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Matrix Operations, Determinants and Gaussian Elimination Notes for GATE DA

Matrix Operations, Determinants and Gaussian Elimination notes for GATE DA: 18 study cards covering concepts, formulas, shortcuts and exam traps, plus solved

matrix operations determinants and gaussian elimination notes

Chapter Roadmap: Matrix Operations, Determinants and Gaussian Elimination

Chapter Journey 1 Gaussian Elimination & Complexity Counting additions/multiplications and seeing how matrix structure changes cost. CURRENT TOPIC 2 Determinants of Matrix Expressions Using determinant properties to simplify matrix expressions. Target: Build the operation-counting habit first. The structural thinking here supports determinant shortcuts later.

Topic Hero: Gaussian Elimination as a Costed Algorithm

Two-Phase Process for A x = b Phase 1: Forward Elimination Create zeros below pivots Convert to U x = c Phase 2: Back Substitution Solve from last variable upward Uses upper triangular form Why count operations? The cost matters as much as the answer. We track additions, subtractions, multiplications, and divisions to understand scaling with n . Core intuition: Gaussian elimination is cheap when the matrix already has the zeros that elimination would otherwise create.

One Elimination Step and the Active Submatrix

Pivot Step k : Active Work Only For each lower row i > k, remove entry a_{ik} using multiplier: m_{ik} = \frac{a _{ik}}{a _{k k}} Row update: R_{i} \leftarrow R_{i} - m_{ik} R_{k} Rows updated: n - k Entries per row: n - k + 1 Key idea: Never count operations on entries already zero or processed. Count only the active submatrix. This shrinking region enables clean summation.

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Question 1

Level 1: Warm-up

Consider the following Assertion (A) and Reason (R) regarding the solution of linear systems with multiple right-hand sides: Assertion (A): When solving A x = b for m different right-hand side vectors, the total computational cost is O (n^{3}) + m \cdot O (n^{2}) . Reason (R): The forward elimination phase must be repeated m times, once for each right-hand side vector, while back substitution is performed only once.

Question 2

Level 1: Warm-up

Consider the following Assertion (A) and Reason (R) regarding the determinant of A + t I : Assertion (A): For any 2 \times 2 matrix A, the determinant det (A + t I) can be computed using the formula t^{2} + tr (A) t + det (A) . Reason (R): The determinant is additive, and the scalar t contributes t to the determinant, so det (A + t I) = det (A) + t .

Question 3

Level 1: Warm-up

Consider the following statements regarding a 3 \times 3 upper triangular matrix A with diagonal entries 1, 2, 3: det (A + 2 I) = 3 \times 4 \times 5 = 60 . det (A + 2 I) = det (A) + det (2 I) = 6 + 8 = 14 . Which of the statements is/are TRUE?

Question 4

Level 1: Warm-up

Consider the back substitution phase for solving an n \times n upper triangular system. Which of the following statements is TRUE regarding the exact operation counts?

Question 5

Level 1: Warm-up

Consider the following statements regarding the determinant of a matrix polynomial: det (A^{2} + 5 A) = det (A) det (A + 5 I) det (A^{2} + B^{2}) = det (A + B) det (A - B) Which of the statements is/are TRUE for all square matrices A and B of the same size?

Question 6

Level 1: Warm-up

According to the core strategy for determinant expressions, the fastest solutions begin with algebraic simplification rather than direct expansion. If a student applies this principle to evaluate det (A^{3} - 2 A^{2}) for a 4 \times 4 matrix A, they will factor the expression first. Given that det (A) = 3 and det (A - 2 I) = - 4, what is the exact value of det (A^{3} - 2 A^{2}) ?

Question 7

Level 1: Warm-up

Let A be a square matrix such that det (A^{2} - 9 I) = 0 . If it is known that det (A - 3 I) = 5, what is the minimum possible value of det (A + 3 I) ?

Question 8

Level 1: Warm-up

Consider the following Assertion (A) and Reason (R) regarding the determinant of a 2 \times 2 matrix polynomial: Assertion (A): For any 2 \times 2 matrix A, det (A^{2} + 3 A) = det (A) det (A + 3 I) . Reason (R): The determinant of a sum of matrices is the sum of their determinants, so det (A^{2} + 3 A) = det (A^{2}) + det (3 A) .

Question 9

Level 1: Warm-up

A student is asked to evaluate det (A^{2} - 4 A) for a 2 \times 2 matrix A . They are given det (A) = 3 and det (A - 4 I) = - 2 . If the student correctly factors the matrix polynomial and uses the multiplicative property of determinants, what is the exact value they will obtain?

Question 10

Level 1: Warm-up

During the forward elimination of an n \times n dense matrix, the number of active entries per row updated at pivot step k is given by n - k + 1 . What is the maximum possible value of this quantity over all valid pivot steps k ?

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Matrix Operations, Determinants and Gaussian Elimination Notes for GATE DA

Matrix Operations, Determinants and Gaussian Elimination notes for GATE DA: 18 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Chapter Roadmap: Matrix Operations, Determinants and Gaussian Elimination

Chapter Journey

Gaussian Elimination & Complexity

Counting additions/multiplications and seeing how matrix structure changes cost.

CURRENT TOPIC

Determinants of Matrix Expressions

Using determinant properties to simplify matrix expressions.

Target: Build the operation-counting habit first. The structural thinking here supports determinant shortcuts later.

Topic Hero: Gaussian Elimination as a Costed Algorithm

Two-Phase Process for $A x = b$

Phase 1: Forward Elimination

Create zeros below pivots
Convert to $U x = c$

Phase 2: Back Substitution

Solve from last variable upward
Uses upper triangular form

Why count operations? The cost matters as much as the answer. We track additions, subtractions, multiplications, and divisions to understand scaling with

n

Core intuition: Gaussian elimination is cheap when the matrix already has the zeros that elimination would otherwise create.

One Elimination Step and the Active Submatrix

Pivot Step $k$ : Active Work Only

For each lower row $i > k$ , remove entry $a_{ik}$ using multiplier:

m_{ik} = \frac{a _{ik}}{a _{k k}}

Row update:

R_{i} \leftarrow R_{i} - m_{ik} R_{k}

Rows updated:
$n - k$

Entries per row:
$n - k + 1$

Key idea: Never count operations on entries already zero or processed. Count only the active submatrix. This shrinking region enables clean summation.

Counting Forward Elimination Operations

Forward Elimination: Exact Counts

Component	Count at step $k$
Rows updated	$n - k$
Active entries/row (incl. RHS)	$n - k + 1$
Multipliers (divisions)	$n - k$

Multiplications & Additions at step $k$ :

M_{k} = A_{k} = (n - k) (n - k + 1)

Summing $k = 1$ to $n - 1$ :

j = 1 \sum n - 1 j (j + 1) = \frac{n ^{3} - n}{3}

Each of additions and multiplications:

\frac{n ^{3} - n}{3}

. Divisions:

\frac{n ( n - 1 )}{2}

. Dominant order:

O (n^{3})

Note: Some sources group divisions into flop count. For questions asking additions/multiplications specifically, keep divisions separate. Order remains cubic.

Matrix Operations, Determinants and Gaussian Elimination: Solved Questions with Step-by-Step Explanations (10 Problems)

Question 1 · Linear Algebra MCQ

Consider the following Assertion (A) and Reason (R) regarding the solution of linear systems with multiple right-hand sides:

Assertion (A): When solving $A x = b$ for $m$ different right-hand side vectors, the total computational cost is $O (n^{3}) + m \cdot O (n^{2})$ .

Reason (R): The forward elimination phase must be repeated $m$ times, once for each right-hand side vector, while back substitution is performed only once.

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.

Correct Answer:

Step-by-Step Solution

Key idea: The expensive factorization/elimination step is a property of the matrix $A$ alone, while back substitution depends on the right-hand side $b$ .

Step 1: Forward elimination transforms $A$ into an upper triangular matrix $U$ . This process uses only the entries of $A$ and does not involve $b$ . Therefore, it is performed exactly once, costing $O (n^{3})$ .

Step 2: Back substitution solves $U x = c$ for a specific right-hand side. Since there are $m$ different vectors $b$ , back substitution must be repeated $m$ times, costing $m \cdot O (n^{2})$ .

Step 3: Assertion (A) correctly states the total cost as $O (n^{3}) + m \cdot O (n^{2})$ .

Step 4: Reason (R) incorrectly states that forward elimination is repeated $m$ times and back substitution is performed once. This is the exact opposite of the truth.

Answer: A is true, but R is false.

Question 2 · Linear Algebra MCQ

Consider the following Assertion (A) and Reason (R) regarding the determinant of $A + t I$ :

Assertion (A): For any $2 \times 2$ matrix $A$ , the determinant $det (A + t I)$ can be computed using the formula $t^{2} + tr (A) t + det (A)$ .

Reason (R): The determinant is additive, and the scalar $t$ contributes $t$ to the determinant, so $det (A + t I) = det (A) + t$ .

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.

Correct Answer:

Step-by-Step Solution

Key idea: The formula for $det (A + t I)$ is derived from the characteristic polynomial, not from additivity.

Step 1: Evaluate Assertion (A). For a $2 \times 2$ matrix, $det (A + t I)$ is indeed $t^{2} + tr (A) t + det (A)$ . This is the characteristic polynomial evaluated at $- t$ . Assertion (A) is TRUE.

Step 2: Evaluate Reason (R). The determinant is NOT additive. Furthermore, the scalar $t$ in $t I$ for a $2 \times 2$ matrix contributes $t^{2}$ to the determinant, not $t$ . Reason (R) is FALSE.

Answer: A is true, but R is false.

Question 3 · Linear Algebra MCQ

Consider the following statements regarding a $3 \times 3$ upper triangular matrix $A$ with diagonal entries 1, 2, 3:

$det (A + 2 I) = 3 \times 4 \times 5 = 60$ .
$det (A + 2 I) = det (A) + det (2 I) = 6 + 8 = 14$ .

Which of the statements is/are TRUE?

A.
1 only
B.
2 only
C.
Both 1 and 2
D.
Neither 1 nor 2

Correct Answer:

Step-by-Step Solution

Key idea: This is a bounding question testing the properties of determinants for triangular matrices and the non-additivity of determinants.

Step 1: Evaluate Statement 1. For an upper triangular matrix $A$ , adding a scalar multiple of the identity matrix $t I$ simply adds $t$ to each diagonal entry. The resulting matrix $A + 2 I$ is also upper triangular.

Step 2: The diagonal entries of $A + 2 I$ are $1 + 2 = 3$ , $2 + 2 = 4$ , and $3 + 2 = 5$ .

Step 3: The determinant of a triangular matrix is the product of its diagonal entries: $3 \times 4 \times 5 = 60$ . Statement 1 is TRUE.

Step 4: Evaluate Statement 2. The determinant is fundamentally multiplicative, not additive. The identity $det (X + Y) = det (X) + det (Y)$ is generally false.

Step 5: Therefore, $det (A + 2 I) \neq = det (A) + det (2 I)$ . Statement 2 is FALSE.

Answer: 1 only

Question 4 · Linear Algebra MCQ

Consider the back substitution phase for solving an $n \times n$ upper triangular system. Which of the following statements is TRUE regarding the exact operation counts?

A.
The total number of divisions is $n (n - 1) /2$ .
B.
The total number of additions is exactly $n^{2}$ .
C.
The total number of multiplications is exactly $n (n - 1) /2$ .
D.
The combined additions and multiplications is $n^{2}$ .

Correct Answer:

Step-by-Step Solution

Key idea: Back substitution solves for variables from bottom to top, and its cost is strictly quadratic. We must recall the exact counts for each operation type.

Step 1: For each variable $x_{i}$ , back substitution requires $n - i$ multiplications, $n - i$ additions, and exactly 1 division.

Step 2: Summing over all $i$ from 1 to $n$ , the total number of multiplications is $\sum_{i = 1}^{n} (n - i) = n (n - 1) /2$ .

Step 3: The total number of additions is also $n (n - 1) /2$ .

Step 4: The total number of divisions is $\sum_{i = 1}^{n} 1 = n$ .

Step 5: The combined additions and multiplications is $n (n - 1) /2 + n (n - 1) /2 = n (n - 1)$ .

Answer: The total number of multiplications is exactly $n (n - 1) /2$ .

Question 5 · Linear Algebra MCQ

Consider the following statements regarding the determinant of a matrix polynomial:

$det (A^{2} + 5 A) = det (A) det (A + 5 I)$
$det (A^{2} + B^{2}) = det (A + B) det (A - B)$

Which of the statements is/are TRUE for all square matrices $A$ and $B$ of the same size?

A.
1 only
B.
2 only
C.
Both 1 and 2
D.
Neither 1 nor 2

Correct Answer:

Step-by-Step Solution

Key idea: Matrix polynomials can be factored only if the factors commute.

Step 1: Evaluate Statement 1. $A^{2} + 5 A = A (A + 5 I)$ . Since $A$ and $A + 5 I$ are both polynomials in $A$ , they commute. Thus, $det (A (A + 5 I)) = det (A) det (A + 5 I)$ . Statement 1 is TRUE.

Step 2: Evaluate Statement 2. The expression $A^{2} + B^{2}$ does not generally factor as $(A + B) (A - B)$ because matrix multiplication is not commutative ( $A B \neq = B A$ in general). Even if it did factor, $det (A^{2} + B^{2})$ is not equal to $det (A^{2} - B^{2})$ . Statement 2 is FALSE.

Answer: 1 only

Question 6 · Linear Algebra MCQ

According to the core strategy for determinant expressions, the fastest solutions begin with algebraic simplification rather than direct expansion. If a student applies this principle to evaluate $det (A^{3} - 2 A^{2})$ for a $4 \times 4$ matrix $A$ , they will factor the expression first. Given that $det (A) = 3$ and $det (A - 2 I) = - 4$ , what is the exact value of $det (A^{3} - 2 A^{2})$ ?

A.
-36
B.
9
C.
-117
D.
-12

Correct Answer:

Step-by-Step Solution

Key idea: This is a matrix polynomial factorization question, recognizable by the presence of powers of $A$ and the instruction to simplify algebraically.

Step 1: Factor the matrix expression inside the determinant. We can factor out $A^{2}$ from $A^{3} - 2 A^{2}$ .

Step 2: The factored form is $A^{2} (A - 2 I)$ . Note that the scalar $2$ must be multiplied by the identity matrix $I$ to maintain dimensional consistency.

Step 3: Use the multiplicative property of determinants: $det (X Y) = det (X) det (Y)$ .

Step 4: Apply this to the factored form: $det (A^{2} (A - 2 I)) = det (A^{2}) det (A - 2 I)$ .

Step 5: Use the power rule $det (A^{m}) = (det (A))^{m}$ . Here, $det (A^{2}) = 3^{2} = 9$ .

Step 6: Multiply the determinants of the factors: $9 \times (- 4) = - 36$ .

Answer: -36

Question 7 · Linear Algebra MCQ

Let $A$ be a square matrix such that $det (A^{2} - 9 I) = 0$ . If it is known that $det (A - 3 I) = 5$ , what is the minimum possible value of $det (A + 3 I)$ ?

A.
0
B.
3
C.
6
D.
-3

Correct Answer:

Step-by-Step Solution

Key idea: This is a contradiction question that relies on the zero product property for determinants of factored matrix polynomials.

Step 1: Factor the matrix expression inside the determinant. The expression $A^{2} - 9 I$ is a difference of squares and factors as $(A - 3 I) (A + 3 I)$ .

Step 2: Apply the multiplicative property of determinants: $det ((A - 3 I) (A + 3 I)) = det (A - 3 I) det (A + 3 I)$ .

Step 3: We are given that this product is 0: $det (A - 3 I) det (A + 3 I) = 0$ .

Step 4: We are also given that $det (A - 3 I) = 5$ . Substitute this into the equation: $5 \times det (A + 3 I) = 0$ .

Step 5: Solve for $det (A + 3 I)$ . The only solution is $det (A + 3 I) = 0$ .

Step 6: Since there is only one possible value, the minimum possible value is 0.

Answer: 0

Question 8 · Linear Algebra MCQ

Consider the following Assertion (A) and Reason (R) regarding the determinant of a $2 \times 2$ matrix polynomial:

Assertion (A): For any $2 \times 2$ matrix $A$ , $det (A^{2} + 3 A) = det (A) det (A + 3 I)$ .

Reason (R): The determinant of a sum of matrices is the sum of their determinants, so $det (A^{2} + 3 A) = det (A^{2}) + det (3 A)$ .

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.

Correct Answer:

Step-by-Step Solution

Key idea: This is a construction question testing the correct algebraic manipulation of matrix polynomials versus the incorrect additive property.

Step 1: Evaluate Assertion (A). The expression $A^{2} + 3 A$ can be factored as $A (A + 3 I)$ . Since $A$ and $A + 3 I$ are polynomials in the same matrix, they commute.

Step 2: Using the multiplicative property, $det (A (A + 3 I)) = det (A) det (A + 3 I)$ . Assertion (A) is TRUE.

Step 3: Evaluate Reason (R). The reason claims that $det (X + Y) = det (X) + det (Y)$ . This is a fundamental misconception; the determinant is multiplicative, not additive.

Step 4: Therefore, $det (A^{2} + 3 A) \neq = det (A^{2}) + det (3 A)$ . Reason (R) is FALSE.

Answer: A is true, but R is false.

Question 9 · Linear Algebra MCQ

A student is asked to evaluate $det (A^{2} - 4 A)$ for a $2 \times 2$ matrix $A$ . They are given $det (A) = 3$ and $det (A - 4 I) = - 2$ . If the student correctly factors the matrix polynomial and uses the multiplicative property of determinants, what is the exact value they will obtain?

A.
-6
B.
-39
C.
-3
D.
1

Correct Answer:

Step-by-Step Solution

Key idea: This is a direct formula application question, recognizable by the presence of a matrix polynomial that can be factored before taking the determinant.

Step 1: Factor the matrix expression inside the determinant. We can factor out $A$ from $A^{2} - 4 A$ .

Step 2: The factored form is $A (A - 4 I)$ . Note that the scalar $4$ must be multiplied by the identity matrix $I$ to maintain dimensional consistency.

Step 3: Use the multiplicative property of determinants: $det (X Y) = det (X) det (Y)$ .

Step 4: Apply this to the factored form: $det (A (A - 4 I)) = det (A) det (A - 4 I)$ .

Step 5: Substitute the given values: $3 \times (- 2) = - 6$ .

Answer: -6

Question 10 · Linear Algebra MCQ

During the forward elimination of an $n \times n$ dense matrix, the number of active entries per row updated at pivot step $k$ is given by $n - k + 1$ . What is the maximum possible value of this quantity over all valid pivot steps $k$ ?

A.
$n - 1$
B.
$n$
C.
$n + 1$
D.
$2 n$

Correct Answer:

Step-by-Step Solution

Key idea: The quantity $n - k + 1$ is a decreasing function of $k$ . To maximize it, we must minimize $k$ within its valid domain.

Step 1: The valid range for the pivot step $k$ in forward elimination is $1 \leq k \leq n - 1$ .

Step 2: The minimum valid value for $k$ is $1$ .

Step 3: Substitute $k = 1$ into the expression: $n - 1 + 1 = n$ .

Step 4: Therefore, the maximum possible value is $n$ .

Answer: $n$

matrix operations determinants and gaussian elimination notes

Chapter Roadmap: Matrix Operations, Determinants and Gaussian Elimination

Chapter Journey

Topic Hero: Gaussian Elimination as a Costed Algorithm

Two-Phase Process for Ax=b

One Elimination Step and the Active Submatrix

Pivot Step k: Active Work Only

Try a question

Login to view the complete notes

Personalised first. High quality throughout.

Built around you, not around a syllabus PDF

Revision that hits your weak spots

Questions calibrated to the real exam

Notes written for recall, not for volume

One place for everything

Honest progress

Unlock the whole course

Matrix Operations, Determinants and Gaussian Elimination Notes for GATE DA

Chapter Roadmap: Matrix Operations, Determinants and Gaussian Elimination

Chapter Journey

Topic Hero: Gaussian Elimination as a Costed Algorithm

Two-Phase Process for Ax=b

One Elimination Step and the Active Submatrix

Pivot Step k: Active Work Only

Counting Forward Elimination Operations

Forward Elimination: Exact Counts

Matrix Operations, Determinants and Gaussian Elimination: Solved Questions with Step-by-Step Explanations (10 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

More notes in this unit

Two-Phase Process for $A x = b$

Pivot Step $k$ : Active Work Only

Two-Phase Process for $A x = b$

Pivot Step $k$ : Active Work Only