Matrix Operations, Determinants and Gaussian Elimination Short Notes for GATE DA
Matrix Operations, Determinants and Gaussian Elimination short notes for GATE DA: 2 study cards covering concepts, formulas, shortcuts and exam traps, plus so
matrix operations determinants and gaussian elimination short notes
Summary: Gaussian Elimination Operation Counts
Revision Summary
DENSE n×n GAUSSIAN ELIMINATION
Phase
Adds/Subs
Mults
Combined
Forward elim.
3n3−n
3n3−n
32(n3−n)
Back sub.
2n(n−1)
2n(n−1)
n(n−1)
Dominant: O(n3)
UPPER TRIANGULAR SOLVE
Forward elimination
0
Back substitution
O(n2)
Complete solve
O(n2)
Dense forward: cubicBack sub: quadraticUpper tri removes forward workSeparate elim-only vs complete-solve
Use triangular, diagonal, or trace-based shortcuts.
Compute remaining determinant directly only when needed.
Key traps
det(A+B)=det(A)+det(B)
det(cA) uses cn, not c
det(p(A))=p(detA)
Singular A⟹det(A(A+cI))=0, but not all polynomials in A.
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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 Ax=b for m different right-hand side vectors, the total computational cost is O(n3)+m⋅O(n2).
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+tI:
Assertion (A): For any 2×2 matrix A, the determinant det(A+tI) can be computed using the formula t2+tr(A)t+det(A).
Reason (R): The determinant is additive, and the scalar t contributes t to the determinant, so det(A+tI)=det(A)+t.
Question 3
Level 1: Warm-up
Consider the following statements regarding a 3×3 upper triangular matrix A with diagonal entries 1, 2, 3:
det(A+2I)=3×4×5=60.
det(A+2I)=det(A)+det(2I)=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×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(A2+5A)=det(A)det(A+5I)
det(A2+B2)=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(A3−2A2) for a 4×4 matrix A, they will factor the expression first. Given that det(A)=3 and det(A−2I)=−4, what is the exact value of det(A3−2A2)?
Question 7
Level 1: Warm-up
Let A be a square matrix such that det(A2−9I)=0. If it is known that det(A−3I)=5, what is the minimum possible value of det(A+3I)?
Question 8
Level 1: Warm-up
Consider the following Assertion (A) and Reason (R) regarding the determinant of a 2×2 matrix polynomial:
Assertion (A): For any 2×2 matrix A, det(A2+3A)=det(A)det(A+3I).
Reason (R): The determinant of a sum of matrices is the sum of their determinants, so det(A2+3A)=det(A2)+det(3A).
Question 9
Level 1: Warm-up
A student is asked to evaluate det(A2−4A) for a 2×2 matrix A. They are given det(A)=3 and det(A−4I)=−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×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 Short Notes for GATE DA
Matrix Operations, Determinants and Gaussian Elimination short notes for GATE DA: 2 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.
Summary: Gaussian Elimination Operation Counts
Revision Summary
DENSE n×n GAUSSIAN ELIMINATION
Phase
Adds/Subs
Mults
Combined
Forward elim.
3n3−n
3n3−n
32(n3−n)
Back sub.
2n(n−1)
2n(n−1)
n(n−1)
Dominant: O(n3)
UPPER TRIANGULAR SOLVE
Forward elimination
0
Back substitution
O(n2)
Complete solve
O(n2)
Dense forward: cubicBack sub: quadraticUpper tri removes forward workSeparate elim-only vs complete-solve
Use triangular, diagonal, or trace-based shortcuts.
Compute remaining determinant directly only when needed.
Key traps
det(A+B)=det(A)+det(B)
det(cA) uses cn, not c
det(p(A))=p(detA)
Singular A⟹det(A(A+cI))=0, but not all polynomials in A.
Matrix Operations, Determinants and Gaussian Elimination: Solved Questions with Step-by-Step Explanations (10 Problems)
Question 1 · Linear AlgebraMCQ
Consider the following Assertion (A) and Reason (R) regarding the solution of linear systems with multiple right-hand sides:
Assertion (A): When solving Ax=b for m different right-hand side vectors, the total computational cost is O(n3)+m⋅O(n2).
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:
C
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(n3).
Step 2: Back substitution solves Ux=c for a specific right-hand side. Since there are m different vectors b, back substitution must be repeated m times, costing m⋅O(n2).
Step 3: Assertion (A) correctly states the total cost as O(n3)+m⋅O(n2).
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 AlgebraMCQ
Consider the following Assertion (A) and Reason (R) regarding the determinant of A+tI:
Assertion (A): For any 2×2 matrix A, the determinant det(A+tI) can be computed using the formula t2+tr(A)t+det(A).
Reason (R): The determinant is additive, and the scalar t contributes t to the determinant, so det(A+tI)=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:
C
Step-by-Step Solution
Key idea: The formula for det(A+tI) is derived from the characteristic polynomial, not from additivity.
Step 1: Evaluate Assertion (A). For a 2×2 matrix, det(A+tI) is indeed t2+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 tI for a 2×2 matrix contributes t2 to the determinant, not t. Reason (R) is FALSE.
Answer: A is true, but R is false.
Question 3 · Linear AlgebraMCQ
Consider the following statements regarding a 3×3 upper triangular matrix A with diagonal entries 1, 2, 3:
det(A+2I)=3×4×5=60.
det(A+2I)=det(A)+det(2I)=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:
A
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 tI simply adds t to each diagonal entry. The resulting matrix A+2I is also upper triangular.
Step 2: The diagonal entries of A+2I 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: 3×4×5=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+2I)=det(A)+det(2I). Statement 2 is FALSE.
Answer: 1 only
Question 4 · Linear AlgebraMCQ
Consider the back substitution phase for solving an n×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 n2.
C.
The total number of multiplications is exactly n(n−1)/2.
D.
The combined additions and multiplications is n2.
Correct Answer:
C
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 xi, 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 ∑i=1n(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 ∑i=1n1=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 AlgebraMCQ
Consider the following statements regarding the determinant of a matrix polynomial:
det(A2+5A)=det(A)det(A+5I)
det(A2+B2)=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:
A
Step-by-Step Solution
Key idea: Matrix polynomials can be factored only if the factors commute.
Step 1: Evaluate Statement 1. A2+5A=A(A+5I). Since A and A+5I are both polynomials in A, they commute. Thus, det(A(A+5I))=det(A)det(A+5I). Statement 1 is TRUE.
Step 2: Evaluate Statement 2. The expression A2+B2 does not generally factor as (A+B)(A−B) because matrix multiplication is not commutative (AB=BA in general). Even if it did factor, det(A2+B2) is not equal to det(A2−B2). Statement 2 is FALSE.
Answer: 1 only
Question 6 · Linear AlgebraMCQ
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(A3−2A2) for a 4×4 matrix A, they will factor the expression first. Given that det(A)=3 and det(A−2I)=−4, what is the exact value of det(A3−2A2)?
A.
-36
B.
9
C.
-117
D.
-12
Correct Answer:
A
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 A2 from A3−2A2.
Step 2: The factored form is A2(A−2I). 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(XY)=det(X)det(Y).
Step 4: Apply this to the factored form: det(A2(A−2I))=det(A2)det(A−2I).
Step 5: Use the power rule det(Am)=(det(A))m. Here, det(A2)=32=9.
Step 6: Multiply the determinants of the factors: 9×(−4)=−36.
Answer: -36
Question 7 · Linear AlgebraMCQ
Let A be a square matrix such that det(A2−9I)=0. If it is known that det(A−3I)=5, what is the minimum possible value of det(A+3I)?
A.
0
B.
3
C.
6
D.
-3
Correct Answer:
A
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 A2−9I is a difference of squares and factors as (A−3I)(A+3I).
Step 2: Apply the multiplicative property of determinants: det((A−3I)(A+3I))=det(A−3I)det(A+3I).
Step 3: We are given that this product is 0: det(A−3I)det(A+3I)=0.
Step 4: We are also given that det(A−3I)=5. Substitute this into the equation: 5×det(A+3I)=0.
Step 5: Solve for det(A+3I). The only solution is det(A+3I)=0.
Step 6: Since there is only one possible value, the minimum possible value is 0.
Answer: 0
Question 8 · Linear AlgebraMCQ
Consider the following Assertion (A) and Reason (R) regarding the determinant of a 2×2 matrix polynomial:
Assertion (A): For any 2×2 matrix A, det(A2+3A)=det(A)det(A+3I).
Reason (R): The determinant of a sum of matrices is the sum of their determinants, so det(A2+3A)=det(A2)+det(3A).
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:
C
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 A2+3A can be factored as A(A+3I). Since A and A+3I are polynomials in the same matrix, they commute.
Step 2: Using the multiplicative property, det(A(A+3I))=det(A)det(A+3I). 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(A2+3A)=det(A2)+det(3A). Reason (R) is FALSE.
Answer: A is true, but R is false.
Question 9 · Linear AlgebraMCQ
A student is asked to evaluate det(A2−4A) for a 2×2 matrix A. They are given det(A)=3 and det(A−4I)=−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:
A
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 A2−4A.
Step 2: The factored form is A(A−4I). 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(XY)=det(X)det(Y).
Step 4: Apply this to the factored form: det(A(A−4I))=det(A)det(A−4I).
Step 5: Substitute the given values: 3×(−2)=−6.
Answer: -6
Question 10 · Linear AlgebraMCQ
During the forward elimination of an n×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.
2n
Correct Answer:
B
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≤k≤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.