1. Algebra — Matrices and Determinants

Solving linear equations in three variables

Q: FAQ 1: When should I use Cramer's Rule versus Matrix Inversion or Gaussian Elimination?

Use Cramer's Rule when asked for a single variable or when evaluating parameter conditions ($k$) for unique/non-unique solutions. Use Matrix Inversion or Gaussian Elimination when solving for all variables with fixed numerical coefficients.

Q: FAQ 2: Why do candidates frequently confuse 'no solution' with 'infinitely many solutions' when $D = 0$?

Candidates often check only $D = 0$ and stop. When $D = 0$, you must check $D_x, D_y, D_z$. If even one of them is non-zero, the system has zero solutions. If all are zero, it typically has infinite solutions (subject to rank confirmation).

Q: FAQ 3: How can I save time on 3x3 determinant calculations during the exam?

Master row/column operations to create at least two zeros in a row or column before expanding, or practice direct Sarrus expansion to avoid arithmetic errors under exam pressure.

This section covers the simultaneous solution of three linear equations using matrices, determinants, and Cramer's Rule.

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

📖 Topic progress

○ Mark as complete

Definition and statement

Cramer's Rule and Matrix Rank Theorem for Three Variables

Consider a system of three linear equations in three unknowns $x, y, z$: $$a_1x + b_1y + c_1z = d_1$$ $$a_2x + b_2y + c_2z = d_2$$ $$a_3x + b_3y + c_3z = d_3$$ Let $A$ be the coefficient matrix, $X$ the variable column vector, and $B$ the constant column vector. The system has a unique solution if and only if $\det(A) \neq 0$. If $\det(A) = 0$, the system is consistent with infinitely many solutions or inconsistent depending on the ranks of $A$ and the augmented matrix $[A \mid B]$.

📐 Geometric Interpretation

Each linear equation represents a plane in $\mathbb{R}^3$. A unique solution means the three planes intersect at a single distinct point. Infinitely many solutions mean they intersect along a common line or coincide. No solution means the planes are parallel or form a triangular prism with no common intersection.

⚙️ Algebraic & Structural Meaning

Translates geometric intersections of planes into linear algebra through matrix invertibility, determinant non-vanishing, and rank-nullity relations encapsulated by the Kronecker-Capelli theorem.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Non-zero Determinant (Unique Solution)	$\det(A) \neq 0$	Strictly Mandatory / Necessary	If $\det(A) = 0$, Cramer's Rule fails completely (division by zero), and the system either has no solution or infinitely many solutions.
Rank Consistency (Kronecker-Capelli Condition)	`$\text{rank}(A) = \text{rank}([A \mid B])`	Strictly Mandatory for Consistency	If $\text{rank}(A) \neq \text{rank}([A \mid B])$ (specifically $\text{rank}(A) < \text{rank}([A \mid B])$), the system is inconsistent and possesses zero solutions.
Parameter Dependency for Infinitely Many Solutions	$\text{rank}(A) = \text{rank}([A \mid B]) < 3$	Essential condition	If rank condition is met but rank equals 3, the solution is unique. If rank is less than 3, degrees of freedom emerge, leading to parametric infinite solutions.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Cramer's Rule Formulae for $x, y, z$

$$x = \frac{D_x}{D}, \quad y = \frac{D_y}{D}, \quad z = \frac{D_z}{D}$$

Rule: When finding only one variable (e.g., $y$), immediately construct $D_y$ by replacing the $y$-column with $B$, and divide by $D = \det(A)$ without solving the full system.

Application: Frequently asked in UP PGT to find the value of a single variable or parameter $k$ for unique solutions.

Homogeneous System Non-Trivial Solution

$$D = \det(A) = 0$$

Rule: For a homogeneous system ($d_1 = d_2 = d_3 = 0$), a non-trivial (non-zero) solution exists if and only if the determinant of the coefficient matrix vanishes.

Application: A staple question type in UP TGT and KVS PGT asking for the value of parameter $\lambda$ for which the system has non-trivial solutions.

Condition for Infinitely Many Solutions in Homogeneous Systems

$$D = 0 \text{ and at least one } 2 \times 2 minor \neq 0$$

Rule: If $D = 0$ and all $2 \times 2$ determinants of $A$ also vanish, the matrix rank is 1, meaning two equations are scalar multiples of the third.

Symmetric Equations Sum Trick

$$x + y + z = \frac{\sum d_i}{\sum a_i} \text{ (when coefficients sum equally)}$$

Rule: If the sum of coefficients of $x, y, z$ in each equation is identical, add all three equations together to instantly get a single linear relation in $(x + y + z)$.

Determinant Expansion via Sarrus Rule or Properties

$$|A| = a_1(b_2c_3 - b_3c_2) - b_1(a_2c_3 - a_3c_2) + c_1(a_2b_3 - a_3b_2)$$

Rule: Use row/column operations ($R_i \to R_i + kR_j$) to create zeros before expanding to cut calculation time by 70%.

Option Substitution Method

$$Ax = B$$

Rule: Substitute the given options $(x, y, z)$ directly into the equations. The correct option must satisfy all three equations simultaneously.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Assuming $D = 0$ Always Means Infinitely Many Solutions

Common Pitfall: Candidates jump to the conclusion that $\det(A) = 0$ automatically implies infinite solutions, forgetting the inconsistent case (zero solutions).

Counterexample: $x + y + z = 1$, $x + y + z = 2$, $x + y + z = 3$

Why it fails: Here $D = 0$, but the planes are parallel and distinct, resulting in $\text{rank}(A) = 1 \neq \text{rank}([A \mid B]) = 2$. Thus, there are zero solutions.

🛡️ Exam Defense: Always check the augmented matrix rank or consistency check ($D_x, D_y, D_z$ values) when $D = 0$ before declaring infinite solutions.

⚠️ COMMON ERROR #02

Ignoring Sign Errors in Cofactor Expansion

Common Pitfall: Missing the alternating negative sign in the second term during $3 \times 3$ determinant expansion ($-\begin{vmatrix} \cdot \end{vmatrix}$).

Counterexample: $-\begin{vmatrix} a_2 & c_2 \\ a_3 & c_3 \\ \end{vmatrix}$ written as $+ \dots$

Why it fails: Leads to an incorrect determinant sign, yielding a wrong parameter value that is invariably placed as Option (A) or (B).

🛡️ Exam Defense: Use the standard sign checker matrix $\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}$ mentally during expansion.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Unique Solution	$D \neq 0$	$x = D_x/D, y = D_y/D, z = D_z/D$	Three planes intersect at a unique point in $\mathbb{R}^3$	Non-zero determinant, independent equations
Infinite Solutions	$D = 0$ and $D_x = D_y = D_z = 0$	Rank$(A) = \text{Rank}([A\|B]) < 3$	Three planes intersect along a common line or coincide	Dependent equations, consistent system with parameter
No Solution (Inconsistent)	$D = 0$ and at least one of $D_x, D_y, D_z \neq 0$	Rank$(A) \neq \text{Rank}([A\|B])$	Planes form a triangular prism or are parallel	Parallel planes or conflicting constant terms
Non-Trivial Homogeneous	$D = 0$ for $AX = 0$	Non-zero vector $X$ exists	Planes pass through the origin and intersect along a line/plane	Homogeneous system with non-zero solutions requested

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 $x + y + z = 1$, $x + 2y + 3z = 2$, $x + 2y + ky^2 = 4$ (Wait, standard exam variant: $x + 2y + 3z = 2$, $x + 2y + kz = 4$) fail to have a unique solution?

(A) $k = 3$(B) $k \neq 3$(C) $k = 0$(D) $k = 1$

📚 Method A: Formal Textbook Derivation

Step 1: Form coefficient matrix $A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & k \end{pmatrix}$. Step 2: Compute $\det(A) = 1(2k - 6) - 1(k - 3) + 1(2 - 2) = 2k - 6 - k + 3 = k - 3$. Step 3: For unique solution, $\det(A) \neq 0 \implies k \neq 3$. Therefore, uniqueness fails when $k = 3$.

⚡ Method B: 25-Second Exam Shortcut

Set determinant equal to zero for non-unique solution: $k - 3 = 0 \implies k = 3$.

💡 Tip: Look out for wording like 'fails to have a unique solution' vs 'has no solution'. Fails to have unique means $D = 0$.

CANONICAL PROBLEM 02 Dual-Method Breakdown

If the system $kx + y + z = 1$, $x + ky + z = 1$, $x + y + kz = 1$ has no solution, then the value of $k$ is:

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

📚 Method A: Formal Textbook Derivation

Step 1: Find determinant $D = \begin{vmatrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{vmatrix} = (k-1)^2(k+2)$. Step 2: Set $D = 0 \implies k = 1$ or $k = -2$. Step 3: Test $k = 1$. All equations become $x+y+z=1$, which gives infinite solutions. Step 4: Test $k = -2$. The sum of the first three equations gives $-(x+y+z) = 3$, while substituting in $D_x$ reveals inconsistency. Thus $k = -2$ yields no solution.

⚡ Method B: 25-Second Exam Shortcut

Use the property for cyclic symmetric determinants: $D = (k+1+1)(k-1)^2$ or similar factorization. For $k = -2$, $D=0$ and rows create contradiction.

💡 Tip: When $k=1$, the equations degenerate into a single plane, giving infinite solutions. Always check root multiplicity.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: The system $x + y + z = 3$, $x + 2y + 3z = 4$, $x + 4y + kz = 6$ has a unique solution if:

A $k \neq 7$

B $k = 7$

C $k \neq 3$

D $k = 3$

Derivation & Analysis:

The coefficient determinant is $\begin{vmatrix}1&1&1\\1&2&3\\1&4&k\end{vmatrix}=k-7$. A unique solution exists when this determinant is non-zero, so $k \neq 7$.

MCQ 2: For what value of $\lambda$ do the equations $x + y - z = 1$, $2x + 3y + \lambda z = 2$, and $x + 2y + 3z = 3$ fail to have a unique solution?

A $2$

B $-5$

C $3$

D $0$

Derivation & Analysis:

The coefficient determinant is $\begin{vmatrix}1&1&-1\\2&3&\lambda\\1&2&3\end{vmatrix}=2-\lambda$. The system fails to have a unique solution when this determinant is zero, so $\lambda=2$.

MCQ 3: If the homogeneous system $ax + y + z = 0$, $x + by + z = 0$, $x + y + cz = 0$ has a non-trivial solution, where $a,b,c \neq 1$, then $\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c}$ equals:

A $0$

B $1$

C $-1$

D $\frac{1}{abc}$

Derivation & Analysis:

Non-trivial solution requires $D = 0$. Using algebraic manipulation on the determinant yields the required sum identity equal to 1.

MCQ 4: The number of solutions of the system $x + 2y + 3z = 1$, $2x + 4y + 6z = 3$, $3x + 6y + 9z = 5$ is:

A 0

B 1

C 2

D Infinite

Derivation & Analysis:

Notice that the left-hand sides are proportional while the right-hand sides are not in the same proportion, indicating parallel planes and zero solutions.

MCQ 5: If $x = a(y + z)$, $y = b(z + x)$, and $z = c(x + y)$ has a non-trivial solution, where $a,b,c \neq -1$, then $\frac{a}{1+a} + \frac{b}{1+b} + \frac{c}{1+c}$ is:

A 0

B 1

C 2

D 3

Derivation & Analysis:

Rewrite as $x - ay - az = 0$, $-bx + y - bz = 0$, $-cx - cy + z = 0$. Set determinant to zero and simplify to get 1.

MCQ 6: Cramer's rule is applicable to solve a system of three linear equations when the coefficient matrix is:

A Singular

B Non-singular

C Symmetric

D Skew-symmetric

Derivation & Analysis:

Cramer's rule requires division by $D = \det(A)$, which is valid only when $A$ is non-singular ($D \neq 0$).

MCQ 7: The system $kx + y + z = 0$, $x + ky + z = 0$, $x + y + kz = 0$ has non-trivial solutions for $k =

A 1 or -2

B -1 or 2

C 1 only

D -2 only

Derivation & Analysis:

Determinant is $(k-1)^2(k+2) = 0$, giving $k = 1$ and $k = -2$.

MCQ 8: If $\det(A) = 0$ and $D_x = D_y = D_z = 0$ for a system of three linear equations, the system necessarily has:

A Unique solution

B No solution

C Either no solution or infinitely many solutions; more information is needed

D Trivial solution only

Derivation & Analysis:

When $D=0$, Cramer’s rule does not decide consistency. For example, a singular system can be inconsistent or can have infinitely many solutions, so the given determinants are not enough to distinguish those cases.

MCQ 9: What is the geometric interpretation of three planes intersecting along a common line?

A Unique solution

B No solution

C Infinitely many solutions

D Inconsistent system

Derivation & Analysis:

An entire line of intersection contains infinitely many points, each representing a valid $(x, y, z)$ solution.

Q10

MCQ 10: For the system $2x + y + z = 1$, $x + 2y + z = 2$, $x + y + 2z = 3$, the value of $x$ is:

A $-1/2$

B $1/4$

C $1/2$

D $3/4$

Derivation & Analysis:

Adding the three equations gives $4(x+y+z)=6$, so $x+y+z=3/2$. The first equation gives $x+(x+y+z)=1$, hence $x=-1/2$.

Common questions

Use Cramer's Rule when asked for a single variable or when evaluating parameter conditions ($k$) for unique/non-unique solutions. Use Matrix Inversion or Gaussian Elimination when solving for all variables with fixed numerical coefficients.

Candidates often check only $D = 0$ and stop. When $D = 0$, you must check $D_x, D_y, D_z$. If even one of them is non-zero, the system has zero solutions. If all are zero, it typically has infinite solutions (subject to rank confirmation).

Master row/column operations to create at least two zeros in a row or column before expanding, or practice direct Sarrus expansion to avoid arithmetic errors under exam pressure.

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