Systems of Linear Equations and LU Decomposition PYQs for GATE CS

Solve 4+ Systems of Linear Equations and LU Decomposition previous year questions for GATE CS with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Systems of Linear Equations

Journey through this chapter

Step 1 · Consistency of Parameterized Systems

Decide unique, infinite, or no solution when parameters appear.

Step 2 · LU Decomposition Properties

Why

P = LU

exists and how it reorganises solving.

Step 3 · LU Factorization & Substitution

Compute

L

and

U

, then solve

L Y = B

U X = Y

Goal: Classify any linear system and solve it using LU decomposition in two triangular sweeps.

Why Parameterized Systems Matter

A parameterized linear system is a system $A X = B$ where some entries of $A$ or $B$ are not fixed numbers but parameters (letters such as $k$ , $a$ , $b$ , $λ$ ).

The central question: for which values of the parameter does the system have a unique solution, infinitely many solutions, or no solution at all?

Why this is asked in GATE:

Parameters model design variables, physical constants, or tuning knobs.
The exam tests whether you can read the structure of a system instead of blindly solving it.
It is the cleanest way to test your understanding of rank, determinant, and consistency in one question.

Behaviour	Meaning
Unique solution	Exactly one $X$ satisfies every equation
Infinitely many	A whole family of $X$ satisfies every equation
No solution	No $X$ can satisfy all equations simultaneously

Systems of Linear Equations and LU Decomposition: Solved Questions with Step-by-Step Explanations (4 Problems)

Question 1 · Engineering Mathematics · 2026_Set2 NAT

Consider the system of linear equations given below.

a x + y = b

16 x + a y = 24

Suppose the values of

a

and

b

are chosen such that the system of linear equations produce multiple solutions. Then the product of

a

and

b

is __________. (answer in integer)

Correct Answer:

Step-by-Step Solution

Key idea: This is a parameterized system question asking for infinite solutions. It is recognisable because it uses parameters $a$ and $b$ in the coefficients and constants, and explicitly states the system has "multiple solutions".

Step 1: Write down the condition for infinite solutions in a 2x2 system.

For the system $a_{1} x + b_{1} y = c_{1}$ and $a_{2} x + b_{2} y = c_{2}$ to have infinitely many solutions, the two equations must represent the same line. This means their coefficients and constants must be strictly proportional:

$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ .

Step 2: Apply the proportionality condition to the given system.

The system is:

$a x + y = b$

$16 x + a y = 24$

So, $\frac{a}{16} = \frac{1}{a} = \frac{b}{24}$ .

Step 3: Solve for $a$ .

From the first equality: $\frac{a}{16} = \frac{1}{a} ⟹ a^{2} = 16 ⟹ a = 4$ or $a = - 4$ .

Step 4: Solve for $b$ in both cases.

Case 1: $a = 4$ .

$\frac{1}{4} = \frac{b}{24} ⟹ b = 6$ .

The product $ab = 4 \times 6 = 24$ .

Case 2: $a = - 4$ .

$\frac{1}{- 4} = \frac{b}{24} ⟹ b = - 6$ .

The product $ab = (- 4) \times (- 6) = 24$ .

Step 5: Conclude the final answer.

In both valid scenarios, the product $ab$ is exactly 24.

Answer: 24

Question 2 · Engineering Mathematics · 2025_Set2 MSQ

Consider a system of linear equations

P X = Q

where

P \in R^{3 \times 3}

and

Q \in R^{3 \times 1}

. Suppose

P

has an LU decomposition,

P = LU

, where

L = 1 l_{21} l_{31} 01 l_{32} 001 and U = u_{11} 00 u_{12} u_{22} 0 u_{13} u_{23} u_{33} .

Which of the following statement(s) is/are TRUE?

A.
The system $P X = Q$ can be solved by first solving $L Y = Q$ and then $U X = Y$ .
B.
If $P$ is invertible, then both $L$ and $U$ are invertible.
C.
If $P$ is singular, then at least one of the diagonal elements of $U$ is zero.
D.
If $P$ is symmetric, then both $L$ and $U$ are symmetric.

Correct Answer:

["A","B","C"]

Step-by-Step Solution

Key idea: This is a theoretical properties question on LU decomposition. It is recognisable because it asks about the invertibility, singularity, and symmetry of matrices $P$ , $L$ , and $U$ without requiring numerical computation.

Step 1: Analyse Option A (Solution Strategy).

The system is $P X = Q$ . Given $P = LU$ , we substitute to get $LU X = Q$ .

Let $U X = Y$ . Then the equation becomes $L Y = Q$ .

Since $L$ is lower triangular, $L Y = Q$ is easily solved by forward substitution.

Once $Y$ is found, we solve $U X = Y$ by backward substitution.

Thus, Option A is TRUE.

Step 2: Analyse Option B (Invertibility).

We know $det (P) = det (L) det (U)$ .

By Doolittle's convention, $L$ is a unit lower triangular matrix, so its diagonal entries are all $1$ . Thus, $det (L) = 1$ .

Therefore, $det (P) = det (U)$ .

If $P$ is invertible, $det (P) \neq = 0$ , which implies $det (U) \neq = 0$ .

Since $det (L) = 1 \neq = 0$ , both $L$ and $U$ are invertible.

Thus, Option B is TRUE.

Step 3: Analyse Option C (Singularity).

Using the same determinant property: $det (P) = det (U)$ .

The determinant of an upper triangular matrix $U$ is the product of its diagonal entries: $det (U) = u_{11} u_{22} u_{33}$ .

If $P$ is singular, $det (P) = 0$ , which means $det (U) = 0$ .

For the product $u_{11} u_{22} u_{33}$ to be zero, at least one of the diagonal elements $u_{ii}$ must be zero.

Thus, Option C is TRUE.

Step 4: Analyse Option D (Symmetry).

Suppose $P$ is symmetric ( $P = P^{T}$ ). Does this mean $L$ and $U$ are symmetric?

$L$ is lower triangular. For $L$ to be symmetric, it must be diagonal.

$U$ is upper triangular. For $U$ to be symmetric, it must be diagonal.

But a general symmetric matrix $P$ does not factor into diagonal $L$ and $U$ matrices.

Counterexample: Let $P = (2112)$ .

$L = (1 1/2 01)$ , $U = (20 1 3/2)$ .

Neither $L$ nor $U$ is symmetric.

Thus, Option D is FALSE.

Answer: A, B, C

Question 3 · Engineering Mathematics · 2025_Set1 MSQ

Consider the given system of linear equations for variables

x

and

y

, where

k

is a real-valued constant. Which of the following option(s) is/are CORRECT?

x + k y = 1 \kx + y = - 1

A.
There is exactly one value of $k$ for which the above system of equations has no solution.
B.
There exist an infinite number of values of $k$ for which the system of equations has no solution.
C.
There exists exactly one value of $k$ for which the system of equations has exactly one solution.
D.
There exists exactly one value of $k$ for which the system of equations has an infinite number of solutions.

Correct Answer:

["A","D"]

Step-by-Step Solution

Key idea: This is a parameterized linear system consistency question. It is recognisable because the coefficients contain an unknown parameter $k$ , and the question asks about the existence and uniqueness of solutions.

Step 1: Write the system in matrix form $A X = B$ .

The coefficient matrix is $A = (1 k k 1)$ and the constant vector is $B = (1 - 1)$ .

Step 2: Calculate the determinant of $A$ .

$det (A) = (1) (1) - (k) (k) = 1 - k^{2}$ .

Step 3: Analyse the case where $det (A) \neq = 0$ .

If $1 - k^{2} \neq = 0$ , which means $k \neq = 1$ and $k \neq = - 1$ , the matrix $A$ is invertible. In this case, the system has exactly one unique solution. Since there are infinitely many such values of $k$ , Option C is false.

Step 4: Analyse the case $k = 1$ .

Substitute $k = 1$ into the system:

$x + y = 1$

$x + y = - 1$

These two equations represent parallel lines. They are inconsistent, meaning there is no solution. This happens for exactly one value of $k$ (which is $k = 1$ ). Thus, Option A is true and Option B is false.

Step 5: Analyse the case $k = - 1$ .

Substitute $k = - 1$ into the system:

$x - y = 1$

$- x + y = - 1 ⟹ x - y = 1$

Both equations are identical. They represent the same line, meaning there are infinitely many solutions. This happens for exactly one value of $k$ (which is $k = - 1$ ). Thus, Option D is true.

Answer: A, D

Question 4 · Engineering Mathematics · 2022 MCQ

Consider solving the following system of simultaneous equations using LU decomposition.

$x_{1} + x_{2} - 2 x_{3} x_{1} + 3 x_{2} - x_{3} 2 x_{1} + x_{2} - 5 x_{3} = 4 = 7 = 7$

where $L$ and $U$ are denoted as

$L = L_{11} L_{21} L_{31} 0 L_{22} L_{32} 00 L_{33}, U = U_{11} 00 U_{12} U_{22} 0 U_{13} U_{23} U_{33}$

Which one of the following is the correct combination of values for $L_{32}$ , $U_{33}$ , and $x_{1}$ ?

A.
$L_{32} = 2, U_{33} = - \frac{1}{2}, x_{1} = - 1$
B.
$L_{32} = 2, U_{33} = 2, x_{1} = - 1$
C.
$L_{32} = - \frac{1}{2}, U_{33} = 2, x_{1} = 0$
D.
$L_{32} = - \frac{1}{2}, U_{33} = - \frac{1}{2}, x_{1} = 0$

Correct Answer:

Step-by-Step Solution

Key idea: This is an LU decomposition computation question. It is recognisable because it provides a specific system of equations and asks for specific entries of the $L$ and $U$ matrices, along with a variable value, using the standard Doolittle convention.

Step 1: Identify the matrix $A$ and vector $b$ .

$A = 112131 - 2 - 1 - 5$ , $b = 477$ .

Step 2: Apply Doolittle's method to find $L$ and $U$ .

By convention, $L$ has $1$ s on its main diagonal.

Row 1 of $U$ equals Row 1 of $A$ :

$u_{11} = 1, u_{12} = 1, u_{13} = - 2$ .

Column 1 of $L$ :

$l_{21} = a_{21} / u_{11} = 1/1 = 1$ .

$l_{31} = a_{31} / u_{11} = 2/1 = 2$ .

Row 2 of $U$ :

$u_{22} = a_{22} - l_{21} u_{12} = 3 - (1) (1) = 2$ .

$u_{23} = a_{23} - l_{21} u_{13} = - 1 - (1) (- 2) = 1$ .

Column 2 of $L$ :

$l_{32} = (a_{32} - l_{31} u_{12}) / u_{22} = (1 - (2) (1)) /2 = - 1/2$ .

Row 3 of $U$ :

$u_{33} = a_{33} - l_{31} u_{13} - l_{32} u_{23} = - 5 - (2) (- 2) - (- 1/2) (1) = - 5 + 4 + 1/2 = - 1/2$ .

Step 3: Solve $L Y = b$ using forward substitution.

$y_{1} = 4$ .

$l_{21} y_{1} + y_{2} = 7 ⟹ (1) (4) + y_{2} = 7 ⟹ y_{2} = 3$ .

$l_{31} y_{1} + l_{32} y_{2} + y_{3} = 7 ⟹ (2) (4) + (- 1/2) (3) + y_{3} = 7 ⟹ 8 - 1.5 + y_{3} = 7 ⟹ y_{3} = 0.5$ .

Step 4: Solve $U X = Y$ using backward substitution.

$u_{33} x_{3} = y_{3} ⟹ (- 1/2) x_{3} = 0.5 ⟹ x_{3} = - 1$ .

$u_{22} x_{2} + u_{23} x_{3} = y_{2} ⟹ 2 x_{2} + (1) (- 1) = 3 ⟹ 2 x_{2} = 4 ⟹ x_{2} = 2$ .

$u_{11} x_{1} + u_{12} x_{2} + u_{13} x_{3} = y_{1} ⟹ (1) x_{1} + (1) (2) + (- 2) (- 1) = 4 ⟹ x_{1} + 4 = 4 ⟹ x_{1} = 0$ .

Step 5: Match with the options.

We found $L_{32} = - 1/2$ , $U_{33} = - 1/2$ , and $x_{1} = 0$ . This matches Option D.

Answer: D

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Systems of Linear Equations and LU Decomposition PYQs for GATE CS

Chapter Roadmap: Systems of Linear Equations

Why Parameterized Systems Matter

Systems of Linear Equations and LU Decomposition: Solved Questions with Step-by-Step Explanations (4 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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