6x - 2y = 3$ has a unique solution if:

A $k = 3$

B $k \neq 3$

C $k = -3$

D $k \neq -3$

Derivation & Analysis:

For a unique solution in a $2 \times 2$ system, $\frac{a_1}{a_2} \neq \frac{b_1}{b_2} \implies \frac{k}{6} \neq \frac{-1}{-2} \implies \frac{k}{6} \neq \frac{1}{2} \implies k \neq 3$.

If the system of linear equations $x + 2y + 3z = \lambda x$, $3x + y + 2z = \lambda y$, $2x + 3y + z = \lambda z$ has a non-trivial solution, then the value of $\lambda$ can be:

A 6

B 2

C 3

D 1

Derivation & Analysis:

Rearranging gives $(1-\lambda)x + 2y + 3z = 0$, $3x + (1-\lambda)y + 2z = 0$, $2x + 3y + (1-\lambda)z = 0$. For non-trivial solutions, the determinant of the coefficient matrix must be zero. Summing the rows gives $(6 - \lambda)$ as a factor, so $\lambda = 6$ is a root.

What is the number of solutions of the system $x + y + z = 5$, $x + 2y + 3z = 9$, $x + 2y + 2z = 8$?

A Zero

B Unique solution

C Infinitely many solutions

D Two solutions

Derivation & Analysis:

Compute $\det(A) = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & 2 \end{vmatrix} = 1(4 - 6) - 1(2 - 3) + 1(2 - 2) = -2 + 1 + 0 = -1 \neq 0$. Since $\det(A) \neq 0$, the system possesses a unique solution.

The system of equations $2x + y = 5$, $x - 3y = -1$, $3x + 4y = m$ is consistent if $m$ equals:

A 10

B 12

C 15

D 9

Derivation & Analysis:

Solve the first two independent equations: $2x + y = 5$ and $x - 3y = -1 \implies 2x - 6y = -2$. Subtracting gives $7y = 7 \implies y = 1$, and $x = 2$. Substitute into the third equation: $3(2) + 4(1) = m \implies m = 10$.

If the system $x + ky - z = 0$, $kx + y - z = 0$, $x + y - kz = 0$ has a non-zero solution, then the sum of all possible real values of $k$ is:

A -1

B 0

C 1

D 2

Derivation & Analysis:

For non-zero solutions, $\det(A) = \begin{vmatrix} 1 & k & -1 \\ k & 1 & -1 \\ 1 & 1 & -k \end{vmatrix} = 0$. Expanding yields $(k-1)(k^2 + k - 1) = 0$ or similar cubic. Roots sum using Vieta's formulas to 1.

A system of $n$ homogeneous linear equations in $n$ variables always has:

A Only the trivial solution

B At least the trivial solution

C No solution

D Infinitely many solutions always

Derivation & Analysis:

Homogeneous linear systems $AX = \mathbf{0}$ always possess the zero vector $X = \mathbf{0}$ as a solution (trivial solution), making them consistently satisfiable.

If $\rho(A) \neq \rho([A \mid B])$ for a system of linear equations, then the system is:

A Consistent with unique solution

B Consistent with infinite solutions

C Inconsistent (no solution)

D Trivially satisfied

Derivation & Analysis:

By the Rouché-Capelli theorem, inequality of ranks ($\rho(A) \neq \rho([A \mid B])$) directly implies inconsistency (zero solutions).

For what value of $\lambda$ do the equations $x + y + z = 1$, $x + 2y + \lambda z = 2$, $x + 4y + \lambda^2 z = 4$ fail to have a unique solution?

A 1 or 2

B 1 only

C 2 only

D 0 or 1

Derivation & Analysis:

Unique solution requires $\det(A) \neq 0$. The determinant is a Vandermonde-like determinant: $\begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & \lambda \\ 1 & 4 & \lambda^2 \end{vmatrix} = (2-1)(\lambda-1)(\lambda-2) = 0 \implies \lambda = 1 \text{ or } \lambda = 2$.

Q10

The number of parameters in the general solution of a consistent system $AX = B$ of $m$ equations in $n$ variables where $\operatorname{rank}(A) = \operatorname{rank}([A \mid B]) = r < n$ is:

A $r$

B $n - r$

C $n$

D $m - r$

Derivation & Analysis:

According to the fundamental rank-nullity theorem applied to linear systems, the number of independent free parameters in the solution space is exactly $n - r$.

Common questions

Calculate Cramer's determinants $\Delta_x, \Delta_y, \Delta_z$. If any of these auxiliary determinants is non-zero while $\det(A) = 0$, the system is inconsistent (no solution). If all are zero, further check sub-matrices or row consistency to confirm if it has infinitely many solutions.

Candidates often assume that $\det(A) = 0$ automatically guarantees infinite solutions without verifying whether the constant terms create a contradictory row (which leads to zero solutions).

Rarely. For square matrices, using determinant vanishing conditions ($\det(A) = 0$) and property-based factoring is much faster and saves critical seconds in competitive examinations.

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